MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Vector
Differentiation
Question Bank - Set 5
Liberty University
Question 1
Question
Let u= (3x2−y, 2xy + 1) and v= (x2−2y, 3xy). Find the gradient of the dot
product u·v.
Solution
To find the gradient of the dot product u·v, we first need to find the dot product
of uand v.
Step 1: Find the dot product of u and v The dot product of two vectors
u= (u1, u2) and v= (v1, v2) is given by u·v=u1v1+u2v2. In this case, we
have:
u·v= (3x2−y)(x2−2y) + (2xy + 1)(3xy)
Expanding and simplifying, we get:
u·v= 3x4−6x2y−xy + 6x2y+ 3xy2
u·v= 3x4+ 3xy2−xy
Step 2: Find the gradient of u ·vThe gradient of a scalar function
fis given by ∇f= (∂f
∂x ,∂f
∂y ). In this case, the scalar function is u·v=
3x4+ 3xy2−xy. Therefore, the gradient of u·vis:
∇(u·v) = ∂
∂x (3x4+ 3xy2−xy),∂
∂y (3x4+ 3xy2−xy)
Calculating the partial derivatives, we get:
∇(u·v) = (12x3+ 3y2−y, 6xy)
Question 2
Question
Let u=
x2
2y
−z
and v=
3x
y
z2
. Find d
dx (u·v).
Solution
Step 1: Find the dot product of uand v.
u·v= (x2)(3x) + (2y)(y)+(−z)(z2)
= 3x3+ 2y2−z3
Step 2: Differentiate the dot product with respect to x.
d
dx (u·v) = d
dx (3x3+ 2y2−z3)
= 9x2+ 0 −0
= 9x2
Therefore, d
dx (u·v)=9x2.
Question 3
Question
Let v=x2
exbe a vector function. Find the gradient of vwith respect to x.
Solution
To find the gradient of vwith respect to x, we need to differentiate each com-
ponent of vwith respect to x.
Step 1: Differentiate the first component x2with respect to x.
d
dx (x2)=2x
Step 2: Differentiate the second component exwith respect to x.
d
dx (ex) = ex
Step 3: The gradient of vwith respect to xis the vector of these derivatives:
∇v=d
dx (x2)
d
dx (ex)=2x
ex
Therefore, the gradient of vwith respect to xis 2x
ex.
2
Question 4
Question
Let v= 3x2i+ 4y3j. Compute dv
dx and dv
dy .
Solution
Step 1: To find dv
dx , we differentiate each component of vwith respect to x.
dv
dx =d
dx (3x2i) + d
dx (4y3j)
= (6x)i+ 0
= 6xi
Step 2: Next, to find dv
dy , we differentiate each component of vwith respect
to y.
dv
dy =d
dy (3x2i) + d
dy (4y3j)
= 0 + (12y2)j
= 12y2j
Therefore, dv
dx = 6xiand dv
dy = 12y2j.
Question 5
Question
Let v=3x2+y
x+ 2y3. Find ∂v
∂x .
Solution
Step 1: To find ∂v
∂x , we differentiate each component of vwith respect to xwhile
treating yas a constant.
Step 2: Differentiating the first component of v, we get
∂
∂x (3x2+y)=6x.
Step 3: Differentiating the second component of v, we get
∂
∂x (x+ 2y3)=1.
Step 4: Putting these results together, we have
∂v
∂x =6x
1.
3
Question 6
Question
Let v=
3x2
2 sin(x)
ex
. Calculate the derivative of vwith respect to x.
Solution
To find the derivative of vwith respect to x, we need to differentiate each
component of vwith respect to x.
Step 1: Differentiate the first component 3x2:
d
dx (3x2)=6x
Therefore, the first component of the derivative of vis 6x.
Step 2: Differentiate the second component 2 sin(x):
d
dx (2 sin(x)) = 2 cos(x)
The second component of the derivative of vis 2 cos(x).
Step 3: Differentiate the third component ex:
d
dx (ex) = ex
The third component of the derivative of vis ex.
Therefore, the derivative of vwith respect to xis:
dv
dx =
6x
2 cos(x)
ex
Question 7
Question
Let u(t) = 2t
3t2and v(t) = et
cos(t). Find d
dt (u(t)·v(t)).
Solution
Step 1: First, we calculate the dot product u(t)·v(t).
u(t)·v(t) = 2t
3t2·et
cos(t)= 2tet+ 3t2cos(t)
4
Step 2: Next, we differentiate u(t)·v(t) with respect to t.
d
dt (u(t)·v(t)) = d
dt (2tet+ 3t2cos(t))
Step 3: Using the product rule of differentiation, we have:
d
dt (2tet+ 3t2cos(t)) = 2et+ 2tet+ 6tcos(t)−3t2sin(t)
Therefore, d
dt (u(t)·v(t)) = 2et+ 2tet+ 6tcos(t)−3t2sin(t).
Question 8
Question
Let u=
x2
2xy
y2
and v=
3x
3y
−3z
. Find d
dt (u·v) where uand vare functions
of t.
Solution
Step 1: Find u·v.
u·v= (x2)(3x) + (2xy)(3y)+(y2)(−3z)=3x3+ 6xy2−3y2z
Step 2: Differentiate the dot product with respect to t.
d
dt (u·v) = d
dt (3x3+ 6xy2−3y2z)
Step 3: Differentiate each term of the dot product using the chain rule.
d
dt (3x3)=9x2dx
dt ,d
dt (6xy2)=6xdy
dt + 12y2dx
dt ,d
dt (−3y2z) = −3y2dz
dt
Step 4: Put it all together.
d
dt (u·v)=9x2dx
dt + 6xdy
dt + 12y2dx
dt −3y2dz
dt
Therefore, d
dt (u·v)=9x2dx
dt + 6xdy
dt + 12y2dx
dt −3y2dz
dt .
Question 9
Question
Let u= 2i−3j+ 4kand v= 5i+ 7j−2k. Find the derivative of the vector
function F(t) = e2tu+ 3 ln(t)vwith respect to t.
5
Solution
Step 1: Calculate the derivative of F(t) using the properties of vector differen-
tiation. dF(t)
dt =d
dt (e2tu+ 3 ln(t)v)
=d
dt (e2tu) + d
dt (3 ln(t)v)
=d
dt (e2t)u+e2tdu
dt +d
dt (3 ln(t))v+ 3 ln(t)dv
dt
= 2e2tu+ 0 ·du
dt +3
tv+ 3 ln(t)dv
dt
= 2e2tu+3
tv+ 3 ln(t)dv
dt
Step 2: Find dv
dt .
dv
dt =d
dt (5i+ 7j−2k)
= 5 di
dt + 7 dj
dt −2dk
dt
= 0i+ 0j+ 0k
=0
Therefore, the derivative of the vector function F(t) = e2tu+ 3 ln(t)vwith
respect to tis
dF(t)
dt = 2e2tu+3
tv
Question 10
Question
Let u= 2i−3j+ 5kand v= 4i−2j+ 3k.
Find d
dt (u·v) where u·vdenotes the dot product of vectors uand v.
Solution
To find d
dt (u·v), we need to differentiate the dot product of the vectors uand
vwith respect to t.
Recall that the dot product u·vof two vectors u=u1i+u2j+u3kand
v=v1i+v2j+v3kis given by
u·v=u1v1+u2v2+u3v3
Therefore, for vectors u= 2i−3j+ 5kand v= 4i−2j+ 3k, we have:
u·v= (2)(4) + (−3)(−2) + (5)(3) = 8 + 6 + 15 = 29
6
To differentiate u·vwith respect to t, we take the derivative of 29 with
respect to twhich is 0.
Therefore, d
dt (u·v) = 0.
Question 11
Question
Let v=
3x2+ 2y
xy2−4z
5xyz
. Find ∇ · v.
Solution
Step 1: The divergence of a vector field v=
f(x, y, z)
g(x, y, z)
h(x, y, z)
is defined as ∇ · v=
∂f
∂x +∂g
∂y +∂h
∂z .
Step 2: Given v=
3x2+ 2y
xy2−4z
5xyz
, we have f(x, y, z)=3x2+ 2y,g(x, y, z) =
xy2−4z, and h(x, y, z)=5xyz.
Step 3: Compute the partial derivatives:
∂f
∂x =∂
∂x (3x2+ 2y)=6x,
∂g
∂y =∂
∂y (xy2−4z) = x(2y)=2xy,
∂h
∂z =∂
∂z (5xyz)=5xy.
Step 4: Sum the partial derivatives to find the divergence:
∇ · v=∂f
∂x +∂g
∂y +∂h
∂z = 6x+ 2xy + 5xy = 6x+ 7xy.
Therefore, ∇ · v= 6x+ 7xy.
Question 12
Question
Let u=3
2and v=−2
5. Given that f(x) = u·x+v·x, where x=x
y,
find ∇f(x, y).
7
Solution
Step 1: Compute ∇f(x, y) by finding the partial derivatives of f(x) with respect
to xand y.
Step 2: We have f(x) = u·x+v·x. Computing these dot products, we get
f(x) = (3x+ 2y)+(−2x+ 5y)
f(x) = x+ 7y
Step 3: Now, let’s find the partial derivative of fwith respect to x:
∂f
∂x =∂
∂x (x+ 7y) = 1
Step 4: Next, find the partial derivative of fwith respect to y:
∂f
∂y =∂
∂y (x+ 7y) = 7
Step 5: Therefore, the gradient of f(x, y) is
∇f(x, y) = "∂f
∂x
∂f
∂y #=1
7
Question 13
Question
Let v=
x2
ex
ln(x)
. Find dv
dx .
Solution
Step 1: Write vas a column vector.
v=
x2
ex
ln(x)
Step 2: Differentiate each component of vwith respect to x.
dv
dx =
d
dx (x2)
d
dx (ex)
d
dx (ln(x))
Step 3: Find the derivatives.
dv
dx =
2x
ex
1
x
8
Therefore, dv
dx =
2x
ex
1
x
.
