MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Vector
Differentiation
Question Bank - Set 1
Liberty University
Question 1
Question
Let u= 3i−4j+ 2kand v= 2i+ 5j−k. Find d
dt (u·v).
Solution
Step 1: Recall that the dot product of two vectors a=a1i+a2j+a3kand
b=b1i+b2j+b3kis given by a·b=a1b1+a2b2+a3b3.
Step 2: Calculate u·vby taking the dot product of uand v.
u·v= (3)(2) + (−4)(5) + (2)(−1) = 6 −20 −2 = −16
Step 3: Differentiate u·vwith respect to t.
d
dt (u·v) = d
dt (−16) = 0
Therefore, d
dt (u·v) = 0.
Question 2
Question
Let v=
3x2z
2y3
xz
, where x, y, z are functions of t. Find
dt.
Solution
Step 1: Find
dtbydifferentiatingeachcomponentofvwithrespecttot.
dt =
d
dt (3x2z)
d
dt (2y3)
d
dt (xz)
Step 2: Differentiate each component of vusing the product and chain rules.
dt =
3(2xdx
dt z+ 3x2dz
dt )
6y2dy
dt
dx
dt z+xdz
dt
Step 3: Simplify the expression.
dt =
6xdx
dt z+ 9x2dz
dt
6y2dy
dt
dx
dt z+xdz
dt
Therefore,
dt =
6xdx
dt z+ 9x2dz
dt
6y2dy
dt
dx
dt z+xdz
dt
.
Question 3
Question
Let u=
2
−1
3
and v=
1
4
−2
. Find the derivative of the dot product u·v
with respect to t, where uand vare functions of t.
Solution
To find the derivative of the dot product u·vwith respect to t, we first find
the dot product of the derivatives of uand v.
Step 1: Find dudt and dv
dt
du
dt =
d
dt 2
d
dt (−1)
d
dt 3
=
0
0
0
2
dv
dt =
d
dt 1
d
dt 4
d
dt (−2)
=
0
0
0
Step 2: Find ddt(u·v)using chain rule
d
dt (u·v) = du
dt ·v+u·dv
dt
d
dt (u·v) =
0
0
0
·
1
4
−2
+
2
−1
3
·
0
0
0
= 0
Therefore, the derivative of the dot product u·vwith respect to tis 0.
Question 4
Question
Let u=
3x2
2y
z
and v=
x3
4y2
z2
. Find ∂
∂u(u·v).
Solution
Step 1: Calculate the dot product u·v:
u·v=
3x2
2y
z
·
x3
4y2
z2
= (3x2)(x3) + (2y)(4y2)+(z)(z2)=3x5+ 8y3+z3
Step 2: Differentiate the dot product with respect to uby treating uas a
constant vector: ∂
∂u(u·v) = ∂
∂u(3x5+ 8y3+z3)
Step 3: Since uis treated as a constant vector, the partial derivative with
respect to ueliminates terms with u:
∂
∂u(u·v) = 0
Therefore, ∂
∂u(u·v) = 0.
Question 5
Question
Let v=
3t2
2t+ 1
t
. Find dv
dt .
3
Solution
To find dv
dt , we need to differentiate each component of vwith respect to t.
Step 1: Differentiate the first component
d
dt (3t2)=6t
Step 2: Differentiate the second component
d
dt (2t+ 1) = 2
Step 3: Differentiate the third component
d
dt (t)=1
Therefore, dv
dt =
6t
2
1
.
Question 6
Question
Let v= 2i−3j+√5kand u= 4i+ 2j−2k. Find the derivative of v·uwith
respect to t, where tis a scalar parameter.
Solution
To find the derivative of v·uwith respect to t, we first need to find the expres-
sions for vand uin terms of t.
Step 1: Express vin terms of t.
v= 2i−3j+√5k
Step 2: Let’s add a tcomponent to each unit vector. Let v= 2ti−3tj+
√5tk.
Now, express uin terms of t:
u= 4i+ 2j−2k
Step 3: Let’s add a tcomponent to each unit vector. Let u= 4ti+2tj−2tk.
Now, we can find the derivative of v·uwith respect to t.
Step 4: Calculate the dot product v·u:
v·u= (2t)(4t)+(−3t)(2t)+(√5t)(−2t)
v·u= 8t2−6t2−2√5t2
4
v·u= (8 −6−2√5)t2
v·u= (2 −2√5)t2
Step 5: Differentiate with respect to t:
d
dt (v·u) = 2(2 −2√5)t
d
dt (v·u)=4−4√5)t
Therefore, the derivative of v·uwith respect to tis (4 −4√5)t.
Question 7
Question
Let u= 3i−2j+kand v= 2i+ 5j−4k. If w=u·(v×u), find ∇ · w.
Solution
Step 1: Find v×u.
v×u=
i j k
2 5 −4
3−2 1
= (5 ·1+4·2)i−(2 ·1+3· −4)j+ (2 · −2−5·3)k
= 13i+ 10j−16k
Step 2: Find u·(v×u).
u·(v×u) = (3)(13) + (−2)(10) + (1)(−16)
= 39 −20 −16
= 3
Step 3: Find ∇ · w.
∇ · w=∇ · (3) = 0
Therefore, ∇ · w= 0.
Question 8
Question
Let u=x+y
2x−yand v=3x
4y. Compute the gradient of the dot product
u·vwith respect to x.
5
Solution
Step 1: Compute the dot product u·v:
u·v=x+y
2x−y·3x
4y= (x+y)(3x) + (2x−y)(4y)
= 3x2+ 3xy + 8xy −4y2
= 3x2+ 11xy −4y2
Step 2: Compute the gradient of u·vwith respect to x: To find the gradient
of a scalar function with respect to a vector, we need to take the partial deriva-
tives of the function with respect to each component of the vector and arrange
them as a column vector.
∇x(u·v) = ∂
∂x (3x2+ 11xy −4y2)
∂
∂y (3x2+ 11xy −4y2)
Step 3: Find the partial derivatives:
∂
∂x (3x2+ 11xy −4y2)=6x+ 11y
∂
∂y (3x2+ 11xy −4y2) = 11x−8y
Step 4: Write the gradient as a column vector:
∇x(u·v) = 6x+ 11y
11x−8y
Therefore, the gradient of the dot product u·vwith respect to xis 6x+ 11y
11x−8y.
Question 9
Question
Let u=3
−4and v=5
2. Find d
dt (u·v).
Solution
Step 1: Recall that the dot product of two vectors u=u1
u2and v=v1
v2is
given by u·v=u1v1+u2v2.
Step 2: Let w(t) = u·v= 3 ·5+(−4) ·2.
Step 3: Differentiate w(t) with respect to tto find d
dt (u·v).
Step 4: d
dt (u·v) = d
dt (15 −8).
Step 5: Simplify to find d
dt (u·v) = d
dt (7).
Step 6: Finally, d
dt (u·v) = 0 .
6
Question 10
Question
Let u= 3i+ 4j−2kand v= 2i−j+ 5k. If f(r) = u·rand g(r) = v·r, find
d
dr(f(r)×g(r)).
Solution
Step 1: Let’s first find expressions for f(r) and g(r):
f(r) = u·r= (3i+ 4j−2k)·(xi+yj+zk)=3x+ 4y−2z
g(r) = v·r= (2i−j+ 5k)·(xi+yj+zk)=2x−y+ 5z
Step 2: Next, let’s find the cross product of f(r) and g(r):
f(r)×g(r) =
i j k
3 4 −2
2−1 5
= (4(5)−(−2)(−1))i−(3(5)−(−2)(2))j+(3(−1)−4(2))k
= 22i−19j−11k
Step 3: Finally, we find the derivative with respect to r:
d
dr(f(r)×g(r)) = d
dr(22i−19j−11k) =
22
−19
−11
Question 11
Question
Let f(t) =
2t2
3t
et
be a vector-valued function. Find df
dt .
Solution
Step 1: The function f(t) can be written as f(t) =
2t2
3t
et
.
Step 2: To differentiate f(t) with respect to t, we differentiate each compo-
nent separately.
Step 3: The derivative of the first component is d
dt (2t2)=4t.
Step 4: The derivative of the second component is d
dt (3t) = 3.
Step 5: The derivative of the third component is d
dt (et) = et.
7
Step 6: Therefore, the derivative of f(t) with respect to t,df
dt , is
df
dt =
4t
3
et
.
Question 12
Question
Let r(t) = t3i+tj+t2kbe a vector function, where i,j, and kare the standard
unit vectors in three-dimensional space. Find dr
dt .
Solution
Step 1: We can find dr
dt by differentiating each component of r(t) with respect
to t.
Step 2: Differentiating the x-component t3iwith respect to tgives d
dt (t3)i=
3t2i.
Step 3: Differentiating the y-component tjwith respect to tgives d
dt (t)j=j.
Step 4: Differentiating the z-component t2kwith respect to tgives d
dt (t2)k=
2tk.
Step 5: Therefore, combining the differentiated components, we have dr
dt =
d
dt (t3i+tj+t2k)=3t2i+j+ 2tk.
Question 13
Question
Let u= 3i−2j+ 4kand v= 2i+j−3k. Find d
dt (u·v).
Solution
Step 1: Compute u·v.
u·v= (3i−2j+ 4k)·(2i+j−3k)
= 3 ·2+(−2) ·1+4·(−3)
= 6 −2−12
=−8
Step 2: Differentiate u·vwith respect to t.
d
dt (u·v) = d
dt (−8)
= 0
Therefore, d
dt (u·v) = 0.
8
Question 14
Question
Let u= 3i+ 4jand v= 2i−j. Find d
dt (u·v).
Solution
Step 1: Let u=u1i+u2jand v=v1i+v2j, where u1= 3, u2= 4, v1= 2, and
v2=−1.
Step 2: The dot product of uand vis given by u·v=u1v1+u2v2.
Step 3: Taking the derivative with respect to ton both sides, we have
d
dt (u·v) = d
dt (u1v1+u2v2).
Step 4: Now, differentiate each term separately using the product rule. We
have d
dt (u1v1) + d
dt (u2v2).
Step 5: Substituting our values of u1,v1,u2, and v2, we get
d
dt (3 ·2) + d
dt (4 · −1).
Step 6: Simplifying, we have
6d
dt (1) −4d
dt (1).
Step 7: Since the derivative of a constant is 0, we are left with
6(0) −4(0).