Question 14
Question
Let u=x2
exand v=sin(y)
y3. Find d
dy (u·v), where u·vdenotes the dot
product of vectors uand v.
Solution
Step 1: The dot product of two vectors u=u1
u2and v=v1
v2is given by
u·v=u1v1+u2v2.
Step 2: Calculate d
dy (u·v) using the chain rule: d
dy (u·v) = d
dy (x2sin(y) +
exy3).
Step 3: Apply the product rule for differentiation: d
dy (x2sin(y) + exy3) =
x2d
dy (sin(y)) + sin(y)d
dy (x2) + exd
dy (y3).
Step 4: Calculate the derivatives: d
dy (sin(y)) = cos(y), d
dy (x2) = 0, and
d
dy (y3)=3y2.
Step 5: Substituting the derivatives back, we get: d
dy (x2sin(y) + exy3) =
x2cos(y)+0+3exy2.
Therefore, d
dy (u·v) = x2cos(y)+3exy2.
Question 15
Question
Let v=
3x2−2y
4y3
5z
. Find ∂v
∂y .
Solution
Step 1: To find ∂v
∂y , we differentiate each component of vwith respect to y.
Step 2: Differentiating the first component 3x2−2ywith respect to ygives
−2.
Step 3: Differentiating the second component 4y3with respect to ygives
12y2.
Step 4: Differentiating the third component 5zwith respect to ygives 0.
9
Step 5: Therefore, ∂v
∂y =
−2
12y2
0
.
Question 16
Question
Let u=
x2
ex
sin(x)
and v=
x
ln(x)
cos(x)
. Find d(u
·v)dx, where ·denotes the dot
product of two vectors.
Solution
Step 1: Calculate u·v
u·v=x2·x+ex·ln(x) + sin(x)·cos(x) = x3+xex+ sin(x) cos(x)
Step 2: Differentiate u·vwith respect to x
d(u
·v)dx =d(x3)
dx +d(xex)
dx +d(sin(x) cos(x))
dx
Step 3: Calculate the derivatives
d(x3)
dx = 3x2
d(xex)
dx =ex+xex
d(sin(x) cos(x))
dx = cos2(x)−sin2(x)
Step 4: Combine the derivatives
d(u
·v)dx = 3x2+ex+xex+ cos2(x)−sin2(x)
Therefore, d(u
·v)dx = 3x2+ex+xex+ cos2(x)−sin2(x).
Question 17
Question
Let uand vbe two vectors where u=⟨x2,sin(x), ex⟩and v=⟨ln(x), x cos(x),3x⟩.
Find d(u·v)
dx .
10
Solution
Step 1: Calculate the dot product of uand v.
u·v=x2ln(x) + sin(x)·xcos(x) + ex·3x
Step 2: Differentiate u·vwith respect to xusing the product rule.
d(u·v)
dx =d(x2ln(x))
dx +d(xsin(x) cos(x))
dx +d(3xex)
dx
Step 3: Apply the product rule for each term.
d(u·v)
dx = 2xln(x) + x·1
x+xsin(x)(−sin(x)) + cos(x)x+ 3ex+ 3xd(ex)
dx
Step 4: Simplify the derivatives.
d(u·v)
dx = 2xln(x)+1−xsin2(x) + xcos(x)+3ex+ 3ex
Step 5: Combine like terms.
d(u·v)
dx = 2xln(x)+1−xsin2(x) + xcos(x)+6ex
Therefore, d(u·v)
dx = 2xln(x)+1−xsin2(x) + xcos(x)+6ex.
Question 18
Question
Let v= (2t2+ 1)i+ (t3−2)jbe a vector function, where tis a scalar variable.
Find dv
dt .
Solution
Step 1: To differentiate vwith respect to t, we simply need to differentiate each
component of the vector separately.
Step 2: Differentiating the x-component:
d
dt ((2t2+ 1)i) = d
dt (2t2+ 1)i= (4t)i
Step 3: Differentiating the y-component:
d
dt ((t3−2)j) = d
dt (t3−2)j= (3t2)j
Step 4: Putting the differentiated components together, we have:
dv
dt = (4t)i+ (3t2)j
11
Question 19
Question
Let u=3
−2and v=4
7. Find d
dt (u·v).
Solution
Step 1: We start by finding the dot product of vectors uand v:
u·v=3
−2·4
7= 3 ·4+(−2) ·7 = 12 −14 = −2.
Step 2: Next, we differentiate the dot product with respect to t:
d
dt (u·v) = d
dt (−2) = 0 .
Therefore, d
dt (u·v) = 0.
Question 20
Question
Let v=
x2
ex
cos(x)
. Find
dx.
Solution
To find the derivative of the vector vwith respect to x, we simply take the
derivative of each component of v. So, v=
x2
ex
cos(x)
, we have:
dx =
d
dx (x2)
d
dx (ex)
d
dx (cos(x))
Step 1: Compute d
dx (x2):
d
dx (x2)=2x
Step 2: Compute d
dx (ex):
d
dx (ex) = ex
12
Step 3: Compute d
dx (cos(x)):
d
dx (cos(x)) = −sin(x)
Therefore, the derivative of vwith respect to xis:
dx =
2x
ex
−sin(x)
Question 21
Question
Let v= (2x3y, 3xy2) be a vector-valued function. Find the derivative of vwith
respect to x, denoted by dv
dx .
Solution
To find the derivative of vwith respect to x, we will differentiate each component
of vwith respect to xseparately.
Step 1: Differentiate the first component of vwith respect to x. Let
f(x, y)=2x3y. To find ∂f
∂x , we differentiate fwith respect to xtreating y
as a constant. ∂f
∂x =∂
∂x (2x3y) = 6x2y
Step 2: Differentiate the second component of vwith respect to x. Let
g(x, y) = 3xy2. To find ∂g
∂x , we differentiate gwith respect to xtreating yas a
constant. ∂g
∂x =∂
∂x (3xy2)=3y2
Step 3: Combine the derivatives of the components to find dv
dx . Therefore,
the derivative of vwith respect to x, denoted by dv
dx , is:
dv
dx = (∂f
∂x ,∂g
∂x ) = (6x2y, 3y2)
Question 22
Question
Let u= (3x2−2y)i+ (4y3−6x)jbe a vector field. Find ∇ · u.
13
Solution
Step 1: The divergence of a vector field u=P(x, y)i+Q(x, y)jis defined as
∇ · u=∂P
∂x +∂Q
∂y .
Step 2: In this case, P(x, y)=3x2−2yand Q(x, y)=4y3−6x.
Step 3: We need to find ∂P
∂x and ∂Q
∂y .
Step 4:
∂P
∂x =∂
∂x (3x2−2y)=6x,
∂Q
∂y =∂
∂y (4y3−6x) = 12y2.
Step 5: Now, we can find the divergence:
∇ · u= 6x+ 12y2= 6x+ 12y2.
Question 23
Question
Let v=
x2
xey
y3
. Find the gradient of v.
Solution
To find the gradient of v, we need to differentiate each component of vwith
respect to their corresponding variables.
Step 1: Differentiate the first component:
∂
∂x (x2)=2x
Step 2: Differentiate the second component:
∂
∂x (xey) = eyand ∂
∂y (xey) = xey
Therefore, the gradient of the second component is ey
xey.
Step 3: Differentiate the third component:
∂
∂y (y3)=3y2
Putting it all together, the gradient of vis:
∇v=
2x
ey
3y2
14
Question 24
Question
Let u= 5i−3j+ 2kand v= 2i+ 4j−k. Find d
dt (u·v) where uand vare
functions of t.
Solution
Step 1: We first need to find the dot product of uand v:
u·v= (5i−3j+ 2k)·(2i+ 4j−k)
= 5 ·2+(−3) ·4+2·(−1) = 10 −12 −2 = −4
Step 2: Next, we differentiate the dot product with respect to tusing the
product rule:
d
dt (u·v) = d
dt (−4)
Since −4 is a constant, its derivative with respect to tis 0.
Therefore, d
dt (u·v) = 0 .
Question 25
Question
Let a=
3
−2
1
and b=
1
4
−2
. Find ∇(a·b).
Solution
Step 1: Recall the dot product formula:
a·b=a1b1+a2b2+a3b3
where a=
a1
a2
a3
and b=
b1
b2
b3
.
Step 2: Calculate a·b:
a·b= (3)(1) + (−2)(4) + (1)(−2) = 3 −8−2 = −7
Step 3: Next, find the gradient of the dot product:
∇(a·b) =
∂
∂x
∂
∂y
∂
∂z
(−7)
15
Step 4: Since the dot product is a scalar, we only need to differentiate the
scalar component -7 with respect to each variable:
∇(a·b) =
∂
∂x (−7)
∂
∂y (−7)
∂
∂z (−7)
Step 5: Therefore, the gradient of a·bis:
∇(a·b) =
0
0
0
Question 26
Question
Let f(x) = x·(x×y), where x=
x1
x2
x3
and y=
y1
y2
y3
are vectors. Find ∂f
∂x.
Solution
f(x) = x·(x×y)
=
x1
x2
x3
·
x1
x2
x3
×
y1
y2
y3
Step 1: Perform the cross product x×y.
x×y=
i j k
x1x2x3
y1y2y3
= (x2y3−x3y2)i−(x1y3−x3y1)j+ (x1y2−x2y1)k
Step 2: Calculate the dot product x·(x×y).
x·(x×y) =
x1
x2
x3
·((x2y3−x3y2)i−(x1y3−x3y1)j+ (x1y2−x2y1)k)
=x1(x2y3−x3y2) + x2(x1y3−x3y1) + x3(x1y2−x2y1)
Step 3: Find ∂f
∂x.
∂f
∂x=∂f1
∂x1
∂f1
∂x2
∂f1
∂x3
=
x2y3−x3y2
x1y3−x3y1
x1y2−x2y1
16
Therefore, ∂f
∂x=
x2y3−x3y2
x1y3−x3y1
x1y2−x2y1
.
Question 27
Question
Let u= 3i−4j+ 2kand v= 5i+ 2j−k. Find the derivative of the vector
function f(t) = e3tu+ ln(t)vwith respect to t.
Solution
Step 1: Find the derivative of f(t) with respect to tby differentiating each term
separately.
df
dt =d
dt e3tu+ ln(t)v
Step 2: Take the derivative of e3tuwith respect to t.
d
dt e3tu=d
dt e3tu+e3tdu
dt
Step 3: We have d
dt e3t= 3e3tand du
dt =0. Therefore,
d
dt e3tu= 3e3tu
Step 4: Take the derivative of ln(t)vwith respect to t.
d
dt (ln(t)v) = d
dt (ln(t)) v+ ln(t)dv
dt
Step 5: We have d
dt (ln(t)) = 1
tand dv
dt =0. Thus,
d
dt (ln(t)v) = 1
tv
Step 6: Putting the results together, we get the derivative of f(t) with respect
to t.df
dt = 3e3tu+1
tv
Therefore, the derivative of the vector function f(t) = e3tu+ ln(t)vwith
respect to tis df
dt = 3e3tu+1
tv.
Question 28
Question
Let v= 4i−2j+kand w= 2i+ 3j−5k. Find d(v
·w)dt.
17
Solution
To differentiate the dot product v·wwith respect to t, we first need to find the
dot product of vand w.
Step 1: Find v·w:
v·w= (4i−2j+k)·(2i+ 3j−5k)
v·w= (4 ·2) + (−2·3) + (1 ·(−5))
v·w= 8 −6−5
v·w=−3
Step 2: Differentiate v·wwith respect to t:
d(v
·w)dt =d(−3)
dt
d(v
·w)dt = 0
Therefore, d(v
·w)dt = 0.
Question 29
Question
Find the gradient of the following vector-valued function: f(r) = r
|r|, where
r=xi+yj+zk.
Solution
Step 1: Calculate the magnitude of r:
|r|=px2+y2+z2
Step 2: Rewrite f(r) using the magnitude of r:
f(r) = xi+yj+zk
px2+y2+z2
Step 3: To find the gradient of f(r), we need to differentiate each component
with respect to x,y, and z.
Step 4: Differentiate the x-component:
∂
∂x x
px2+y2+z2!=1
px2+y2+z2−x2
(x2+y2+z2)3/2
18
Step 5: Differentiate the y-component:
∂
∂y y
px2+y2+z2!=1
px2+y2+z2−y2
(x2+y2+z2)3/2
Step 6: Differentiate the z-component:
∂
∂z z
px2+y2+z2!=1
px2+y2+z2−z2
(x2+y2+z2)3/2
Step 7: The gradient of f(r) is:
∇f(r) = ∂
∂x x
|r|i+∂
∂y y
|r|j+∂
∂z z
|r|k
= 1
px2+y2+z2−x2
(x2+y2+z2)3/2!i+ 1
px2+y2+z2−y2
(x2+y2+z2)3/2!j+ 1
px2+y2+z2−z2
(x2+y2+z2)3/2!k
Question 30
Question
Let v=
3x2
4xy
5yz
be a vector valued function. Find the derivative of vwith
respect to x,dv
dx .
Solution
To find the derivative of vector valued function vwith respect to x, we differ-
entiate each component of vseparately with respect to x.
Step 1: For the first component 3x2, we differentiate with respect to x:
d
dx (3x2)=6x
Step 2: For the second component 4xy, we differentiate with respect to x:
d
dx (4xy)=4y
Step 3: For the third component 5yz, we differentiate with respect to x:
d
dx (5yz)=0
Step 4: Putting all the components together, the derivative of vwith respect
to x,dv
dx , is:
dv
dx =
6x
4y
0
19
Question 31
Question
Let a= 2i−3jand b= 5i+ 4j. Find the derivative of f=a·bwith respect
to t.
Solution
Step 1: First, we compute the dot product f=a·b.
f= (2i−3j)·(5i+ 4j)
f= 2 ·5+(−3) ·4 = 10 −12 = −2
Step 2: Next, we express aand bin terms of t.
a= 2i−3j= 2i(t)−3j(t)
b= 5i+ 4j= 5i(t)+4j(t)
Step 3: We differentiate fwith respect to tusing the product rule for vectors.
df
dt =d(2i(t))
dt ·(5i+4j)+(2i−3j)·d(5i(t))
dt +d(−3j(t))
dt ·(5i+4j)+(2i−3j)·d(4j(t))
dt
Step 4: Evaluating each term using the chain rule and product rule for
differentiation, we get:
df
dt = 2di
dt ·5i+ 2i·5di
dt −3dj
dt ·5i−3j·4dj
dt
Step 5: Simplifying further, we have:
df
dt = 2j·5i+ 2i·5j−3j·5i−3i·4j
df
dt = 10j+ 10i−15j−12i
Step 6: Combining like terms, we get the derivative of fwith respect to t:
df
dt =−2i−5j
Question 32
Question
Let u= 2i−3j+kand v=i+ 4j−2k. Find the derivative of u·vwith respect
to t, where i,j,kare the standard basis vectors and trepresents time.
20
Solution
Step 1: The dot product of two vectors u=u1i+u2j+u3kand v=v1i+v2j+v3k
is given by:
u·v=u1v1+u2v2+u3v3
Step 2: Given that u= 2i−3j+kand v=i+ 4j−2k, we have:
u·v= (2)(1) + (−3)(4) + (1)(−2)
Step 3: Calculating the dot product, we get:
u·v= 2 −12 −2 = −12
Step 4: To find the derivative of u·vwith respect to t, we differentiate the
dot product with respect to t:
d
dt (u·v) = d
dt (−12) = 0
Step 5: Therefore, the derivative of u·vwith respect to tis 0 .
Question 33
Question
Let aand bbe constant vectors. If r(t) = t2a+ sin(t)b, find dr
dt .
Solution
To find dr
dt , we differentiate each component of r(t) with respect to t.
Step 1: Differentiate the first component t2a:
d
dt (t2a)=2ta
Step 2: Differentiate the second component sin(t)b:
d
dt (sin(t)b) = cos(t)b
Therefore, the derivative of r(t) with respect to tis:
dr
dt = 2ta+ cos(t)b
Question 34
Question
Let u=ex
ln(y)and v=sin(z)
cos(t). Determine d
dt (u·v).
21
Solution
Step 1: Find the dot product u·v.
u·v= (ex)(sin(z)) + (ln(y))(cos(t))
Step 2: Differentiate with respect to t.
d
dt (u·v) = d
dt [(ex)(sin(z)) + (ln(y))(cos(t))]
Step 3: Apply the product rule for differentiation.
d
dt (u·v) = d(ex)
dt (sin(z)) + (ex)d(sin(z))
dt +d(ln(y))
dt (cos(t)) + (ln(y))d(cos(t))
dt
Step 4: Simplify by evaluating each derivative.
d
dt (u·v) = (ex)(0)(sin(z)) + (ex)(cos(z))dz
dt + 0(cos(t)) + 1
y
dy
dt (cos(t))
Step 5: Express the final answer.
d
dt (u·v)=(ex)(cos(z))dz
dt +cos(t)
y
dy
dt
Question 35
Question
Let v= (x2+y2)i+ 2xyj. Find dv
dx .
Solution
Step 1: To differentiate a vector with respect to x, we differentiate each com-
ponent of the vector separately. Let v= (f(x, y))i+ (g(x, y))j, where f(x, y) =
x2+y2and g(x, y)=2xy.
Step 2: Differentiating the first component of the vector:
∂f
∂x =∂
∂x (x2+y2)=2x
Step 3: Differentiating the second component of the vector:
∂g
∂x =∂
∂x (2xy)=2y
Step 4: Putting it all together, we get:
dv
dx =d
dx (x2+y2)i+ 2xyj=d
dx (x2+y2)i+d
dx (2xy)j= 2xi+ 2yj
22
Question 2
Question
Let u=
x2
2y
−z
and v=
3x
y
z2
. Find d
dx (u·v).
Solution
Step 1: Find the dot product of uand v.
u·v= (x2)(3x) + (2y)(y)+(−z)(z2)
= 3x3+ 2y2−z3
Step 2: Differentiate the dot product with respect to x.
d
dx (u·v) = d
dx (3x3+ 2y2−z3)
= 9x2+ 0 −0
= 9x2
Therefore, d
dx (u·v)=9x2.
Question 3
Question
Let v=x2
exbe a vector function. Find the gradient of vwith respect to x.
Solution
To find the gradient of vwith respect to x, we need to differentiate each com-
ponent of vwith respect to x.
Step 1: Differentiate the first component x2with respect to x.
d
dx (x2)=2x
Step 2: Differentiate the second component exwith respect to x.
d
dx (ex) = ex
Step 3: The gradient of vwith respect to xis the vector of these derivatives:
∇v=d
dx (x2)
d
dx (ex)=2x
ex
Therefore, the gradient of vwith respect to xis 2x
ex.
2
Question 4
Question
Let v= 3x2i+ 4y3j. Compute dv
dx and dv
dy .
Solution
Step 1: To find dv
dx , we differentiate each component of vwith respect to x.
dv
dx =d
dx (3x2i) + d
dx (4y3j)
= (6x)i+ 0
= 6xi
Step 2: Next, to find dv
dy , we differentiate each component of vwith respect
to y.
dv
dy =d
dy (3x2i) + d
dy (4y3j)
= 0 + (12y2)j
= 12y2j
Therefore, dv
dx = 6xiand dv
dy = 12y2j.
Question 5
Question
Let v=3x2+y
x+ 2y3. Find ∂v
∂x .
Solution
Step 1: To find ∂v
∂x , we differentiate each component of vwith respect to xwhile
treating yas a constant.
Step 2: Differentiating the first component of v, we get
∂
∂x (3x2+y)=6x.
Step 3: Differentiating the second component of v, we get
∂
∂x (x+ 2y3)=1.
Step 4: Putting these results together, we have
∂v
∂x =6x
1.
3
Question 6
Question
Let v=
3x2
2 sin(x)
ex
. Calculate the derivative of vwith respect to x.