Step 8: Thus, the final answer is 0 .
Question 15
Question
Let r(t) = t2
sin(t)be a vector function. Find dr
dt and d2r
dt2.
Solution
Step 1: To find dr
dt , differentiate each component of r(t) with respect to t.
dr
dt =d
dt (t2)
d
dt (sin(t))
9
=2t
cos(t)
Step 2: To find d2r
dt2, differentiate each component of dr
dt with respect to t.
d2r
dt2=d
dt (2t)
d
dt (cos(t))
=2
−sin(t)
Question 16
Question
Let v=
x2
xy
y2
be a vector function. Find dv
dt , where x=t2and y=√t.
Solution
Step 1: Express vin terms of tusing the given values of xand y.
v=
(t2)2
(t2)√t
(√t)2
=
t4
t5/2
t
Step 2: Differentiate each component of vwith respect to t.
dv
dt =
d(t4)
dt
d(t5/2)
dt
d(t)
dt
=
4t3
5
2t3/2
1
Therefore, dv
dt =
4t3
5
2t3/2
1
.
Question 17
Question
Let r(t) = ti+ 2 cos(t)j−3 sin(t)kbe a vector-valued function. Find dr
dt .
10
Solution
Step 1: Given r(t) = ti+ 2 cos(t)j−3 sin(t)k, we need to find dr
dt .
Step 2: To find dr
dt , we differentiate each component of r(t) with respect to
tseparately. Let’s start by finding d
dt (ti).
Step 3:
d
dt (ti) = d
dt (t)i
= 1 ·i
=i
Step 4: Next, let’s find d
dt (2 cos(t)j).
d
dt (2 cos(t)j) = d
dt (2 cos(t))j
=−2 sin(t)j
Step 5: Finally, we will find d
dt (−3 sin(t)k).
d
dt (−3 sin(t)k) = d
dt (−3 sin(t))k
=−3 cos(t)k
Step 6: Combining the results, we have:
dr
dt =i−2 sin(t)j−3 cos(t)k
Question 18
Question
Find the gradient of the scalar field f(x, y, z) = x2y+y2z+z2x.
Solution
To find the gradient of a scalar field, we need to compute the partial derivatives
of the function with respect to each variable.
Step 1: Find ∂f ∂x
∂f
∂x =∂
∂x (x2y) + ∂
∂x (y2z) + ∂
∂x (z2x)
∂f
∂x = 2xy +z2
Step 2: Find ∂f ∂y
∂f
∂y =x2+ 2yz +z2x
11
Step 3: Find ∂f ∂z
∂f
∂z =y2+ 2xz
Step 4: Gradient of fThe gradient of fis the vector:
∇f=∂f
∂x ,∂f
∂y ,∂f
∂z
Substitute the previously calculated partial derivatives:
∇f=2xy +z2, x2+ 2yz +z2x, y2+ 2xz
Therefore, the gradient of the scalar field f(x, y, z) = x2y+y2z+z2xis ∇f=
(2xy +z2, x2+ 2yz +z2x, y2+ 2xz).
Question 19
Question
Let uand vbe two vectors in R3, where u= 3i−2j+ 4kand v= 2i+ 5j−k.
Find d
dt (u·v).
Solution
Step 1: Recall that the dot product of two vectors uand vis given by u·v=
u1v1+u2v2+u3v3, where u=u1i+u2j+u3kand v=v1i+v2j+v3k.
Step 2: We can compute the dot product of uand vas follows:
u·v= (3i−2j+ 4k)·(2i+ 5j−k)
= 3(2) + (−2)(5) + 4(−1)
= 6 −10 −4
=−8
Step 3: Now, differentiate both sides of the equation with respect to t:
d
dt (u·v) = d
dt (−8)
= 0
Step 4: Therefore, d
dt (u·v) = 0 .
Question 20
Question
Let u= 3i−2j+ 4kand v= 2i+ 5j−k. Find the derivative of u·vwith
respect to t.
12
Solution
To find the derivative of u·vwith respect to t, we first need to find the expres-
sions for u·vand its derivative.
Step 1: Calculate u ·v
u·v= (3i−2j+ 4k)·(2i+ 5j−k)
u·v= 3(2) + (−2)(5) + 4(−1)
u·v= 6 −10 −4
u·v=−8
Step 2: Find the derivative Let f(t) = u·v. We want to find df
dt .
df
dt =d
dt (−8)
df
dt = 0
Therefore, the derivative of u·vwith respect to tis 0.
Question 21
Question
Let uand vbe constant vectors, and let r(t)=2tu−3tv+t2u·v. Find dr
dt .
Solution
Step 1: Find dr
dt using the rules of vector differentiation.
dr
dt =d
dt (2tu)−d
dt (3tv) + d
dt (t2u·v)
Step 2: Differentiate each term with respect to t.
dr
dt = 2u−3v+ 2tu·v
Step 3: Simplify the expression.
dr
dt = 2u−3v+ 2tu·v
Therefore, the derivative of r(t) with respect to tis 2u−3v+ 2tu·v.
13
Question 22
Question
Let u=3
5and v=−2
4. Compute the derivative of the dot product u·v
with respect to t, where uand vare both functions of t.
Solution
To find the derivative of the dot product u·vwith respect to t, we first need
to express uand vas functions of t. Let u(t) = 3t
5tand v(t) = −2t
4t.
Step 1: Compute u(t)·v(t).
u(t)·v(t) = 3t
5t·−2t
4t= (3t)(−2t) + (5t)(4t) = −6t2+ 20t2= 14t2
Step 2: Find d
dt (u(t)·v(t)) using the product rule.
d
dt (u(t)·v(t)) = d
dt (14t2)
= 28t
Therefore, the derivative of the dot product u·vwith respect to tis 28t.
Question 23
Question
Let u=
3
−4
5
and v=
−2
7
1
. Determine d
dt (u·v) where u·vrepresents the
dot product of vectors uand v.
Solution
To find the derivative of the dot product of two vectors with respect to t, we
can use the formula d
dt (u·v) = du
dt ·v+u·dv
dt .
Given u=
3
−4
5
and v=
−2
7
1
, we can find du
dt and dv
dt : - du
dt =
d
dt (3)
d
dt (−4)
d
dt (5)
=
0
0
0
-dv
dt =
d
dt (−2)
d
dt (7)
d
dt (1)
=
0
0
0
14
Therefore, d
dt (u·v) =
0
0
0
·
−2
7
1
+
3
−4
5
·
0
0
0
= 0.
Hence, the derivative of u·vwith respect to tis 0.
Question 24
Question
Let u=3
−2and v=−1
4. Compute the following derivative: d
dt (3u−tv+
2t2u).
Solution
Step 1: Use the properties of vector calculus to differentiate each term sepa-
rately. Step 2: Let’s differentiate 3uwith respect to tfirst.
d
dt (3u)=3du
dt
Step 3: Now, differentiate uwith respect to t.
du
dt =du1
dt
du2
dt =0
0
Therefore, d
dt (3u)=30
0=0
0. Step 4: Moving on to the second term −tv.
d
dt (−tv) = −v
Step 5: Next, differentiate 2t2uwith respect to t.
d
dt (2t2u)=2ud
dt (t2)
Step 6: Compute the derivative of t2with respect to t.
d
dt (t2)=2t
Therefore, d
dt (2t2u)=23
−2(2t) = 12t
−8t. Step 7: Finally, sum up all the
derivatives. d
dt (3u−tv+ 2t2u) = 0
0−−1
4+12t
−8t
=12t+ 1
−8t−4
15
Question 25
Question
Let v = 3x2ˆ
i+xsin(2y)ˆ
jbe a vector field. Find ∇ ·v.
Solution
Step 1: Recall that the divergence of a vector field v =P(x, y)ˆ
i+Q(x, y)ˆ
jis
given by the expression ∇ ·v =∂P
∂x +∂Q
∂y .
Step 2: In this case, P(x, y)=3x2and Q(x, y) = xsin(2y).
Step 3: Calculate the partial derivatives:
∂P
∂x =∂
∂x (3x2)=6x
and ∂Q
∂y =∂
∂y (xsin(2y)) = x(2 cos(2y)) = 2xcos(2y)
Step 4: Add the partial derivatives to find the divergence:
∇ ·v =∂P
∂x +∂Q
∂y = 6x+ 2xcos(2y)=2x(3 + cos(2y))
Step 5: Therefore, the divergence of the vector field v is ∇ · v = 2x(3 +
cos(2y)).
Question 26
Question
Let u= 3i−2j+ 4kand v= 2i+ 5j−3k. Find d
dt (u·v).
Solution
To find d
dt (u·v), we first need to compute the dot product of uand v: Step 1:
Calculate u·v.
u·v= (3i−2j+ 4k)·(2i+ 5j−3k)
u·v= 3(2) + (−2)(5) + 4(−3)
u·v= 6 −10 −12
u·v=−16
Step 2: Differentiate the dot product with respect to t.
d
dt (u·v) = d
dt (−16)
d
dt (u·v)=0
Therefore, d
dt (u·v) = 0.
16
Question 27
Question
Let u= 3i−2j+kand v= 2i+ 4j−k. Find the derivative of u·vwith respect
to t, where u=
3
−2
1
,v=
2
4
−1
and tis a scalar variable.
Solution
Step 1: Recall that the dot product of two vectors u=
u1
u2
u3
and v=
v1
v2
v3
is given by u·v=u1v1+u2v2+u3v3.
Step 2: Given u=
3
−2
1
and v=
2
4
−1
, we can find the dot product as
follows: u·v= (3)(2) + (−2)(4) + (1)(−1).
Step 3: Simplifying the dot product, we get u·v= 6 −8−1 = −3.
Step 4: Now, let’s differentiate u·vwith respect to t. Since u=
3
−2
1
and
v=
2
4
−1
are functions of time t, the derivative can be found using the chain
rule.
Step 5: Let f(t) = u·v. Then, df
dt =d(u·v)
dt .
Step 6: By the chain rule, we have df
dt =d(u·v)
du·du
dt +d(u·v)
dv·dv
dt .
Step 7: Since the dot product is a scalar, d(u·v)
du=vand d(u·v)
dv=u.
Step 8: Therefore, df
dt =v·du
dt +u·dv
dt .
Step 9: Substituting the given values for uand v, we have df
dt =
2
4
−1
·
d
dt (3)
d
dt (−2)
d
dt (1)
+
3
−2
1
·
d
dt (2)
d
dt (4)
d
dt (−1)
.