Solution
To find the derivative of vwith respect to x, we need to differentiate each
component of vwith respect to x.
Step 1: Differentiate the first component 3x2:
d
dx (3x2)=6x
Therefore, the first component of the derivative of vis 6x.
Step 2: Differentiate the second component 2 sin(x):
d
dx (2 sin(x)) = 2 cos(x)
The second component of the derivative of vis 2 cos(x).
Step 3: Differentiate the third component ex:
d
dx (ex) = ex
The third component of the derivative of vis ex.
Therefore, the derivative of vwith respect to xis:
dv
dx =
6x
2 cos(x)
ex
Question 7
Question
Let u(t) = 2t
3t2and v(t) = et
cos(t). Find d
dt (u(t)·v(t)).
Solution
Step 1: First, we calculate the dot product u(t)·v(t).
u(t)·v(t) = 2t
3t2·et
cos(t)= 2tet+ 3t2cos(t)
4
Step 2: Next, we differentiate u(t)·v(t) with respect to t.
d
dt (u(t)·v(t)) = d
dt (2tet+ 3t2cos(t))
Step 3: Using the product rule of differentiation, we have:
d
dt (2tet+ 3t2cos(t)) = 2et+ 2tet+ 6tcos(t)−3t2sin(t)
Therefore, d
dt (u(t)·v(t)) = 2et+ 2tet+ 6tcos(t)−3t2sin(t).
Question 8
Question
Let u=
x2
2xy
y2
and v=
3x
3y
−3z
. Find d
dt (u·v) where uand vare functions
of t.
Solution
Step 1: Find u·v.
u·v= (x2)(3x) + (2xy)(3y)+(y2)(−3z)=3x3+ 6xy2−3y2z
Step 2: Differentiate the dot product with respect to t.
d
dt (u·v) = d
dt (3x3+ 6xy2−3y2z)
Step 3: Differentiate each term of the dot product using the chain rule.
d
dt (3x3)=9x2dx
dt ,d
dt (6xy2)=6xdy
dt + 12y2dx
dt ,d
dt (−3y2z) = −3y2dz
dt
Step 4: Put it all together.
d
dt (u·v)=9x2dx
dt + 6xdy
dt + 12y2dx
dt −3y2dz
dt
Therefore, d
dt (u·v)=9x2dx
dt + 6xdy
dt + 12y2dx
dt −3y2dz
dt .
Question 9
Question
Let u= 2i−3j+ 4kand v= 5i+ 7j−2k. Find the derivative of the vector
function F(t) = e2tu+ 3 ln(t)vwith respect to t.
5
Solution
Step 1: Calculate the derivative of F(t) using the properties of vector differen-
tiation. dF(t)
dt =d
dt (e2tu+ 3 ln(t)v)
=d
dt (e2tu) + d
dt (3 ln(t)v)
=d
dt (e2t)u+e2tdu
dt +d
dt (3 ln(t))v+ 3 ln(t)dv
dt
= 2e2tu+ 0 ·du
dt +3
tv+ 3 ln(t)dv
dt
= 2e2tu+3
tv+ 3 ln(t)dv
dt
Step 2: Find dv
dt .
dv
dt =d
dt (5i+ 7j−2k)
= 5 di
dt + 7 dj
dt −2dk
dt
= 0i+ 0j+ 0k
=0
Therefore, the derivative of the vector function F(t) = e2tu+ 3 ln(t)vwith
respect to tis
dF(t)
dt = 2e2tu+3
tv
Question 10
Question
Let u= 2i−3j+ 5kand v= 4i−2j+ 3k.
Find d
dt (u·v) where u·vdenotes the dot product of vectors uand v.
Solution
To find d
dt (u·v), we need to differentiate the dot product of the vectors uand
vwith respect to t.
Recall that the dot product u·vof two vectors u=u1i+u2j+u3kand
v=v1i+v2j+v3kis given by
u·v=u1v1+u2v2+u3v3
Therefore, for vectors u= 2i−3j+ 5kand v= 4i−2j+ 3k, we have:
u·v= (2)(4) + (−3)(−2) + (5)(3) = 8 + 6 + 15 = 29
6
To differentiate u·vwith respect to t, we take the derivative of 29 with
respect to twhich is 0.
Therefore, d
dt (u·v) = 0.
Question 11
Question
Let v=
3x2+ 2y
xy2−4z
5xyz
. Find ∇ · v.
Solution
Step 1: The divergence of a vector field v=
f(x, y, z)
g(x, y, z)
h(x, y, z)
is defined as ∇ · v=
∂f
∂x +∂g
∂y +∂h
∂z .
Step 2: Given v=
3x2+ 2y
xy2−4z
5xyz
, we have f(x, y, z)=3x2+ 2y,g(x, y, z) =
xy2−4z, and h(x, y, z)=5xyz.
Step 3: Compute the partial derivatives:
∂f
∂x =∂
∂x (3x2+ 2y)=6x,
∂g
∂y =∂
∂y (xy2−4z) = x(2y)=2xy,
∂h
∂z =∂
∂z (5xyz)=5xy.
Step 4: Sum the partial derivatives to find the divergence:
∇ · v=∂f
∂x +∂g
∂y +∂h
∂z = 6x+ 2xy + 5xy = 6x+ 7xy.
Therefore, ∇ · v= 6x+ 7xy.
Question 12
Question
Let u=3
2and v=−2
5. Given that f(x) = u·x+v·x, where x=x
y,
find ∇f(x, y).
7
Solution
Step 1: Compute ∇f(x, y) by finding the partial derivatives of f(x) with respect
to xand y.
Step 2: We have f(x) = u·x+v·x. Computing these dot products, we get
f(x) = (3x+ 2y)+(−2x+ 5y)
f(x) = x+ 7y
Step 3: Now, let’s find the partial derivative of fwith respect to x:
∂f
∂x =∂
∂x (x+ 7y) = 1
Step 4: Next, find the partial derivative of fwith respect to y:
∂f
∂y =∂
∂y (x+ 7y) = 7
Step 5: Therefore, the gradient of f(x, y) is
∇f(x, y) = "∂f
∂x
∂f
∂y #=1
7
Question 13
Question
Let v=
x2
ex
ln(x)
. Find dv
dx .
Solution
Step 1: Write vas a column vector.
v=
x2
ex
ln(x)
Step 2: Differentiate each component of vwith respect to x.
dv
dx =
d
dx (x2)
d
dx (ex)
d
dx (ln(x))
Step 3: Find the derivatives.
dv
dx =
2x
ex
1
x
8
Therefore, dv
dx =
2x
ex
1
x
.
Question 14
Question
Let u=x2
exand v=sin(y)
y3. Find d
dy (u·v), where u·vdenotes the dot
product of vectors uand v.
Solution
Step 1: The dot product of two vectors u=u1
u2and v=v1
v2is given by
u·v=u1v1+u2v2.
Step 2: Calculate d
dy (u·v) using the chain rule: d
dy (u·v) = d
dy (x2sin(y) +
exy3).
Step 3: Apply the product rule for differentiation: d
dy (x2sin(y) + exy3) =
x2d
dy (sin(y)) + sin(y)d
dy (x2) + exd
dy (y3).
Step 4: Calculate the derivatives: d
dy (sin(y)) = cos(y), d
dy (x2) = 0, and
d
dy (y3)=3y2.
Step 5: Substituting the derivatives back, we get: d
dy (x2sin(y) + exy3) =
x2cos(y)+0+3exy2.
Therefore, d
dy (u·v) = x2cos(y)+3exy2.
Question 15
Question
Let v=
3x2−2y
4y3
5z
. Find ∂v
∂y .
Solution
Step 1: To find ∂v
∂y , we differentiate each component of vwith respect to y.
Step 2: Differentiating the first component 3x2−2ywith respect to ygives
−2.
Step 3: Differentiating the second component 4y3with respect to ygives
12y2.
Step 4: Differentiating the third component 5zwith respect to ygives 0.
9
Step 5: Therefore, ∂v
∂y =
−2
12y2
0
.
Question 16
Question
Let u=
x2
ex
sin(x)
and v=
x
ln(x)
cos(x)
. Find d(u
·v)dx, where ·denotes the dot
product of two vectors.
Solution
Step 1: Calculate u·v
u·v=x2·x+ex·ln(x) + sin(x)·cos(x) = x3+xex+ sin(x) cos(x)
Step 2: Differentiate u·vwith respect to x
d(u
·v)dx =d(x3)
dx +d(xex)
dx +d(sin(x) cos(x))
dx
Step 3: Calculate the derivatives
d(x3)
dx = 3x2
d(xex)
dx =ex+xex
d(sin(x) cos(x))
dx = cos2(x)−sin2(x)
Step 4: Combine the derivatives
d(u
·v)dx = 3x2+ex+xex+ cos2(x)−sin2(x)
Therefore, d(u
·v)dx = 3x2+ex+xex+ cos2(x)−sin2(x).
Question 17
Question
Let uand vbe two vectors where u=⟨x2,sin(x), ex⟩and v=⟨ln(x), x cos(x),3x⟩.
Find d(u·v)
dx .
10
Solution
Step 1: Calculate the dot product of uand v.
u·v=x2ln(x) + sin(x)·xcos(x) + ex·3x
Step 2: Differentiate u·vwith respect to xusing the product rule.
d(u·v)
dx =d(x2ln(x))
dx +d(xsin(x) cos(x))
dx +d(3xex)
dx
Step 3: Apply the product rule for each term.
d(u·v)
dx = 2xln(x) + x·1
x+xsin(x)(−sin(x)) + cos(x)x+ 3ex+ 3xd(ex)
dx
Step 4: Simplify the derivatives.
d(u·v)
dx = 2xln(x)+1−xsin2(x) + xcos(x)+3ex+ 3ex
Step 5: Combine like terms.
d(u·v)
dx = 2xln(x)+1−xsin2(x) + xcos(x)+6ex
Therefore, d(u·v)
dx = 2xln(x)+1−xsin2(x) + xcos(x)+6ex.
Question 18
Question
Let v= (2t2+ 1)i+ (t3−2)jbe a vector function, where tis a scalar variable.
Find dv
dt .
Solution
Step 1: To differentiate vwith respect to t, we simply need to differentiate each
component of the vector separately.
Step 2: Differentiating the x-component:
d
dt ((2t2+ 1)i) = d
dt (2t2+ 1)i= (4t)i
Step 3: Differentiating the y-component:
d
dt ((t3−2)j) = d
dt (t3−2)j= (3t2)j
Step 4: Putting the differentiated components together, we have:
dv
dt = (4t)i+ (3t2)j
11
Question 19
Question
Let u=3
−2and v=4
7. Find d
dt (u·v).
Solution
Step 1: We start by finding the dot product of vectors uand v:
u·v=3
−2·4
7= 3 ·4+(−2) ·7 = 12 −14 = −2.
Step 2: Next, we differentiate the dot product with respect to t:
d
dt (u·v) = d
dt (−2) = 0 .
Therefore, d
dt (u·v) = 0.
Question 20
Question
Let v=
x2
ex
cos(x)
. Find
dx.
Solution
To find the derivative of the vector vwith respect to x, we simply take the
derivative of each component of v. So, v=
x2
ex
cos(x)
, we have:
dx =
d
dx (x2)
d
dx (ex)
d
dx (cos(x))
Step 1: Compute d
dx (x2):
d
dx (x2)=2x
Step 2: Compute d
dx (ex):
d
dx (ex) = ex
12
Step 3: Compute d
dx (cos(x)):
d
dx (cos(x)) = −sin(x)
Therefore, the derivative of vwith respect to xis:
dx =
2x
ex
−sin(x)
Question 21
Question
Let v= (2x3y, 3xy2) be a vector-valued function. Find the derivative of vwith
respect to x, denoted by dv
dx .
Solution
To find the derivative of vwith respect to x, we will differentiate each component
of vwith respect to xseparately.
Step 1: Differentiate the first component of vwith respect to x. Let
f(x, y)=2x3y. To find ∂f
∂x , we differentiate fwith respect to xtreating y
as a constant. ∂f
∂x =∂
∂x (2x3y) = 6x2y
Step 2: Differentiate the second component of vwith respect to x. Let
g(x, y) = 3xy2. To find ∂g
∂x , we differentiate gwith respect to xtreating yas a
constant. ∂g
∂x =∂
∂x (3xy2)=3y2
Step 3: Combine the derivatives of the components to find dv
dx . Therefore,
the derivative of vwith respect to x, denoted by dv
dx , is:
dv
dx = (∂f
∂x ,∂g
∂x ) = (6x2y, 3y2)
Question 22
Question
Let u= (3x2−2y)i+ (4y3−6x)jbe a vector field. Find ∇ · u.
13
Solution
Step 1: The divergence of a vector field u=P(x, y)i+Q(x, y)jis defined as
∇ · u=∂P
∂x +∂Q
∂y .
Step 2: In this case, P(x, y)=3x2−2yand Q(x, y)=4y3−6x.
Step 3: We need to find ∂P
∂x and ∂Q
∂y .
Step 4:
∂P
∂x =∂
∂x (3x2−2y)=6x,
∂Q
∂y =∂
∂y (4y3−6x) = 12y2.
Step 5: Now, we can find the divergence:
∇ · u= 6x+ 12y2= 6x+ 12y2.
Question 23
Question
Let v=
x2
xey
y3
. Find the gradient of v.
Solution
To find the gradient of v, we need to differentiate each component of vwith
respect to their corresponding variables.
Step 1: Differentiate the first component:
∂
∂x (x2)=2x
Step 2: Differentiate the second component:
∂
∂x (xey) = eyand ∂
∂y (xey) = xey
Therefore, the gradient of the second component is ey
xey.
Step 3: Differentiate the third component:
∂
∂y (y3)=3y2
Putting it all together, the gradient of vis:
∇v=
2x
ey
3y2
14
Question 24
Question
Let u= 5i−3j+ 2kand v= 2i+ 4j−k. Find d
dt (u·v) where uand vare
functions of t.
Solution
Step 1: We first need to find the dot product of uand v:
u·v= (5i−3j+ 2k)·(2i+ 4j−k)
= 5 ·2+(−3) ·4+2·(−1) = 10 −12 −2 = −4
Step 2: Next, we differentiate the dot product with respect to tusing the
product rule:
d
dt (u·v) = d
dt (−4)
Since −4 is a constant, its derivative with respect to tis 0.
Therefore, d
dt (u·v) = 0 .
Question 25
Question
Let a=
3
−2
1
and b=
1
4
−2
. Find ∇(a·b).
Solution
Step 1: Recall the dot product formula:
a·b=a1b1+a2b2+a3b3
where a=
a1
a2
a3
and b=
b1
b2
b3
.
Step 2: Calculate a·b:
a·b= (3)(1) + (−2)(4) + (1)(−2) = 3 −8−2 = −7
Step 3: Next, find the gradient of the dot product:
∇(a·b) =
∂
∂x
∂
∂y
∂
∂z
(−7)
15
Step 4: Since the dot product is a scalar, we only need to differentiate the
scalar component -7 with respect to each variable:
∇(a·b) =
∂
∂x (−7)
∂
∂y (−7)
∂
∂z (−7)
Step 5: Therefore, the gradient of a·bis:
∇(a·b) =
0
0
0
Question 26
Question
Let f(x) = x·(x×y), where x=
x1
x2
x3
and y=
y1
y2
y3
are vectors. Find ∂f
∂x.
Solution
f(x) = x·(x×y)
=
x1
x2
x3
·
x1
x2
x3
×
y1
y2
y3
Step 1: Perform the cross product x×y.
x×y=
i j k
x1x2x3
y1y2y3
= (x2y3−x3y2)i−(x1y3−x3y1)j+ (x1y2−x2y1)k
Step 2: Calculate the dot product x·(x×y).
x·(x×y) =
x1
x2
x3
·((x2y3−x3y2)i−(x1y3−x3y1)j+ (x1y2−x2y1)k)
=x1(x2y3−x3y2) + x2(x1y3−x3y1) + x3(x1y2−x2y1)
Step 3: Find ∂f
∂x.
∂f
∂x=∂f1
∂x1
∂f1
∂x2
∂f1
∂x3
=
x2y3−x3y2
x1y3−x3y1
x1y2−x2y1
16
Therefore, ∂f
∂x=
x2y3−x3y2
x1y3−x3y1
x1y2−x2y1
.
Question 27
Question
Let u= 3i−4j+ 2kand v= 5i+ 2j−k. Find the derivative of the vector
function f(t) = e3tu+ ln(t)vwith respect to t.
Solution
Step 1: Find the derivative of f(t) with respect to tby differentiating each term
separately.
df
dt =d
dt e3tu+ ln(t)v
Step 2: Take the derivative of e3tuwith respect to t.
d
dt e3tu=d
dt e3tu+e3tdu
dt
Step 3: We have d
dt e3t= 3e3tand du
dt =0. Therefore,
d
dt e3tu= 3e3tu
Step 4: Take the derivative of ln(t)vwith respect to t.
d
dt (ln(t)v) = d
dt (ln(t)) v+ ln(t)dv
dt
Step 5: We have d
dt (ln(t)) = 1
tand dv
dt =0. Thus,
d
dt (ln(t)v) = 1
tv
Step 6: Putting the results together, we get the derivative of f(t) with respect
to t.df
dt = 3e3tu+1
tv
Therefore, the derivative of the vector function f(t) = e3tu+ ln(t)vwith
respect to tis df
dt = 3e3tu+1
tv.
Question 28
Question
Let v= 4i−2j+kand w= 2i+ 3j−5k. Find d(v
·w)dt.
17
Solution
To differentiate the dot product v·wwith respect to t, we first need to find the
dot product of vand w.
Step 1: Find v·w:
v·w= (4i−2j+k)·(2i+ 3j−5k)
v·w= (4 ·2) + (−2·3) + (1 ·(−5))
v·w= 8 −6−5
v·w=−3
Step 2: Differentiate v·wwith respect to t:
d(v
·w)dt =d(−3)
dt
d(v
·w)dt = 0
Therefore, d(v
·w)dt = 0.
Question 29
Question
Find the gradient of the following vector-valued function: f(r) = r
|r|, where
r=xi+yj+zk.
Solution
Step 1: Calculate the magnitude of r:
|r|=px2+y2+z2
Step 2: Rewrite f(r) using the magnitude of r:
f(r) = xi+yj+zk
px2+y2+z2
Step 3: To find the gradient of f(r), we need to differentiate each component
with respect to x,y, and z.
Step 4: Differentiate the x-component:
∂
∂x x
px2+y2+z2!=1
px2+y2+z2−x2
(x2+y2+z2)3/2
18
Step 5: Differentiate the y-component:
∂
∂y y
px2+y2+z2!=1
px2+y2+z2−y2
(x2+y2+z2)3/2
Step 6: Differentiate the z-component:
∂
∂z z
px2+y2+z2!=1
px2+y2+z2−z2
(x2+y2+z2)3/2
Step 7: The gradient of f(r) is:
∇f(r) = ∂
∂x x
|r|i+∂
∂y y
|r|j+∂
∂z z
|r|k
= 1
px2+y2+z2−x2
(x2+y2+z2)3/2!i+ 1
px2+y2+z2−y2
(x2+y2+z2)3/2!j+ 1
px2+y2+z2−z2
(x2+y2+z2)3/2!k
Question 30
Question
Let v=
3x2
4xy
5yz
be a vector valued function. Find the derivative of vwith
respect to x,dv
dx .
Solution
To find the derivative of vector valued function vwith respect to x, we differ-
entiate each component of vseparately with respect to x.