Step 10: Simplifying, we get df
dt =
Question 28
Question
Let f(x, y)=(excos(y), exsin(y)) be a vector-valued function. Find ∂f
∂x and ∂f
∂y .
17
Solution
Step 1: To find ∂f
∂x , we differentiate each component of fwith respect to x.
∂f
∂x =∂
∂x excos(y),∂
∂x exsin(y)
Step 2: Taking the partial derivative of excos(y) with respect to xgives:
∂
∂x excos(y) = excos(y)
Step 3: Taking the partial derivative of exsin(y) with respect to xgives:
∂
∂x exsin(y) = exsin(y)
Step 4: Therefore, we have ∂f
∂x = (excos(y), exsin(y)).
Step 5: Next, to find ∂f
∂y , we differentiate each component of fwith respect
to y.
∂f
∂y =∂
∂y excos(y),∂
∂y exsin(y)
Step 6: Taking the partial derivative of excos(y) with respect to ygives:
∂
∂y excos(y) = −exsin(y)
Step 7: Taking the partial derivative of exsin(y) with respect to ygives:
∂
∂y exsin(y) = excos(y)
Step 8: Therefore, we have ∂f
∂y = (−exsin(y), excos(y)).
Question 29
Question
Let v= 3x2i−2y2j+zkbe a vector field. Find the gradient of v.
Solution
To find the gradient of v, we need to find the partial derivatives of each com-
ponent with respect to x,y, and z.
Step 1: Find ∂
∂x :
∂
∂x (3x2)=6x, ∂
∂x (−2y2)=0,∂
∂x z= 0
18
So, the x-component of the gradient is 6xi.
Step 2: Find ∂
∂y :
∂
∂y (3x2) = 0,∂
∂y (−2y2) = −4y, ∂
∂y z= 0
Therefore, the y-component of the gradient is −4yj.
Step 3: Find ∂
∂z :
∂
∂z (3x2)=0,∂
∂z (−2y2) = 0,∂
∂z z= 1
Hence, the z-component of the gradient is 1k.
Thus, the gradient of vis ∇ · v= 6xi−4yj+ 1k.
Question 30
Question
Let u= 3i+ 2j−kand v= 2i−4j+ 5k. Compute the derivative of the vector
u·vwith respect to t, where uand vare both functions of t.
Solution
To find the derivative of u·vwith respect to t, we first find their dot product
and then differentiate with respect to t.
Step 1: Find u ·v
u·v= (3i+ 2j−k)·(2i−4j+ 5k) = 3(2) + 2(−4) + (−1)(5) = 6 −8−5 = −7
Step 2: Differentiate with respect to tLet u=u1i+u2j+u3kand
v=v1i+v2j+v3k. Then,
d
dt (u·v) = d
dt (u1v1+u2v2+u3v3)
Step 3: Compute the derivative Since uand vare functions of t, we
have: d
dt (u·v) = d
dt (3 ·2+2·(−4) + (−1) ·5) = d
dt (−7) = 0
Question 31
Question
Let u=x2
3xand v=2xy
y2. Calculate d(u·v)
dx .
19
Solution
Step 1: We start by finding u·v:
u·v= (x2)(2xy) + (3x)(y2)=2x3y+ 3xy2
Step 2: Next, we differentiate u·vwith respect to x:
d(u·v)
dx =d(2x3y+ 3xy2)
dx
Step 3: Applying the differentiation rules:
d(u·v)
dx =d(2x3y)
dx +d(3xy2)
dx
Step 4: Using the product rule for differentiation, we have:
d(u·v)
dx = 2(3x2)y+ 2x3dy
dx + 3y2+ 3xdy
dx
Step 5: Simplifying further gives:
d(u·v)
dx = 6x2y+ 2x3dy
dx + 3y2+ 3xdy
dx
Therefore, d(u·v)
dx = 6x2y+ 2x3dy
dx + 3y2+ 3xdy
dx .
Question 32
Question
Let v= 3x2i+ 4xyj−2zk. Find ∇ · v.
Solution
Step 1: Compute ∇ · vby taking the dot product of the vector differential
operator (∇) and the vector field v.
∇ · v=∂
∂x (3x2) + ∂
∂y (4xy) + ∂
∂z (−2z)
Step 2: Calculate the partial derivatives.
∂
∂x (3x2)=6x, ∂
∂y (4xy) = 4x, ∂
∂z (−2z) = −2
Step 3: Substitute the partial derivatives back into the expression for ∇·v.
∇ · v= 6x+ 4x−2 = 10x−2
Therefore, the divergence of the vector field vis 10x−2 .
20
Question 33
Question
Let u= 3i−2j+ 4kand v= 2i+ 5j−1k. Compute the derivative of u·vwith
respect to t, where uand vare functions of t.
Solution
To find the derivative of u·vwith respect to t, we first need to find expressions
for uand vin terms of t. Given that uand vare vectors, we can express them
as:
u(t)=3i−2j+ 4k
v(t)=2i+ 5j−1k
Now, to find the derivative of u·vwith respect to t, we will use the property
that d
dt (u·v) = u·
dt +
dt ·v.
Step 1: Find dudt and
dt
dt = ddt(3i−2j+4k)=0i+0j+0k=0
dt = ddt(2i+5j−1k)=0i+0j+0k=0
Step 2: Compute u·
dtand
dt ·v
u·
dt = (3i- 2j+ 4k)·0= 0
dt ·v=0·(2i+ 5j−1k)=0
Step 3: Find the derivative of u ·vHaving computed the components,
we see that both terms are zero. Hence, the derivative of u·vwith respect to
tis 0.
Question 34
Question
Let vand wbe two differentiable vectors given by
v=
2t3
t2
5
and w=
4t
3t
2t
.
Find the derivative of v·wwith respect to t.
21
Solution
To find the derivative of v·w, we use the product rule for differentiation of
vectors. The dot product of two vectors aand bis given by a·b=a1b1+
a2b2+a3b3.
Step 1: Find v·w:
v·w= (2t3)(4t)+(t2)(3t) + (5)(2t) = 8t4+ 3t3+ 10t.
Step 2: Differentiate v·wwith respect to tusing the product rule:
d
dt (v·w) = d
dt (8t4+ 3t3+ 10t)
= 32t3+ 9t2+ 10.
Thus, the derivative of v·wwith respect to tis 32t3+ 9t2+ 10 .
Question 35
Question
Let v=1
x
ln(x+y)and u=exy
sin(xy). Find d
dx (v·u).
Solution
To find d
dx (v·u), we first need to find the dot product of vand u:
v·u=1
x
ln(x+y)·exy
sin(xy)=1
xexy + ln(x+y) sin(xy)
Now, we will differentiate v·uwith respect to xusing the product rule:
d
dx (v·u) = d
dx 1
xexy + ln(x+y) sin(xy)
Step 1: Differentiate the first term:
d
dx 1
xexy=−1
x2exy +1
xyexy
Step 2: Differentiate the second term:
d
dx (ln(x+y) sin(xy)) = 1
x+y·d
dx (x+y) sin(xy) + ln(x+y) cos(xy)·d
dx (xy)
Step 3: Simplify the derivative:
d
dx (v·u) = −1
x2exy +1
xyexy +1
x+yexy sin(xy) + ln(x+y) cos(xy)
Therefore, d
dx (v·u) = −1
x2exy +1
xyexy +1
x+yexy sin(xy) + ln(x+y) cos(xy).
22
Solution
Step 1: Find
dtbydif ferentiatingeachcomponentof vwithrespecttot.
dt =
d
dt (3x2z)
d
dt (2y3)
d
dt (xz)
Step 2: Differentiate each component of vusing the product and chain rules.
dt =
3(2xdx
dt z+ 3x2dz
dt )
6y2dy
dt
dx
dt z+xdz
dt
Step 3: Simplify the expression.
dt =
6xdx
dt z+ 9x2dz
dt
6y2dy
dt
dx
dt z+xdz
dt
Therefore,
dt =
6xdx
dt z+ 9x2dz
dt
6y2dy
dt
dx
dt z+xdz
dt
.
Question 3
Question
Let u=
2
−1
3
and v=
1
4
−2
. Find the derivative of the dot product u·v
with respect to t, where uand vare functions of t.
Solution
To find the derivative of the dot product u·vwith respect to t, we first find
the dot product of the derivatives of uand v.
Step 1: Find dudt and dv
dt
du
dt =
d
dt 2
d
dt (−1)
d
dt 3
=
0
0
0
2
dv
dt =
d
dt 1
d
dt 4
d
dt (−2)
=
0
0
0
Step 2: Find ddt(u·v)using chain rule
d
dt (u·v) = du
dt ·v+u·dv
dt
d
dt (u·v) =
0
0
0
·
1
4
−2
+
2
−1
3
·
0
0
0
= 0
Therefore, the derivative of the dot product u·vwith respect to tis 0.
Question 4
Question
Let u=
3x2
2y
z
and v=
x3
4y2
z2
. Find ∂
∂u(u·v).
Solution
Step 1: Calculate the dot product u·v:
u·v=
3x2
2y
z
·
x3
4y2
z2
= (3x2)(x3) + (2y)(4y2)+(z)(z2)=3x5+ 8y3+z3
Step 2: Differentiate the dot product with respect to uby treating uas a
constant vector: ∂
∂u(u·v) = ∂
∂u(3x5+ 8y3+z3)
Step 3: Since uis treated as a constant vector, the partial derivative with
respect to ueliminates terms with u:
∂
∂u(u·v) = 0
Therefore, ∂
∂u(u·v) = 0.
Question 5
Question
Let v=
3t2
2t+ 1
t
. Find dv
dt .
3
Solution
To find dv
dt , we need to differentiate each component of vwith respect to t.
Step 1: Differentiate the first component
d
dt (3t2)=6t
Step 2: Differentiate the second component
d
dt (2t+ 1) = 2
Step 3: Differentiate the third component
d
dt (t)=1
Therefore, dv
dt =
6t
2
1
.
Question 6
Question
Let v= 2i−3j+√5kand u= 4i+ 2j−2k. Find the derivative of v·uwith
respect to t, where tis a scalar parameter.
Solution
To find the derivative of v·uwith respect to t, we first need to find the expres-
sions for vand uin terms of t.