Step 1: For the first component 3x2, we differentiate with respect to x:
d
dx (3x2)=6x
Step 2: For the second component 4xy, we differentiate with respect to x:
d
dx (4xy)=4y
Step 3: For the third component 5yz, we differentiate with respect to x:
d
dx (5yz)=0
Step 4: Putting all the components together, the derivative of vwith respect
to x,dv
dx , is:
dv
dx =
6x
4y
0
19
Question 31
Question
Let a= 2i−3jand b= 5i+ 4j. Find the derivative of f=a·bwith respect
to t.
Solution
Step 1: First, we compute the dot product f=a·b.
f= (2i−3j)·(5i+ 4j)
f= 2 ·5+(−3) ·4 = 10 −12 = −2
Step 2: Next, we express aand bin terms of t.
a= 2i−3j= 2i(t)−3j(t)
b= 5i+ 4j= 5i(t)+4j(t)
Step 3: We differentiate fwith respect to tusing the product rule for vectors.
df
dt =d(2i(t))
dt ·(5i+4j)+(2i−3j)·d(5i(t))
dt +d(−3j(t))
dt ·(5i+4j)+(2i−3j)·d(4j(t))
dt
Step 4: Evaluating each term using the chain rule and product rule for
differentiation, we get:
df
dt = 2di
dt ·5i+ 2i·5di
dt −3dj
dt ·5i−3j·4dj
dt
Step 5: Simplifying further, we have:
df
dt = 2j·5i+ 2i·5j−3j·5i−3i·4j
df
dt = 10j+ 10i−15j−12i
Step 6: Combining like terms, we get the derivative of fwith respect to t:
df
dt =−2i−5j
Question 32
Question
Let u= 2i−3j+kand v=i+ 4j−2k. Find the derivative of u·vwith respect
to t, where i,j,kare the standard basis vectors and trepresents time.
20
Solution
Step 1: The dot product of two vectors u=u1i+u2j+u3kand v=v1i+v2j+v3k
is given by:
u·v=u1v1+u2v2+u3v3
Step 2: Given that u= 2i−3j+kand v=i+ 4j−2k, we have:
u·v= (2)(1) + (−3)(4) + (1)(−2)
Step 3: Calculating the dot product, we get:
u·v= 2 −12 −2 = −12
Step 4: To find the derivative of u·vwith respect to t, we differentiate the
dot product with respect to t:
d
dt (u·v) = d
dt (−12) = 0
Step 5: Therefore, the derivative of u·vwith respect to tis 0 .
Question 33
Question
Let aand bbe constant vectors. If r(t) = t2a+ sin(t)b, find dr
dt .
Solution
To find dr
dt , we differentiate each component of r(t) with respect to t.
Step 1: Differentiate the first component t2a:
d
dt (t2a)=2ta
Step 2: Differentiate the second component sin(t)b:
d
dt (sin(t)b) = cos(t)b
Therefore, the derivative of r(t) with respect to tis:
dr
dt = 2ta+ cos(t)b
Question 34
Question
Let u=ex
ln(y)and v=sin(z)
cos(t). Determine d
dt (u·v).
21
Solution
Step 1: Find the dot product u·v.
u·v= (ex)(sin(z)) + (ln(y))(cos(t))
Step 2: Differentiate with respect to t.
d
dt (u·v) = d
dt [(ex)(sin(z)) + (ln(y))(cos(t))]
Step 3: Apply the product rule for differentiation.
d
dt (u·v) = d(ex)
dt (sin(z)) + (ex)d(sin(z))
dt +d(ln(y))
dt (cos(t)) + (ln(y))d(cos(t))
dt
Step 4: Simplify by evaluating each derivative.
d
dt (u·v) = (ex)(0)(sin(z)) + (ex)(cos(z))dz
dt + 0(cos(t)) + 1
y
dy
dt (cos(t))
Step 5: Express the final answer.
d
dt (u·v)=(ex)(cos(z))dz
dt +cos(t)
y
dy
dt
Question 35
Question
Let v= (x2+y2)i+ 2xyj. Find dv
dx .
Solution
Step 1: To differentiate a vector with respect to x, we differentiate each com-
ponent of the vector separately. Let v= (f(x, y))i+ (g(x, y))j, where f(x, y) =
x2+y2and g(x, y)=2xy.
Step 2: Differentiating the first component of the vector:
∂f
∂x =∂
∂x (x2+y2)=2x
Step 3: Differentiating the second component of the vector:
∂g
∂x =∂
∂x (2xy)=2y
Step 4: Putting it all together, we get:
dv
dx =d
dx (x2+y2)i+ 2xyj=d
dx (x2+y2)i+d
dx (2xy)j= 2xi+ 2yj
22
Question 2
Question
Let u=
x2
2y
−z
and v=
3x
y
z2
. Find d
dx (u·v).
Solution
Step 1: Find the dot product of uand v.
u·v= (x2)(3x) + (2y)(y)+(−z)(z2)
= 3x3+ 2y2−z3
Step 2: Differentiate the dot product with respect to x.
d
dx (u·v) = d
dx (3x3+ 2y2−z3)
= 9x2+ 0 −0
= 9x2
Therefore, d
dx (u·v)=9x2.
Question 3
Question
Let v=x2
exbe a vector function. Find the gradient of vwith respect to x.
Solution
To find the gradient of vwith respect to x, we need to differentiate each com-
ponent of vwith respect to x.
Step 1: Differentiate the first component x2with respect to x.
d
dx (x2)=2x
Step 2: Differentiate the second component exwith respect to x.
d
dx (ex) = ex
Step 3: The gradient of vwith respect to xis the vector of these derivatives:
∇v=d
dx (x2)
d
dx (ex)=2x
ex
Therefore, the gradient of vwith respect to xis 2x
ex.
2
Question 4
Question
Let v= 3x2i+ 4y3j. Compute dv
dx and dv
dy .
Solution
Step 1: To find dv
dx , we differentiate each component of vwith respect to x.
dv
dx =d
dx (3x2i) + d
dx (4y3j)
= (6x)i+ 0
= 6xi
Step 2: Next, to find dv
dy , we differentiate each component of vwith respect
to y.
dv
dy =d
dy (3x2i) + d
dy (4y3j)
= 0 + (12y2)j
= 12y2j
Therefore, dv
dx = 6xiand dv
dy = 12y2j.
Question 5
Question
Let v=3x2+y
x+ 2y3. Find ∂v
∂x .
Solution
Step 1: To find ∂v
∂x , we differentiate each component of vwith respect to xwhile
treating yas a constant.
Step 2: Differentiating the first component of v, we get
∂
∂x (3x2+y)=6x.
Step 3: Differentiating the second component of v, we get
∂
∂x (x+ 2y3)=1.
Step 4: Putting these results together, we have
∂v
∂x =6x
1.
3
Question 6
Question
Let v=
3x2
2 sin(x)
ex
. Calculate the derivative of vwith respect to x.
Solution
To find the derivative of vwith respect to x, we need to differentiate each
component of vwith respect to x.
Step 1: Differentiate the first component 3x2:
d
dx (3x2)=6x
Therefore, the first component of the derivative of vis 6x.
Step 2: Differentiate the second component 2 sin(x):
d
dx (2 sin(x)) = 2 cos(x)
The second component of the derivative of vis 2 cos(x).
Step 3: Differentiate the third component ex:
d
dx (ex) = ex
The third component of the derivative of vis ex.
Therefore, the derivative of vwith respect to xis:
dv
dx =
6x
2 cos(x)
ex
Question 7
Question
Let u(t) = 2t
3t2and v(t) = et
cos(t). Find d
dt (u(t)·v(t)).
Solution
Step 1: First, we calculate the dot product u(t)·v(t).
u(t)·v(t) = 2t
3t2·et
cos(t)= 2tet+ 3t2cos(t)
4
Step 2: Next, we differentiate u(t)·v(t) with respect to t.
d
dt (u(t)·v(t)) = d
dt (2tet+ 3t2cos(t))
Step 3: Using the product rule of differentiation, we have:
d
dt (2tet+ 3t2cos(t)) = 2et+ 2tet+ 6tcos(t)−3t2sin(t)
Therefore, d
dt (u(t)·v(t)) = 2et+ 2tet+ 6tcos(t)−3t2sin(t).
Question 8
Question
Let u=
x2
2xy
y2
and v=
3x
3y
−3z
. Find d
dt (u·v) where uand vare functions
of t.
Solution
Step 1: Find u·v.
u·v= (x2)(3x) + (2xy)(3y)+(y2)(−3z)=3x3+ 6xy2−3y2z
Step 2: Differentiate the dot product with respect to t.
d
dt (u·v) = d
dt (3x3+ 6xy2−3y2z)
Step 3: Differentiate each term of the dot product using the chain rule.
d
dt (3x3)=9x2dx
dt ,d
dt (6xy2)=6xdy
dt + 12y2dx
dt ,d
dt (−3y2z) = −3y2dz
dt
Step 4: Put it all together.
d
dt (u·v)=9x2dx
dt + 6xdy
dt + 12y2dx
dt −3y2dz
dt
Therefore, d
dt (u·v)=9x2dx
dt + 6xdy
dt + 12y2dx
dt −3y2dz
dt .
Question 9
Question
Let u= 2i−3j+ 4kand v= 5i+ 7j−2k. Find the derivative of the vector
function F(t) = e2tu+ 3 ln(t)vwith respect to t.
5
Solution
Step 1: Calculate the derivative of F(t) using the properties of vector differen-
tiation. dF(t)
dt =d
dt (e2tu+ 3 ln(t)v)
=d
dt (e2tu) + d
dt (3 ln(t)v)
=d
dt (e2t)u+e2tdu
dt +d
dt (3 ln(t))v+ 3 ln(t)dv
dt
= 2e2tu+ 0 ·du
dt +3
tv+ 3 ln(t)dv
dt
= 2e2tu+3
tv+ 3 ln(t)dv
dt
Step 2: Find dv
dt .
dv
dt =d
dt (5i+ 7j−2k)
= 5 di
dt + 7 dj
dt −2dk
dt
= 0i+ 0j+ 0k
=0
Therefore, the derivative of the vector function F(t) = e2tu+ 3 ln(t)vwith
respect to tis
dF(t)
dt = 2e2tu+3
tv
Question 10
Question
Let u= 2i−3j+ 5kand v= 4i−2j+ 3k.