Step 1: Express vin terms of t.
v= 2i−3j+√5k
Step 2: Let’s add a tcomponent to each unit vector. Let v= 2ti−3tj+
√5tk.
Now, express uin terms of t:
u= 4i+ 2j−2k
Step 3: Let’s add a tcomponent to each unit vector. Let u= 4ti+2tj−2tk.
Now, we can find the derivative of v·uwith respect to t.
Step 4: Calculate the dot product v·u:
v·u= (2t)(4t)+(−3t)(2t)+(√5t)(−2t)
v·u= 8t2−6t2−2√5t2
4
v·u= (8 −6−2√5)t2
v·u= (2 −2√5)t2
Step 5: Differentiate with respect to t:
d
dt (v·u) = 2(2 −2√5)t
d
dt (v·u)=4−4√5)t
Therefore, the derivative of v·uwith respect to tis (4 −4√5)t.
Question 7
Question
Let u= 3i−2j+kand v= 2i+ 5j−4k. If w=u·(v×u), find ∇ · w.
Solution
Step 1: Find v×u.
v×u=
i j k
2 5 −4
3−2 1
= (5 ·1+4·2)i−(2 ·1+3· −4)j+ (2 · −2−5·3)k
= 13i+ 10j−16k
Step 2: Find u·(v×u).
u·(v×u) = (3)(13) + (−2)(10) + (1)(−16)
= 39 −20 −16
= 3
Step 3: Find ∇ · w.
∇ · w=∇ · (3) = 0
Therefore, ∇ · w= 0.
Question 8
Question
Let u=x+y
2x−yand v=3x
4y. Compute the gradient of the dot product
u·vwith respect to x.
5
Solution
Step 1: Compute the dot product u·v:
u·v=x+y
2x−y·3x
4y= (x+y)(3x) + (2x−y)(4y)
= 3x2+ 3xy + 8xy −4y2
= 3x2+ 11xy −4y2
Step 2: Compute the gradient of u·vwith respect to x: To find the gradient
of a scalar function with respect to a vector, we need to take the partial deriva-
tives of the function with respect to each component of the vector and arrange
them as a column vector.
∇x(u·v) = ∂
∂x (3x2+ 11xy −4y2)
∂
∂y (3x2+ 11xy −4y2)
Step 3: Find the partial derivatives:
∂
∂x (3x2+ 11xy −4y2)=6x+ 11y
∂
∂y (3x2+ 11xy −4y2) = 11x−8y
Step 4: Write the gradient as a column vector:
∇x(u·v) = 6x+ 11y
11x−8y
Therefore, the gradient of the dot product u·vwith respect to xis 6x+ 11y
11x−8y.
Question 9
Question
Let u=3
−4and v=5
2. Find d
dt (u·v).
Solution
Step 1: Recall that the dot product of two vectors u=u1
u2and v=v1
v2is
given by u·v=u1v1+u2v2.
Step 2: Let w(t) = u·v= 3 ·5+(−4) ·2.
Step 3: Differentiate w(t) with respect to tto find d
dt (u·v).
Step 4: d
dt (u·v) = d
dt (15 −8).
Step 5: Simplify to find d
dt (u·v) = d
dt (7).
Step 6: Finally, d
dt (u·v) = 0 .
6
Question 10
Question
Let u= 3i+ 4j−2kand v= 2i−j+ 5k. If f(r) = u·rand g(r) = v·r, find
d
dr(f(r)×g(r)).
Solution
Step 1: Let’s first find expressions for f(r) and g(r):
f(r) = u·r= (3i+ 4j−2k)·(xi+yj+zk)=3x+ 4y−2z
g(r) = v·r= (2i−j+ 5k)·(xi+yj+zk)=2x−y+ 5z
Step 2: Next, let’s find the cross product of f(r) and g(r):
f(r)×g(r) =
i j k
3 4 −2
2−1 5
= (4(5)−(−2)(−1))i−(3(5)−(−2)(2))j+(3(−1)−4(2))k
= 22i−19j−11k
Step 3: Finally, we find the derivative with respect to r:
d
dr(f(r)×g(r)) = d
dr(22i−19j−11k) =
22
−19
−11
Question 11
Question
Let f(t) =
2t2
3t
et
be a vector-valued function. Find df
dt .
Solution
Step 1: The function f(t) can be written as f(t) =
2t2
3t
et
.
Step 2: To differentiate f(t) with respect to t, we differentiate each compo-
nent separately.
Step 3: The derivative of the first component is d
dt (2t2)=4t.
Step 4: The derivative of the second component is d
dt (3t) = 3.
Step 5: The derivative of the third component is d
dt (et) = et.
7
Step 6: Therefore, the derivative of f(t) with respect to t,df
dt , is
df
dt =
4t
3
et
.
Question 12
Question
Let r(t) = t3i+tj+t2kbe a vector function, where i,j, and kare the standard
unit vectors in three-dimensional space. Find dr
dt .
Solution
Step 1: We can find dr
dt by differentiating each component of r(t) with respect
to t.
Step 2: Differentiating the x-component t3iwith respect to tgives d
dt (t3)i=
3t2i.
Step 3: Differentiating the y-component tjwith respect to tgives d
dt (t)j=j.
Step 4: Differentiating the z-component t2kwith respect to tgives d
dt (t2)k=
2tk.
Step 5: Therefore, combining the differentiated components, we have dr
dt =
d
dt (t3i+tj+t2k)=3t2i+j+ 2tk.
Question 13
Question
Let u= 3i−2j+ 4kand v= 2i+j−3k. Find d
dt (u·v).
Solution
Step 1: Compute u·v.
u·v= (3i−2j+ 4k)·(2i+j−3k)
= 3 ·2+(−2) ·1+4·(−3)
= 6 −2−12
=−8
Step 2: Differentiate u·vwith respect to t.
d
dt (u·v) = d
dt (−8)
= 0
Therefore, d
dt (u·v) = 0.
8
Question 14
Question
Let u= 3i+ 4jand v= 2i−j. Find d
dt (u·v).
Solution
Step 1: Let u=u1i+u2jand v=v1i+v2j, where u1= 3, u2= 4, v1= 2, and
v2=−1.
Step 2: The dot product of uand vis given by u·v=u1v1+u2v2.
Step 3: Taking the derivative with respect to ton both sides, we have
d
dt (u·v) = d
dt (u1v1+u2v2).
Step 4: Now, differentiate each term separately using the product rule. We
have d
dt (u1v1) + d
dt (u2v2).
Step 5: Substituting our values of u1,v1,u2, and v2, we get
d
dt (3 ·2) + d
dt (4 · −1).
Step 6: Simplifying, we have
6d
dt (1) −4d
dt (1).
Step 7: Since the derivative of a constant is 0, we are left with
6(0) −4(0).
Step 8: Thus, the final answer is 0 .
Question 15
Question
Let r(t) = t2
sin(t)be a vector function. Find dr
dt and d2r
dt2.
Solution
Step 1: To find dr
dt , differentiate each component of r(t) with respect to t.
dr
dt =d
dt (t2)
d
dt (sin(t))
9
=2t
cos(t)
Step 2: To find d2r
dt2, differentiate each component of dr
dt with respect to t.
d2r
dt2=d
dt (2t)
d
dt (cos(t))
=2
−sin(t)
Question 16
Question
Let v=
x2
xy
y2
be a vector function. Find dv
dt , where x=t2and y=√t.
Solution
Step 1: Express vin terms of tusing the given values of xand y.
v=
(t2)2
(t2)√t
(√t)2
=
t4
t5/2
t
Step 2: Differentiate each component of vwith respect to t.
dv
dt =
d(t4)
dt
d(t5/2)
dt
d(t)
dt
=
4t3
5
2t3/2
1
Therefore, dv
dt =
4t3
5
2t3/2
1
.
Question 17
Question
Let r(t) = ti+ 2 cos(t)j−3 sin(t)kbe a vector-valued function. Find dr
dt .
10
Solution
Step 1: Given r(t) = ti+ 2 cos(t)j−3 sin(t)k, we need to find dr
dt .
Step 2: To find dr
dt , we differentiate each component of r(t) with respect to
tseparately. Let’s start by finding d
dt (ti).
Step 3:
d
dt (ti) = d
dt (t)i
= 1 ·i
=i
Step 4: Next, let’s find d
dt (2 cos(t)j).
d
dt (2 cos(t)j) = d
dt (2 cos(t))j
=−2 sin(t)j
Step 5: Finally, we will find d
dt (−3 sin(t)k).
d
dt (−3 sin(t)k) = d
dt (−3 sin(t))k
=−3 cos(t)k
Step 6: Combining the results, we have:
dr
dt =i−2 sin(t)j−3 cos(t)k
Question 18
Question
Find the gradient of the scalar field f(x, y, z) = x2y+y2z+z2x.
Solution
To find the gradient of a scalar field, we need to compute the partial derivatives
of the function with respect to each variable.
Step 1: Find ∂f ∂x
∂f
∂x =∂
∂x (x2y) + ∂
∂x (y2z) + ∂
∂x (z2x)
∂f
∂x = 2xy +z2
Step 2: Find ∂f ∂y
∂f
∂y =x2+ 2yz +z2x
11
Step 3: Find ∂f ∂z
∂f
∂z =y2+ 2xz
Step 4: Gradient of fThe gradient of fis the vector:
∇f=∂f
∂x ,∂f
∂y ,∂f
∂z
Substitute the previously calculated partial derivatives:
∇f=2xy +z2, x2+ 2yz +z2x, y2+ 2xz
Therefore, the gradient of the scalar field f(x, y, z) = x2y+y2z+z2xis ∇f=
(2xy +z2, x2+ 2yz +z2x, y2+ 2xz).
Question 19
Question
Let uand vbe two vectors in R3, where u= 3i−2j+ 4kand v= 2i+ 5j−k.
Find d
dt (u·v).
Solution
Step 1: Recall that the dot product of two vectors uand vis given by u·v=
u1v1+u2v2+u3v3, where u=u1i+u2j+u3kand v=v1i+v2j+v3k.
Step 2: We can compute the dot product of uand vas follows:
u·v= (3i−2j+ 4k)·(2i+ 5j−k)
= 3(2) + (−2)(5) + 4(−1)
= 6 −10 −4
=−8
Step 3: Now, differentiate both sides of the equation with respect to t:
d
dt (u·v) = d
dt (−8)
= 0
Step 4: Therefore, d
dt (u·v) = 0 .