Find d
dt (u·v) where u·vdenotes the dot product of vectors uand v.
Solution
To find d
dt (u·v), we need to differentiate the dot product of the vectors uand
vwith respect to t.
Recall that the dot product u·vof two vectors u=u1i+u2j+u3kand
v=v1i+v2j+v3kis given by
u·v=u1v1+u2v2+u3v3
Therefore, for vectors u= 2i−3j+ 5kand v= 4i−2j+ 3k, we have:
u·v= (2)(4) + (−3)(−2) + (5)(3) = 8 + 6 + 15 = 29
6
To differentiate u·vwith respect to t, we take the derivative of 29 with
respect to twhich is 0.
Therefore, d
dt (u·v) = 0.
Question 11
Question
Let v=
3x2+ 2y
xy2−4z
5xyz
. Find ∇ · v.
Solution
Step 1: The divergence of a vector field v=
f(x, y, z)
g(x, y, z)
h(x, y, z)
is defined as ∇ · v=
∂f
∂x +∂g
∂y +∂h
∂z .
Step 2: Given v=
3x2+ 2y
xy2−4z
5xyz
, we have f(x, y, z)=3x2+ 2y,g(x, y, z) =
xy2−4z, and h(x, y, z)=5xyz.
Step 3: Compute the partial derivatives:
∂f
∂x =∂
∂x (3x2+ 2y)=6x,
∂g
∂y =∂
∂y (xy2−4z) = x(2y)=2xy,
∂h
∂z =∂
∂z (5xyz)=5xy.
Step 4: Sum the partial derivatives to find the divergence:
∇ · v=∂f
∂x +∂g
∂y +∂h
∂z = 6x+ 2xy + 5xy = 6x+ 7xy.
Therefore, ∇ · v= 6x+ 7xy.
Question 12
Question
Let u=3
2and v=−2
5. Given that f(x) = u·x+v·x, where x=x
y,
find ∇f(x, y).
7
Solution
Step 1: Compute ∇f(x, y) by finding the partial derivatives of f(x) with respect
to xand y.
Step 2: We have f(x) = u·x+v·x. Computing these dot products, we get
f(x) = (3x+ 2y)+(−2x+ 5y)
f(x) = x+ 7y
Step 3: Now, let’s find the partial derivative of fwith respect to x:
∂f
∂x =∂
∂x (x+ 7y) = 1
Step 4: Next, find the partial derivative of fwith respect to y:
∂f
∂y =∂
∂y (x+ 7y) = 7
Step 5: Therefore, the gradient of f(x, y) is
∇f(x, y) = "∂f
∂x
∂f
∂y #=1
7
Question 13
Question
Let v=
x2
ex
ln(x)
. Find dv
dx .
Solution
Step 1: Write vas a column vector.
v=
x2
ex
ln(x)
Step 2: Differentiate each component of vwith respect to x.
dv
dx =
d
dx (x2)
d
dx (ex)
d
dx (ln(x))
Step 3: Find the derivatives.
dv
dx =
2x
ex
1
x
8
Therefore, dv
dx =
2x
ex
1
x
.
Question 14
Question
Let u=x2
exand v=sin(y)
y3. Find d
dy (u·v), where u·vdenotes the dot
product of vectors uand v.
Solution
Step 1: The dot product of two vectors u=u1
u2and v=v1
v2is given by
u·v=u1v1+u2v2.
Step 2: Calculate d
dy (u·v) using the chain rule: d
dy (u·v) = d
dy (x2sin(y) +
exy3).
Step 3: Apply the product rule for differentiation: d
dy (x2sin(y) + exy3) =
x2d
dy (sin(y)) + sin(y)d
dy (x2) + exd
dy (y3).
Step 4: Calculate the derivatives: d
dy (sin(y)) = cos(y), d
dy (x2) = 0, and
d
dy (y3)=3y2.
Step 5: Substituting the derivatives back, we get: d
dy (x2sin(y) + exy3) =
x2cos(y)+0+3exy2.
Therefore, d
dy (u·v) = x2cos(y)+3exy2.
Question 15
Question
Let v=
3x2−2y
4y3
5z
. Find ∂v
∂y .
Solution
Step 1: To find ∂v
∂y , we differentiate each component of vwith respect to y.
Step 2: Differentiating the first component 3x2−2ywith respect to ygives
−2.
Step 3: Differentiating the second component 4y3with respect to ygives
12y2.
Step 4: Differentiating the third component 5zwith respect to ygives 0.
9
Step 5: Therefore, ∂v
∂y =
−2
12y2
0
.
Question 16
Question
Let u=
x2
ex
sin(x)
and v=
x
ln(x)
cos(x)
. Find d(u
·v)dx, where ·denotes the dot
product of two vectors.
Solution
Step 1: Calculate u·v
u·v=x2·x+ex·ln(x) + sin(x)·cos(x) = x3+xex+ sin(x) cos(x)
Step 2: Differentiate u·vwith respect to x
d(u
·v)dx =d(x3)
dx +d(xex)
dx +d(sin(x) cos(x))
dx
Step 3: Calculate the derivatives
d(x3)
dx = 3x2
d(xex)
dx =ex+xex
d(sin(x) cos(x))
dx = cos2(x)−sin2(x)
Step 4: Combine the derivatives
d(u
·v)dx = 3x2+ex+xex+ cos2(x)−sin2(x)
Therefore, d(u
·v)dx = 3x2+ex+xex+ cos2(x)−sin2(x).
Question 17
Question
Let uand vbe two vectors where u=⟨x2,sin(x), ex⟩and v=⟨ln(x), x cos(x),3x⟩.
Find d(u·v)
dx .
10
Solution
Step 1: Calculate the dot product of uand v.
u·v=x2ln(x) + sin(x)·xcos(x) + ex·3x
Step 2: Differentiate u·vwith respect to xusing the product rule.
d(u·v)
dx =d(x2ln(x))
dx +d(xsin(x) cos(x))
dx +d(3xex)
dx
Step 3: Apply the product rule for each term.
d(u·v)
dx = 2xln(x) + x·1
x+xsin(x)(−sin(x)) + cos(x)x+ 3ex+ 3xd(ex)
dx
Step 4: Simplify the derivatives.
d(u·v)
dx = 2xln(x)+1−xsin2(x) + xcos(x)+3ex+ 3ex
Step 5: Combine like terms.
d(u·v)
dx = 2xln(x)+1−xsin2(x) + xcos(x)+6ex
Therefore, d(u·v)
dx = 2xln(x)+1−xsin2(x) + xcos(x)+6ex.
Question 18
Question
Let v= (2t2+ 1)i+ (t3−2)jbe a vector function, where tis a scalar variable.
Find dv
dt .
Solution
Step 1: To differentiate vwith respect to t, we simply need to differentiate each
component of the vector separately.
Step 2: Differentiating the x-component:
d
dt ((2t2+ 1)i) = d
dt (2t2+ 1)i= (4t)i
Step 3: Differentiating the y-component:
d
dt ((t3−2)j) = d
dt (t3−2)j= (3t2)j
Step 4: Putting the differentiated components together, we have:
dv
dt = (4t)i+ (3t2)j
11
Question 19
Question
Let u=3
−2and v=4
7. Find d
dt (u·v).
Solution
Step 1: We start by finding the dot product of vectors uand v:
u·v=3
−2·4
7= 3 ·4+(−2) ·7 = 12 −14 = −2.
Step 2: Next, we differentiate the dot product with respect to t:
d
dt (u·v) = d
dt (−2) = 0 .
Therefore, d
dt (u·v) = 0.
Question 20
Question
Let v=
x2
ex
cos(x)
. Find
dx.
Solution
To find the derivative of the vector vwith respect to x, we simply take the
derivative of each component of v. So, v=
x2
ex
cos(x)
, we have:
dx =
d
dx (x2)
d
dx (ex)
d
dx (cos(x))
Step 1: Compute d
dx (x2):
d
dx (x2)=2x
Step 2: Compute d
dx (ex):
d
dx (ex) = ex
12
Step 3: Compute d
dx (cos(x)):
d
dx (cos(x)) = −sin(x)
Therefore, the derivative of vwith respect to xis:
dx =
2x
ex
−sin(x)
Question 21
Question
Let v= (2x3y, 3xy2) be a vector-valued function. Find the derivative of vwith
respect to x, denoted by dv
dx .
Solution
To find the derivative of vwith respect to x, we will differentiate each component
of vwith respect to xseparately.
Step 1: Differentiate the first component of vwith respect to x. Let
f(x, y)=2x3y. To find ∂f
∂x , we differentiate fwith respect to xtreating y
as a constant. ∂f
∂x =∂
∂x (2x3y) = 6x2y
Step 2: Differentiate the second component of vwith respect to x. Let
g(x, y) = 3xy2. To find ∂g
∂x , we differentiate gwith respect to xtreating yas a
constant. ∂g
∂x =∂
∂x (3xy2)=3y2
Step 3: Combine the derivatives of the components to find dv
dx . Therefore,
the derivative of vwith respect to x, denoted by dv
dx , is:
dv
dx = (∂f
∂x ,∂g
∂x ) = (6x2y, 3y2)
Question 22
Question
Let u= (3x2−2y)i+ (4y3−6x)jbe a vector field. Find ∇ · u.
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Solution
Step 1: The divergence of a vector field u=P(x, y)i+Q(x, y)jis defined as
∇ · u=∂P
∂x +∂Q
∂y .
Step 2: In this case, P(x, y)=3x2−2yand Q(x, y)=4y3−6x.
Step 3: We need to find ∂P
∂x and ∂Q
∂y .
Step 4:
∂P
∂x =∂
∂x (3x2−2y)=6x,
∂Q
∂y =∂
∂y (4y3−6x) = 12y2.
Step 5: Now, we can find the divergence:
∇ · u= 6x+ 12y2= 6x+ 12y2.