Question 20
Question
Let u= 3i−2j+ 4kand v= 2i+ 5j−k. Find the derivative of u·vwith
respect to t.
12
Solution
To find the derivative of u·vwith respect to t, we first need to find the expres-
sions for u·vand its derivative.
Step 1: Calculate u ·v
u·v= (3i−2j+ 4k)·(2i+ 5j−k)
u·v= 3(2) + (−2)(5) + 4(−1)
u·v= 6 −10 −4
u·v=−8
Step 2: Find the derivative Let f(t) = u·v. We want to find df
dt .
df
dt =d
dt (−8)
df
dt = 0
Therefore, the derivative of u·vwith respect to tis 0.
Question 21
Question
Let uand vbe constant vectors, and let r(t)=2tu−3tv+t2u·v. Find dr
dt .
Solution
Step 1: Find dr
dt using the rules of vector differentiation.
dr
dt =d
dt (2tu)−d
dt (3tv) + d
dt (t2u·v)
Step 2: Differentiate each term with respect to t.
dr
dt = 2u−3v+ 2tu·v
Step 3: Simplify the expression.
dr
dt = 2u−3v+ 2tu·v
Therefore, the derivative of r(t) with respect to tis 2u−3v+ 2tu·v.
13
Question 22
Question
Let u=3
5and v=−2
4. Compute the derivative of the dot product u·v
with respect to t, where uand vare both functions of t.
Solution
To find the derivative of the dot product u·vwith respect to t, we first need
to express uand vas functions of t. Let u(t) = 3t
5tand v(t) = −2t
4t.
Step 1: Compute u(t)·v(t).
u(t)·v(t) = 3t
5t·−2t
4t= (3t)(−2t) + (5t)(4t) = −6t2+ 20t2= 14t2
Step 2: Find d
dt (u(t)·v(t)) using the product rule.
d
dt (u(t)·v(t)) = d
dt (14t2)
= 28t
Therefore, the derivative of the dot product u·vwith respect to tis 28t.
Question 23
Question
Let u=
3
−4
5
and v=
−2
7
1
. Determine d
dt (u·v) where u·vrepresents the
dot product of vectors uand v.
Solution
To find the derivative of the dot product of two vectors with respect to t, we
can use the formula d
dt (u·v) = du
dt ·v+u·dv
dt .
Given u=
3
−4
5
and v=
−2
7
1
, we can find du
dt and dv
dt : - du
dt =
d
dt (3)
d
dt (−4)
d
dt (5)
=
0
0
0
-dv
dt =
d
dt (−2)
d
dt (7)
d
dt (1)
=
0
0
0
14
Therefore, d
dt (u·v) =
0
0
0
·
−2
7
1
+
3
−4
5
·
0
0
0
= 0.
Hence, the derivative of u·vwith respect to tis 0.
Question 24
Question
Let u=3
−2and v=−1
4. Compute the following derivative: d
dt (3u−tv+
2t2u).
Solution
Step 1: Use the properties of vector calculus to differentiate each term sepa-
rately. Step 2: Let’s differentiate 3uwith respect to tfirst.
d
dt (3u)=3du
dt
Step 3: Now, differentiate uwith respect to t.
du
dt =du1
dt
du2
dt =0
0
Therefore, d
dt (3u)=30
0=0
0. Step 4: Moving on to the second term −tv.
d
dt (−tv) = −v
Step 5: Next, differentiate 2t2uwith respect to t.
d
dt (2t2u)=2ud
dt (t2)
Step 6: Compute the derivative of t2with respect to t.
d
dt (t2)=2t
Therefore, d
dt (2t2u)=23
−2(2t) = 12t
−8t. Step 7: Finally, sum up all the
derivatives. d
dt (3u−tv+ 2t2u) = 0
0−−1
4+12t
−8t
=12t+ 1
−8t−4
15
Question 25
Question
Let v = 3x2ˆ
i+xsin(2y)ˆ
jbe a vector field. Find ∇ ·v.
Solution
Step 1: Recall that the divergence of a vector field v =P(x, y)ˆ
i+Q(x, y)ˆ
jis
given by the expression ∇ ·v =∂P
∂x +∂Q
∂y .
Step 2: In this case, P(x, y)=3x2and Q(x, y) = xsin(2y).
Step 3: Calculate the partial derivatives:
∂P
∂x =∂
∂x (3x2)=6x
and ∂Q
∂y =∂
∂y (xsin(2y)) = x(2 cos(2y)) = 2xcos(2y)
Step 4: Add the partial derivatives to find the divergence:
∇ ·v =∂P
∂x +∂Q
∂y = 6x+ 2xcos(2y)=2x(3 + cos(2y))
Step 5: Therefore, the divergence of the vector field v is ∇ · v = 2x(3 +
cos(2y)).
Question 26
Question
Let u= 3i−2j+ 4kand v= 2i+ 5j−3k. Find d
dt (u·v).
Solution
To find d
dt (u·v), we first need to compute the dot product of uand v: Step 1:
Calculate u·v.
u·v= (3i−2j+ 4k)·(2i+ 5j−3k)
u·v= 3(2) + (−2)(5) + 4(−3)
u·v= 6 −10 −12
u·v=−16
Step 2: Differentiate the dot product with respect to t.
d
dt (u·v) = d
dt (−16)
d
dt (u·v)=0
Therefore, d
dt (u·v) = 0.
16
Question 27
Question
Let u= 3i−2j+kand v= 2i+ 4j−k. Find the derivative of u·vwith respect
to t, where u=
3
−2
1
,v=
2
4
−1
and tis a scalar variable.
Solution
Step 1: Recall that the dot product of two vectors u=
u1
u2
u3
and v=
v1
v2
v3
is given by u·v=u1v1+u2v2+u3v3.
Step 2: Given u=
3
−2
1
and v=
2
4
−1
, we can find the dot product as
follows: u·v= (3)(2) + (−2)(4) + (1)(−1).
Step 3: Simplifying the dot product, we get u·v= 6 −8−1 = −3.
Step 4: Now, let’s differentiate u·vwith respect to t. Since u=
3
−2
1
and
v=
2
4
−1
are functions of time t, the derivative can be found using the chain
rule.
Step 5: Let f(t) = u·v. Then, df
dt =d(u·v)
dt .
Step 6: By the chain rule, we have df
dt =d(u·v)
du·du
dt +d(u·v)
dv·dv
dt .
Step 7: Since the dot product is a scalar, d(u·v)
du=vand d(u·v)
dv=u.
Step 8: Therefore, df
dt =v·du
dt +u·dv
dt .
Step 9: Substituting the given values for uand v, we have df
dt =
2
4
−1
·
d
dt (3)
d
dt (−2)
d
dt (1)
+
3
−2
1
·
d
dt (2)
d
dt (4)
d
dt (−1)
.
Step 10: Simplifying, we get df
dt =
Question 28
Question
Let f(x, y)=(excos(y), exsin(y)) be a vector-valued function. Find ∂f
∂x and ∂f
∂y .
17
Solution
Step 1: To find ∂f
∂x , we differentiate each component of fwith respect to x.
∂f
∂x =∂
∂x excos(y),∂
∂x exsin(y)
Step 2: Taking the partial derivative of excos(y) with respect to xgives:
∂
∂x excos(y) = excos(y)
Step 3: Taking the partial derivative of exsin(y) with respect to xgives:
∂
∂x exsin(y) = exsin(y)
Step 4: Therefore, we have ∂f
∂x = (excos(y), exsin(y)).
Step 5: Next, to find ∂f
∂y , we differentiate each component of fwith respect
to y.
∂f
∂y =∂
∂y excos(y),∂
∂y exsin(y)
Step 6: Taking the partial derivative of excos(y) with respect to ygives:
∂
∂y excos(y) = −exsin(y)
Step 7: Taking the partial derivative of exsin(y) with respect to ygives:
∂
∂y exsin(y) = excos(y)
Step 8: Therefore, we have ∂f
∂y = (−exsin(y), excos(y)).
Question 29
Question
Let v= 3x2i−2y2j+zkbe a vector field. Find the gradient of v.
Solution
To find the gradient of v, we need to find the partial derivatives of each com-
ponent with respect to x,y, and z.
Step 1: Find ∂
∂x :
∂
∂x (3x2)=6x, ∂
∂x (−2y2)=0,∂
∂x z= 0
18
So, the x-component of the gradient is 6xi.
Step 2: Find ∂
∂y :
∂
∂y (3x2) = 0,∂
∂y (−2y2) = −4y, ∂
∂y z= 0
Therefore, the y-component of the gradient is −4yj.
Step 3: Find ∂
∂z :
∂
∂z (3x2)=0,∂
∂z (−2y2) = 0,∂
∂z z= 1
Hence, the z-component of the gradient is 1k.
Thus, the gradient of vis ∇ · v= 6xi−4yj+ 1k.
Question 30
Question
Let u= 3i+ 2j−kand v= 2i−4j+ 5k. Compute the derivative of the vector
u·vwith respect to t, where uand vare both functions of t.
Solution
To find the derivative of u·vwith respect to t, we first find their dot product
and then differentiate with respect to t.
Step 1: Find u ·v
u·v= (3i+ 2j−k)·(2i−4j+ 5k) = 3(2) + 2(−4) + (−1)(5) = 6 −8−5 = −7
Step 2: Differentiate with respect to tLet u=u1i+u2j+u3kand
v=v1i+v2j+v3k. Then,
d
dt (u·v) = d
dt (u1v1+u2v2+u3v3)
Step 3: Compute the derivative Since uand vare functions of t, we
have: d
dt (u·v) = d
dt (3 ·2+2·(−4) + (−1) ·5) = d
dt (−7) = 0
Question 31
Question
Let u=x2
3xand v=2xy
y2. Calculate d(u·v)
dx .
19
Solution
Step 1: We start by finding u·v:
u·v= (x2)(2xy) + (3x)(y2)=2x3y+ 3xy2
Step 2: Next, we differentiate u·vwith respect to x:
d(u·v)
dx =d(2x3y+ 3xy2)
dx
Step 3: Applying the differentiation rules:
d(u·v)
dx =d(2x3y)
dx +d(3xy2)
dx
Step 4: Using the product rule for differentiation, we have:
d(u·v)
dx = 2(3x2)y+ 2x3dy
dx + 3y2+ 3xdy
dx
Step 5: Simplifying further gives:
d(u·v)
dx = 6x2y+ 2x3dy
dx + 3y2+ 3xdy
dx
Therefore, d(u·v)
dx = 6x2y+ 2x3dy
dx + 3y2+ 3xdy
dx .
Question 32
Question
Let v= 3x2i+ 4xyj−2zk. Find ∇ · v.
Solution
Step 1: Compute ∇ · vby taking the dot product of the vector differential
operator (∇) and the vector field v.
∇ · v=∂
∂x (3x2) + ∂
∂y (4xy) + ∂
∂z (−2z)
Step 2: Calculate the partial derivatives.
∂
∂x (3x2)=6x, ∂
∂y (4xy) = 4x, ∂
∂z (−2z) = −2
Step 3: Substitute the partial derivatives back into the expression for ∇·v.
∇ · v= 6x+ 4x−2 = 10x−2
Therefore, the divergence of the vector field vis 10x−2 .
20
Question 33
Question
Let u= 3i−2j+ 4kand v= 2i+ 5j−1k. Compute the derivative of u·vwith
respect to t, where uand vare functions of t.
Solution
To find the derivative of u·vwith respect to t, we first need to find expressions
for uand vin terms of t. Given that uand vare vectors, we can express them
as:
u(t)=3i−2j+ 4k
v(t)=2i+ 5j−1k
Now, to find the derivative of u·vwith respect to t, we will use the property
that d
dt (u·v) = u·
dt +
dt ·v.
Step 1: Find dudt and
dt
dt = ddt(3i−2j+4k)=0i+0j+0k=0
dt = ddt(2i+5j−1k)=0i+0j+0k=0
Step 2: Compute u·
dtand
dt ·v
u·
dt = (3i- 2j+ 4k)·0= 0
dt ·v=0·(2i+ 5j−1k)=0
Step 3: Find the derivative of u ·vHaving computed the components,
we see that both terms are zero. Hence, the derivative of u·vwith respect to
tis 0.
Question 34
Question
Let vand wbe two differentiable vectors given by
v=
2t3
t2
5
and w=
4t
3t
2t
.
Find the derivative of v·wwith respect to t.
21
Solution
To find the derivative of v·w, we use the product rule for differentiation of
vectors. The dot product of two vectors aand bis given by a·b=a1b1+
a2b2+a3b3.
Step 1: Find v·w:
v·w= (2t3)(4t)+(t2)(3t) + (5)(2t) = 8t4+ 3t3+ 10t.
Step 2: Differentiate v·wwith respect to tusing the product rule:
d
dt (v·w) = d
dt (8t4+ 3t3+ 10t)
= 32t3+ 9t2+ 10.
Thus, the derivative of v·wwith respect to tis 32t3+ 9t2+ 10 .
Question 35
Question
Let v=1
x
ln(x+y)and u=exy
sin(xy). Find d
dx (v·u).
Solution
To find d
dx (v·u), we first need to find the dot product of vand u:
v·u=1
x
ln(x+y)·exy
sin(xy)=1
xexy + ln(x+y) sin(xy)
Now, we will differentiate v·uwith respect to xusing the product rule:
d
dx (v·u) = d
dx 1
xexy + ln(x+y) sin(xy)
Step 1: Differentiate the first term:
d
dx 1
xexy=−1
x2exy +1
xyexy
Step 2: Differentiate the second term:
d
dx (ln(x+y) sin(xy)) = 1
x+y·d
dx (x+y) sin(xy) + ln(x+y) cos(xy)·d
dx (xy)
Step 3: Simplify the derivative:
d
dx (v·u) = −1
x2exy +1
xyexy +1
x+yexy sin(xy) + ln(x+y) cos(xy)
Therefore, d
dx (v·u) = −1
x2exy +1
xyexy +1
x+yexy sin(xy) + ln(x+y) cos(xy).
22
Solution
Step 1: Find
dtbydif ferentiatingeachcomponentof vwithrespecttot.
dt =
d
dt (3x2z)
d
dt (2y3)
d
dt (xz)
Step 2: Differentiate each component of vusing the product and chain rules.
dt =
3(2xdx
dt z+ 3x2dz
dt )
6y2dy
dt
dx
dt z+xdz
dt
Step 3: Simplify the expression.
dt =
6xdx
dt z+ 9x2dz
dt
6y2dy
dt
dx
dt z+xdz
dt
Therefore,
dt =
6xdx
dt z+ 9x2dz
dt
6y2dy
dt
dx
dt z+xdz
dt
.
Question 3
Question
Let u=
2
−1
3
and v=
1
4
−2
. Find the derivative of the dot product u·v
with respect to t, where uand vare functions of t.
Solution
To find the derivative of the dot product u·vwith respect to t, we first find
the dot product of the derivatives of uand v.
Step 1: Find dudt and dv
dt
du
dt =
d
dt 2
d
dt (−1)
d
dt 3
=
0
0
0
2
dv
dt =
d
dt 1
d
dt 4
d
dt (−2)
=
0
0
0
Step 2: Find ddt(u·v)using chain rule
d
dt (u·v) = du
dt ·v+u·dv
dt
d
dt (u·v) =
0
0
0
·
1
4
−2
+
2
−1
3
·
0
0
0
= 0
Therefore, the derivative of the dot product u·vwith respect to tis 0.
Question 4
Question
Let u=
3x2
2y
z
and v=
x3
4y2
z2
. Find ∂
∂u(u·v).
Solution
Step 1: Calculate the dot product u·v:
u·v=
3x2
2y
z
·
x3
4y2
z2
= (3x2)(x3) + (2y)(4y2)+(z)(z2)=3x5+ 8y3+z3
Step 2: Differentiate the dot product with respect to uby treating uas a
constant vector: ∂
∂u(u·v) = ∂
∂u(3x5+ 8y3+z3)
Step 3: Since uis treated as a constant vector, the partial derivative with
respect to ueliminates terms with u:
∂
∂u(u·v) = 0
Therefore, ∂
∂u(u·v) = 0.
Question 5
Question
Let v=
3t2
2t+ 1
t
. Find dv
dt .
3
Solution
To find dv
dt , we need to differentiate each component of vwith respect to t.
Step 1: Differentiate the first component
d
dt (3t2)=6t
Step 2: Differentiate the second component
d
dt (2t+ 1) = 2
Step 3: Differentiate the third component
d
dt (t)=1
Therefore, dv
dt =
6t
2
1
.
Question 6
Question
Let v= 2i−3j+√5kand u= 4i+ 2j−2k. Find the derivative of v·uwith
respect to t, where tis a scalar parameter.
Solution
To find the derivative of v·uwith respect to t, we first need to find the expres-
sions for vand uin terms of t.
Step 1: Express vin terms of t.
v= 2i−3j+√5k
Step 2: Let’s add a tcomponent to each unit vector. Let v= 2ti−3tj+
√5tk.
Now, express uin terms of t:
u= 4i+ 2j−2k
Step 3: Let’s add a tcomponent to each unit vector. Let u= 4ti+2tj−2tk.
Now, we can find the derivative of v·uwith respect to t.
Step 4: Calculate the dot product v·u:
v·u= (2t)(4t)+(−3t)(2t)+(√5t)(−2t)
v·u= 8t2−6t2−2√5t2
4
v·u= (8 −6−2√5)t2
v·u= (2 −2√5)t2
Step 5: Differentiate with respect to t:
d
dt (v·u) = 2(2 −2√5)t
d
dt (v·u)=4−4√5)t
Therefore, the derivative of v·uwith respect to tis (4 −4√5)t.
Question 7
Question
Let u= 3i−2j+kand v= 2i+ 5j−4k. If w=u·(v×u), find ∇ · w.
Solution
Step 1: Find v×u.
v×u=
i j k
2 5 −4
3−2 1
= (5 ·1+4·2)i−(2 ·1+3· −4)j+ (2 · −2−5·3)k
= 13i+ 10j−16k
Step 2: Find u·(v×u).
u·(v×u) = (3)(13) + (−2)(10) + (1)(−16)
= 39 −20 −16
= 3
Step 3: Find ∇ · w.
∇ · w=∇ · (3) = 0
Therefore, ∇ · w= 0.
Question 8
Question
Let u=x+y
2x−yand v=3x
4y. Compute the gradient of the dot product
u·vwith respect to x.
5
Solution
Step 1: Compute the dot product u·v:
u·v=x+y
2x−y·3x
4y= (x+y)(3x) + (2x−y)(4y)
= 3x2+ 3xy + 8xy −4y2
= 3x2+ 11xy −4y2
Step 2: Compute the gradient of u·vwith respect to x: To find the gradient
of a scalar function with respect to a vector, we need to take the partial deriva-
tives of the function with respect to each component of the vector and arrange
them as a column vector.
∇x(u·v) = ∂
∂x (3x2+ 11xy −4y2)
∂
∂y (3x2+ 11xy −4y2)
Step 3: Find the partial derivatives:
∂
∂x (3x2+ 11xy −4y2)=6x+ 11y
∂
∂y (3x2+ 11xy −4y2) = 11x−8y
Step 4: Write the gradient as a column vector:
∇x(u·v) = 6x+ 11y
11x−8y
Therefore, the gradient of the dot product u·vwith respect to xis 6x+ 11y
11x−8y.
Question 9
Question
Let u=3
−4and v=5
2. Find d
dt (u·v).
Solution
Step 1: Recall that the dot product of two vectors u=u1
u2and v=v1
v2is
given by u·v=u1v1+u2v2.
Step 2: Let w(t) = u·v= 3 ·5+(−4) ·2.
Step 3: Differentiate w(t) with respect to tto find d
dt (u·v).
Step 4: d
dt (u·v) = d
dt (15 −8).
Step 5: Simplify to find d
dt (u·v) = d
dt (7).
Step 6: Finally, d
dt (u·v) = 0 .
6
Question 10
Question
Let u= 3i+ 4j−2kand v= 2i−j+ 5k. If f(r) = u·rand g(r) = v·r, find
d
dr(f(r)×g(r)).
Solution
Step 1: Let’s first find expressions for f(r) and g(r):
f(r) = u·r= (3i+ 4j−2k)·(xi+yj+zk)=3x+ 4y−2z
g(r) = v·r= (2i−j+ 5k)·(xi+yj+zk)=2x−y+ 5z
Step 2: Next, let’s find the cross product of f(r) and g(r):
f(r)×g(r) =
i j k
3 4 −2
2−1 5
= (4(5)−(−2)(−1))i−(3(5)−(−2)(2))j+(3(−1)−4(2))k
= 22i−19j−11k
Step 3: Finally, we find the derivative with respect to r:
d
dr(f(r)×g(r)) = d
dr(22i−19j−11k) =
22
−19
−11
Question 11
Question
Let f(t) =
2t2
3t
et
be a vector-valued function. Find df
dt .
Solution
Step 1: The function f(t) can be written as f(t) =
2t2
3t
et
.
Step 2: To differentiate f(t) with respect to t, we differentiate each compo-
nent separately.
Step 3: The derivative of the first component is d
dt (2t2)=4t.
Step 4: The derivative of the second component is d
dt (3t) = 3.
Step 5: The derivative of the third component is d
dt (et) = et.
7
Step 6: Therefore, the derivative of f(t) with respect to t,df
dt , is
df
dt =
4t
3
et
.
Question 12
Question
Let r(t) = t3i+tj+t2kbe a vector function, where i,j, and kare the standard
unit vectors in three-dimensional space. Find dr
dt .
Solution
Step 1: We can find dr
dt by differentiating each component of r(t) with respect
to t.
Step 2: Differentiating the x-component t3iwith respect to tgives d
dt (t3)i=
3t2i.
Step 3: Differentiating the y-component tjwith respect to tgives d
dt (t)j=j.
Step 4: Differentiating the z-component t2kwith respect to tgives d
dt (t2)k=
2tk.
Step 5: Therefore, combining the differentiated components, we have dr
dt =
d
dt (t3i+tj+t2k)=3t2i+j+ 2tk.
Question 13
Question
Let u= 3i−2j+ 4kand v= 2i+j−3k. Find d
dt (u·v).
Solution
Step 1: Compute u·v.
u·v= (3i−2j+ 4k)·(2i+j−3k)
= 3 ·2+(−2) ·1+4·(−3)
= 6 −2−12
=−8
Step 2: Differentiate u·vwith respect to t.
d
dt (u·v) = d
dt (−8)
= 0
Therefore, d
dt (u·v) = 0.
8
Question 14
Question
Let u= 3i+ 4jand v= 2i−j. Find d
dt (u·v).
Solution
Step 1: Let u=u1i+u2jand v=v1i+v2j, where u1= 3, u2= 4, v1= 2, and
v2=−1.
Step 2: The dot product of uand vis given by u·v=u1v1+u2v2.
Step 3: Taking the derivative with respect to ton both sides, we have
d
dt (u·v) = d
dt (u1v1+u2v2).
Step 4: Now, differentiate each term separately using the product rule. We
have d
dt (u1v1) + d
dt (u2v2).
Step 5: Substituting our values of u1,v1,u2, and v2, we get
d
dt (3 ·2) + d
dt (4 · −1).
Step 6: Simplifying, we have
6d
dt (1) −4d
dt (1).
Step 7: Since the derivative of a constant is 0, we are left with
6(0) −4(0).
Step 8: Thus, the final answer is 0 .
Question 15
Question
Let r(t) = t2
sin(t)be a vector function. Find dr
dt and d2r
dt2.
Solution
Step 1: To find dr
dt , differentiate each component of r(t) with respect to t.
dr
dt =d
dt (t2)
d
dt (sin(t))
9
=2t
cos(t)
Step 2: To find d2r
dt2, differentiate each component of dr
dt with respect to t.
d2r
dt2=d
dt (2t)
d
dt (cos(t))
=2
−sin(t)
Question 16
Question
Let v=
x2
xy
y2
be a vector function. Find dv
dt , where x=t2and y=√t.
Solution
Step 1: Express vin terms of tusing the given values of xand y.
v=
(t2)2
(t2)√t
(√t)2
=
t4
t5/2
t
Step 2: Differentiate each component of vwith respect to t.
dv
dt =
d(t4)
dt
d(t5/2)
dt
d(t)
dt
=
4t3
5
2t3/2
1
Therefore, dv
dt =
4t3
5
2t3/2
1
.
Question 17
Question
Let r(t) = ti+ 2 cos(t)j−3 sin(t)kbe a vector-valued function. Find dr
dt .
10
Solution
Step 1: Given r(t) = ti+ 2 cos(t)j−3 sin(t)k, we need to find dr
dt .
Step 2: To find dr
dt , we differentiate each component of r(t) with respect to
tseparately. Let’s start by finding d
dt (ti).
Step 3:
d
dt (ti) = d
dt (t)i
= 1 ·i
=i
Step 4: Next, let’s find d
dt (2 cos(t)j).
d
dt (2 cos(t)j) = d
dt (2 cos(t))j
=−2 sin(t)j
Step 5: Finally, we will find d
dt (−3 sin(t)k).
d
dt (−3 sin(t)k) = d
dt (−3 sin(t))k
=−3 cos(t)k
Step 6: Combining the results, we have:
dr
dt =i−2 sin(t)j−3 cos(t)k
Question 18
Question
Find the gradient of the scalar field f(x, y, z) = x2y+y2z+z2x.
Solution
To find the gradient of a scalar field, we need to compute the partial derivatives
of the function with respect to each variable.
Step 1: Find ∂f ∂x
∂f
∂x =∂
∂x (x2y) + ∂
∂x (y2z) + ∂
∂x (z2x)
∂f
∂x = 2xy +z2
Step 2: Find ∂f ∂y
∂f
∂y =x2+ 2yz +z2x
11
Step 3: Find ∂f ∂z
∂f
∂z =y2+ 2xz
Step 4: Gradient of fThe gradient of fis the vector:
∇f=∂f
∂x ,∂f
∂y ,∂f
∂z
Substitute the previously calculated partial derivatives:
∇f=2xy +z2, x2+ 2yz +z2x, y2+ 2xz
Therefore, the gradient of the scalar field f(x, y, z) = x2y+y2z+z2xis ∇f=
(2xy +z2, x2+ 2yz +z2x, y2+ 2xz).
Question 19
Question
Let uand vbe two vectors in R3, where u= 3i−2j+ 4kand v= 2i+ 5j−k.
Find d
dt (u·v).
Solution
Step 1: Recall that the dot product of two vectors uand vis given by u·v=
u1v1+u2v2+u3v3, where u=u1i+u2j+u3kand v=v1i+v2j+v3k.
Step 2: We can compute the dot product of uand vas follows:
u·v= (3i−2j+ 4k)·(2i+ 5j−k)
= 3(2) + (−2)(5) + 4(−1)
= 6 −10 −4
=−8
Step 3: Now, differentiate both sides of the equation with respect to t:
d
dt (u·v) = d
dt (−8)
= 0
Step 4: Therefore, d
dt (u·v) = 0 .
Question 20
Question
Let u= 3i−2j+ 4kand v= 2i+ 5j−k. Find the derivative of u·vwith
respect to t.
12
Solution
To find the derivative of u·vwith respect to t, we first need to find the expres-
sions for u·vand its derivative.
Step 1: Calculate u ·v
u·v= (3i−2j+ 4k)·(2i+ 5j−k)
u·v= 3(2) + (−2)(5) + 4(−1)
u·v= 6 −10 −4
u·v=−8
Step 2: Find the derivative Let f(t) = u·v. We want to find df
dt .
df
dt =d
dt (−8)
df
dt = 0
Therefore, the derivative of u·vwith respect to tis 0.
Question 21
Question
Let uand vbe constant vectors, and let r(t)=2tu−3tv+t2u·v. Find dr
dt .
Solution
Step 1: Find dr
dt using the rules of vector differentiation.
dr
dt =d
dt (2tu)−d
dt (3tv) + d
dt (t2u·v)
Step 2: Differentiate each term with respect to t.
dr
dt = 2u−3v+ 2tu·v
Step 3: Simplify the expression.
dr
dt = 2u−3v+ 2tu·v
Therefore, the derivative of r(t) with respect to tis 2u−3v+ 2tu·v.
13
Question 22
Question
Let u=3
5and v=−2
4. Compute the derivative of the dot product u·v
with respect to t, where uand vare both functions of t.
Solution
To find the derivative of the dot product u·vwith respect to t, we first need
to express uand vas functions of t. Let u(t) = 3t
5tand v(t) = −2t
4t.
Step 1: Compute u(t)·v(t).
u(t)·v(t) = 3t
5t·−2t
4t= (3t)(−2t) + (5t)(4t) = −6t2+ 20t2= 14t2
Step 2: Find d
dt (u(t)·v(t)) using the product rule.
d
dt (u(t)·v(t)) = d
dt (14t2)
= 28t
Therefore, the derivative of the dot product u·vwith respect to tis 28t.
Question 23
Question
Let u=
3
−4
5
and v=
−2
7
1
. Determine d
dt (u·v) where u·vrepresents the
dot product of vectors uand v.
Solution
To find the derivative of the dot product of two vectors with respect to t, we
can use the formula d
dt (u·v) = du
dt ·v+u·dv
dt .
Given u=
3
−4
5
and v=
−2
7
1
, we can find du
dt and dv
dt : - du
dt =
d
dt (3)
d
dt (−4)
d
dt (5)
=
0
0
0
-dv
dt =
d
dt (−2)
d
dt (7)
d
dt (1)
=
0
0
0
14
Therefore, d
dt (u·v) =
0
0
0
·
−2
7
1
+
3
−4
5
·
0
0
0
= 0.
Hence, the derivative of u·vwith respect to tis 0.
Question 24
Question
Let u=3
−2and v=−1
4. Compute the following derivative: d
dt (3u−tv+
2t2u).
Solution
Step 1: Use the properties of vector calculus to differentiate each term sepa-
rately. Step 2: Let’s differentiate 3uwith respect to tfirst.
d
dt (3u)=3du
dt
Step 3: Now, differentiate uwith respect to t.
du
dt =du1
dt
du2
dt =0
0
Therefore, d
dt (3u)=30
0=0
0. Step 4: Moving on to the second term −tv.
d
dt (−tv) = −v
Step 5: Next, differentiate 2t2uwith respect to t.
d
dt (2t2u)=2ud
dt (t2)
Step 6: Compute the derivative of t2with respect to t.
d
dt (t2)=2t
Therefore, d
dt (2t2u)=23
−2(2t) = 12t
−8t. Step 7: Finally, sum up all the
derivatives. d
dt (3u−tv+ 2t2u) = 0
0−−1
4+12t
−8t
=12t+ 1
−8t−4
15
Question 25
Question
Let v = 3x2ˆ
i+xsin(2y)ˆ
jbe a vector field. Find ∇ ·v.
Solution
Step 1: Recall that the divergence of a vector field v =P(x, y)ˆ
i+Q(x, y)ˆ
jis
given by the expression ∇ ·v =∂P
∂x +∂Q
∂y .
Step 2: In this case, P(x, y)=3x2and Q(x, y) = xsin(2y).
Step 3: Calculate the partial derivatives:
∂P
∂x =∂
∂x (3x2)=6x
and ∂Q
∂y =∂
∂y (xsin(2y)) = x(2 cos(2y)) = 2xcos(2y)
Step 4: Add the partial derivatives to find the divergence:
∇ ·v =∂P
∂x +∂Q
∂y = 6x+ 2xcos(2y)=2x(3 + cos(2y))
Step 5: Therefore, the divergence of the vector field v is ∇ · v = 2x(3 +
cos(2y)).
Question 26
Question
Let u= 3i−2j+ 4kand v= 2i+ 5j−3k. Find d
dt (u·v).
Solution
To find d
dt (u·v), we first need to compute the dot product of uand v: Step 1:
Calculate u·v.
u·v= (3i−2j+ 4k)·(2i+ 5j−3k)
u·v= 3(2) + (−2)(5) + 4(−3)
u·v= 6 −10 −12
u·v=−16
Step 2: Differentiate the dot product with respect to t.
d
dt (u·v) = d
dt (−16)
d
dt (u·v)=0
Therefore, d
dt (u·v) = 0.
16
Question 27
Question
Let u= 3i−2j+kand v= 2i+ 4j−k. Find the derivative of u·vwith respect
to t, where u=
3
−2
1
,v=
2
4
−1
and tis a scalar variable.
Solution
Step 1: Recall that the dot product of two vectors u=
u1
u2
u3
and v=
v1
v2
v3
is given by u·v=u1v1+u2v2+u3v3.
Step 2: Given u=
3
−2
1
and v=
2
4
−1
, we can find the dot product as
follows: u·v= (3)(2) + (−2)(4) + (1)(−1).
Step 3: Simplifying the dot product, we get u·v= 6 −8−1 = −3.
Step 4: Now, let’s differentiate u·vwith respect to t. Since u=
3
−2
1
and
v=
2
4
−1
are functions of time t, the derivative can be found using the chain
rule.
Step 5: Let f(t) = u·v. Then, df
dt =d(u·v)
dt .
Step 6: By the chain rule, we have df
dt =d(u·v)
du·du
dt +d(u·v)
dv·dv
dt .
Step 7: Since the dot product is a scalar, d(u·v)
du=vand d(u·v)
dv=u.
Step 8: Therefore, df
dt =v·du
dt +u·dv
dt .
Step 9: Substituting the given values for uand v, we have df
dt =
2
4
−1
·
d
dt (3)
d
dt (−2)
d
dt (1)
+
3
−2
1
·
d
dt (2)
d
dt (4)
d
dt (−1)
.
Step 10: Simplifying, we get df
dt =
Question 28
Question
Let f(x, y)=(excos(y), exsin(y)) be a vector-valued function. Find ∂f
∂x and ∂f
∂y .
17
Solution
Step 1: To find ∂f
∂x , we differentiate each component of fwith respect to x.
∂f
∂x =∂
∂x excos(y),∂
∂x exsin(y)
Step 2: Taking the partial derivative of excos(y) with respect to xgives:
∂
∂x excos(y) = excos(y)
Step 3: Taking the partial derivative of exsin(y) with respect to xgives:
∂
∂x exsin(y) = exsin(y)
Step 4: Therefore, we have ∂f
∂x = (excos(y), exsin(y)).
Step 5: Next, to find ∂f
∂y , we differentiate each component of fwith respect
to y.
∂f
∂y =∂
∂y excos(y),∂
∂y exsin(y)
Step 6: Taking the partial derivative of excos(y) with respect to ygives:
∂
∂y excos(y) = −exsin(y)
Step 7: Taking the partial derivative of exsin(y) with respect to ygives:
∂
∂y exsin(y) = excos(y)
Step 8: Therefore, we have ∂f
∂y = (−exsin(y), excos(y)).
Question 29
Question
Let v= 3x2i−2y2j+zkbe a vector field. Find the gradient of v.
Solution
To find the gradient of v, we need to find the partial derivatives of each com-
ponent with respect to x,y, and z.
Step 1: Find ∂
∂x :
∂
∂x (3x2)=6x, ∂
∂x (−2y2)=0,∂
∂x z= 0
18
So, the x-component of the gradient is 6xi.
Step 2: Find ∂
∂y :
∂
∂y (3x2) = 0,∂
∂y (−2y2) = −4y, ∂
∂y z= 0
Therefore, the y-component of the gradient is −4yj.
Step 3: Find ∂
∂z :
∂
∂z (3x2)=0,∂
∂z (−2y2) = 0,∂
∂z z= 1
Hence, the z-component of the gradient is 1k.
Thus, the gradient of vis ∇ · v= 6xi−4yj+ 1k.
Question 30
Question
Let u= 3i+ 2j−kand v= 2i−4j+ 5k. Compute the derivative of the vector
u·vwith respect to t, where uand vare both functions of t.
Solution
To find the derivative of u·vwith respect to t, we first find their dot product
and then differentiate with respect to t.
Step 1: Find u ·v
u·v= (3i+ 2j−k)·(2i−4j+ 5k) = 3(2) + 2(−4) + (−1)(5) = 6 −8−5 = −7
Step 2: Differentiate with respect to tLet u=u1i+u2j+u3kand
v=v1i+v2j+v3k. Then,
d
dt (u·v) = d
dt (u1v1+u2v2+u3v3)
Step 3: Compute the derivative Since uand vare functions of t, we
have: d
dt (u·v) = d
dt (3 ·2+2·(−4) + (−1) ·5) = d
dt (−7) = 0
Question 31
Question
Let u=x2
3xand v=2xy
y2. Calculate d(u·v)
dx .
19
Solution
Step 1: We start by finding u·v:
u·v= (x2)(2xy) + (3x)(y2)=2x3y+ 3xy2
Step 2: Next, we differentiate u·vwith respect to x:
d(u·v)
dx =d(2x3y+ 3xy2)
dx
Step 3: Applying the differentiation rules:
d(u·v)
dx =d(2x3y)
dx +d(3xy2)
dx
Step 4: Using the product rule for differentiation, we have:
d(u·v)
dx = 2(3x2)y+ 2x3dy
dx + 3y2+ 3xdy
dx
Step 5: Simplifying further gives:
d(u·v)
dx = 6x2y+ 2x3dy
dx + 3y2+ 3xdy
dx
Therefore, d(u·v)
dx = 6x2y+ 2x3dy
dx + 3y2+ 3xdy
dx .
Question 32
Question
Let v= 3x2i+ 4xyj−2zk. Find ∇ · v.
Solution
Step 1: Compute ∇ · vby taking the dot product of the vector differential
operator (∇) and the vector field v.
∇ · v=∂
∂x (3x2) + ∂
∂y (4xy) + ∂
∂z (−2z)
Step 2: Calculate the partial derivatives.
∂
∂x (3x2)=6x, ∂
∂y (4xy) = 4x, ∂
∂z (−2z) = −2
Step 3: Substitute the partial derivatives back into the expression for ∇·v.
∇ · v= 6x+ 4x−2 = 10x−2
Therefore, the divergence of the vector field vis 10x−2 .
20
Question 33
Question
Let u= 3i−2j+ 4kand v= 2i+ 5j−1k. Compute the derivative of u·vwith
respect to t, where uand vare functions of t.
Solution
To find the derivative of u·vwith respect to t, we first need to find expressions
for uand vin terms of t. Given that uand vare vectors, we can express them
as:
u(t)=3i−2j+ 4k
v(t)=2i+ 5j−1k
Now, to find the derivative of u·vwith respect to t, we will use the property
that d
dt (u·v) = u·
dt +
dt ·v.
Step 1: Find dudt and
dt
dt = ddt(3i−2j+4k)=0i+0j+0k=0
dt = ddt(2i+5j−1k)=0i+0j+0k=0
Step 2: Compute u·
dtand
dt ·v
u·
dt = (3i- 2j+ 4k)·0= 0
dt ·v=0·(2i+ 5j−1k)=0
Step 3: Find the derivative of u ·vHaving computed the components,
we see that both terms are zero. Hence, the derivative of u·vwith respect to
tis 0.
Question 34
Question
Let vand wbe two differentiable vectors given by
v=
2t3
t2
5
and w=
4t
3t
2t
.
Find the derivative of v·wwith respect to t.
21
Solution
To find the derivative of v·w, we use the product rule for differentiation of
vectors. The dot product of two vectors aand bis given by a·b=a1b1+
a2b2+a3b3.
Step 1: Find v·w:
v·w= (2t3)(4t)+(t2)(3t) + (5)(2t) = 8t4+ 3t3+ 10t.
Step 2: Differentiate v·wwith respect to tusing the product rule:
d
dt (v·w) = d
dt (8t4+ 3t3+ 10t)
= 32t3+ 9t2+ 10.
Thus, the derivative of v·wwith respect to tis 32t3+ 9t2+ 10 .
Question 35
Question
Let v=1
x
ln(x+y)and u=exy
sin(xy). Find d
dx (v·u).
Solution
To find d
dx (v·u), we first need to find the dot product of vand u:
v·u=1
x
ln(x+y)·exy
sin(xy)=1
xexy + ln(x+y) sin(xy)
Now, we will differentiate v·uwith respect to xusing the product rule:
d
dx (v·u) = d
dx 1
xexy + ln(x+y) sin(xy)
Step 1: Differentiate the first term:
d
dx 1
xexy=−1
x2exy +1
xyexy
Step 2: Differentiate the second term:
d
dx (ln(x+y) sin(xy)) = 1
x+y·d
dx (x+y) sin(xy) + ln(x+y) cos(xy)·d
dx (xy)
Step 3: Simplify the derivative:
d
dx (v·u) = −1
x2exy +1
xyexy +1
x+yexy sin(xy) + ln(x+y) cos(xy)
Therefore, d
dx (v·u) = −1
x2exy +1
xyexy +1
x+yexy sin(xy) + ln(x+y) cos(xy).
22