Question 23
Question
Let v=
x2
xey
y3
. Find the gradient of v.
Solution
To find the gradient of v, we need to differentiate each component of vwith
respect to their corresponding variables.
Step 1: Differentiate the first component:
∂
∂x (x2)=2x
Step 2: Differentiate the second component:
∂
∂x (xey) = eyand ∂
∂y (xey) = xey
Therefore, the gradient of the second component is ey
xey.
Step 3: Differentiate the third component:
∂
∂y (y3)=3y2
Putting it all together, the gradient of vis:
∇v=
2x
ey
3y2
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Question 24
Question
Let u= 5i−3j+ 2kand v= 2i+ 4j−k. Find d
dt (u·v) where uand vare
functions of t.
Solution
Step 1: We first need to find the dot product of uand v:
u·v= (5i−3j+ 2k)·(2i+ 4j−k)
= 5 ·2+(−3) ·4+2·(−1) = 10 −12 −2 = −4
Step 2: Next, we differentiate the dot product with respect to tusing the
product rule:
d
dt (u·v) = d
dt (−4)
Since −4 is a constant, its derivative with respect to tis 0.
Therefore, d
dt (u·v) = 0 .
Question 25
Question
Let a=
3
−2
1
and b=
1
4
−2
. Find ∇(a·b).
Solution
Step 1: Recall the dot product formula:
a·b=a1b1+a2b2+a3b3
where a=
a1
a2
a3
and b=
b1
b2
b3
.
Step 2: Calculate a·b:
a·b= (3)(1) + (−2)(4) + (1)(−2) = 3 −8−2 = −7
Step 3: Next, find the gradient of the dot product:
∇(a·b) =
∂
∂x
∂
∂y
∂
∂z
(−7)
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Step 4: Since the dot product is a scalar, we only need to differentiate the
scalar component -7 with respect to each variable:
∇(a·b) =
∂
∂x (−7)
∂
∂y (−7)
∂
∂z (−7)
Step 5: Therefore, the gradient of a·bis:
∇(a·b) =
0
0
0
Question 26
Question
Let f(x) = x·(x×y), where x=
x1
x2
x3
and y=
y1
y2
y3
are vectors. Find ∂f
∂x.
Solution
f(x) = x·(x×y)
=
x1
x2
x3
·
x1
x2
x3
×
y1
y2
y3
Step 1: Perform the cross product x×y.
x×y=
i j k
x1x2x3
y1y2y3
= (x2y3−x3y2)i−(x1y3−x3y1)j+ (x1y2−x2y1)k
Step 2: Calculate the dot product x·(x×y).
x·(x×y) =
x1
x2
x3
·((x2y3−x3y2)i−(x1y3−x3y1)j+ (x1y2−x2y1)k)
=x1(x2y3−x3y2) + x2(x1y3−x3y1) + x3(x1y2−x2y1)
Step 3: Find ∂f
∂x.
∂f
∂x=∂f1
∂x1
∂f1
∂x2
∂f1
∂x3
=
x2y3−x3y2
x1y3−x3y1
x1y2−x2y1
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Therefore, ∂f
∂x=
x2y3−x3y2
x1y3−x3y1
x1y2−x2y1
.
Question 27
Question
Let u= 3i−4j+ 2kand v= 5i+ 2j−k. Find the derivative of the vector
function f(t) = e3tu+ ln(t)vwith respect to t.
Solution
Step 1: Find the derivative of f(t) with respect to tby differentiating each term
separately.
df
dt =d
dt e3tu+ ln(t)v
Step 2: Take the derivative of e3tuwith respect to t.
d
dt e3tu=d
dt e3tu+e3tdu
dt
Step 3: We have d
dt e3t= 3e3tand du
dt =0. Therefore,
d
dt e3tu= 3e3tu
Step 4: Take the derivative of ln(t)vwith respect to t.
d
dt (ln(t)v) = d
dt (ln(t)) v+ ln(t)dv
dt
Step 5: We have d
dt (ln(t)) = 1
tand dv
dt =0. Thus,
d
dt (ln(t)v) = 1
tv
Step 6: Putting the results together, we get the derivative of f(t) with respect
to t.df
dt = 3e3tu+1
tv
Therefore, the derivative of the vector function f(t) = e3tu+ ln(t)vwith
respect to tis df
dt = 3e3tu+1
tv.
Question 28
Question
Let v= 4i−2j+kand w= 2i+ 3j−5k. Find d(v
·w)dt.
17
Solution
To differentiate the dot product v·wwith respect to t, we first need to find the
dot product of vand w.
Step 1: Find v·w:
v·w= (4i−2j+k)·(2i+ 3j−5k)
v·w= (4 ·2) + (−2·3) + (1 ·(−5))
v·w= 8 −6−5
v·w=−3
Step 2: Differentiate v·wwith respect to t:
d(v
·w)dt =d(−3)
dt
d(v
·w)dt = 0
Therefore, d(v
·w)dt = 0.
Question 29
Question
Find the gradient of the following vector-valued function: f(r) = r
|r|, where
r=xi+yj+zk.
Solution
Step 1: Calculate the magnitude of r:
|r|=px2+y2+z2
Step 2: Rewrite f(r) using the magnitude of r:
f(r) = xi+yj+zk
px2+y2+z2
Step 3: To find the gradient of f(r), we need to differentiate each component
with respect to x,y, and z.
Step 4: Differentiate the x-component:
∂
∂x x
px2+y2+z2!=1
px2+y2+z2−x2
(x2+y2+z2)3/2
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Step 5: Differentiate the y-component:
∂
∂y y
px2+y2+z2!=1
px2+y2+z2−y2
(x2+y2+z2)3/2
Step 6: Differentiate the z-component:
∂
∂z z
px2+y2+z2!=1
px2+y2+z2−z2
(x2+y2+z2)3/2
Step 7: The gradient of f(r) is:
∇f(r) = ∂
∂x x
|r|i+∂
∂y y
|r|j+∂
∂z z
|r|k
= 1
px2+y2+z2−x2
(x2+y2+z2)3/2!i+ 1
px2+y2+z2−y2
(x2+y2+z2)3/2!j+ 1
px2+y2+z2−z2
(x2+y2+z2)3/2!k
Question 30
Question
Let v=
3x2
4xy
5yz
be a vector valued function. Find the derivative of vwith
respect to x,dv
dx .
Solution
To find the derivative of vector valued function vwith respect to x, we differ-
entiate each component of vseparately with respect to x.
Step 1: For the first component 3x2, we differentiate with respect to x:
d
dx (3x2)=6x
Step 2: For the second component 4xy, we differentiate with respect to x:
d
dx (4xy)=4y
Step 3: For the third component 5yz, we differentiate with respect to x:
d
dx (5yz)=0
Step 4: Putting all the components together, the derivative of vwith respect
to x,dv
dx , is:
dv
dx =
6x
4y
0
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Question 31
Question
Let a= 2i−3jand b= 5i+ 4j. Find the derivative of f=a·bwith respect
to t.
Solution
Step 1: First, we compute the dot product f=a·b.
f= (2i−3j)·(5i+ 4j)
f= 2 ·5+(−3) ·4 = 10 −12 = −2
Step 2: Next, we express aand bin terms of t.
a= 2i−3j= 2i(t)−3j(t)
b= 5i+ 4j= 5i(t)+4j(t)
Step 3: We differentiate fwith respect to tusing the product rule for vectors.
df
dt =d(2i(t))
dt ·(5i+4j)+(2i−3j)·d(5i(t))
dt +d(−3j(t))
dt ·(5i+4j)+(2i−3j)·d(4j(t))
dt
Step 4: Evaluating each term using the chain rule and product rule for
differentiation, we get:
df
dt = 2di
dt ·5i+ 2i·5di
dt −3dj
dt ·5i−3j·4dj
dt
Step 5: Simplifying further, we have:
df
dt = 2j·5i+ 2i·5j−3j·5i−3i·4j
df
dt = 10j+ 10i−15j−12i
Step 6: Combining like terms, we get the derivative of fwith respect to t:
df
dt =−2i−5j
Question 32
Question
Let u= 2i−3j+kand v=i+ 4j−2k. Find the derivative of u·vwith respect
to t, where i,j,kare the standard basis vectors and trepresents time.
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Solution
Step 1: The dot product of two vectors u=u1i+u2j+u3kand v=v1i+v2j+v3k
is given by:
u·v=u1v1+u2v2+u3v3
Step 2: Given that u= 2i−3j+kand v=i+ 4j−2k, we have:
u·v= (2)(1) + (−3)(4) + (1)(−2)
Step 3: Calculating the dot product, we get:
u·v= 2 −12 −2 = −12
Step 4: To find the derivative of u·vwith respect to t, we differentiate the
dot product with respect to t:
d
dt (u·v) = d
dt (−12) = 0
Step 5: Therefore, the derivative of u·vwith respect to tis 0 .
Question 33
Question
Let aand bbe constant vectors. If r(t) = t2a+ sin(t)b, find dr
dt .
Solution
To find dr
dt , we differentiate each component of r(t) with respect to t.
Step 1: Differentiate the first component t2a:
d
dt (t2a)=2ta
Step 2: Differentiate the second component sin(t)b:
d
dt (sin(t)b) = cos(t)b
Therefore, the derivative of r(t) with respect to tis:
dr
dt = 2ta+ cos(t)b
Question 34
Question
Let u=ex
ln(y)and v=sin(z)
cos(t). Determine d
dt (u·v).
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Solution
Step 1: Find the dot product u·v.
u·v= (ex)(sin(z)) + (ln(y))(cos(t))
Step 2: Differentiate with respect to t.
d
dt (u·v) = d
dt [(ex)(sin(z)) + (ln(y))(cos(t))]
Step 3: Apply the product rule for differentiation.
d
dt (u·v) = d(ex)
dt (sin(z)) + (ex)d(sin(z))
dt +d(ln(y))
dt (cos(t)) + (ln(y))d(cos(t))
dt
Step 4: Simplify by evaluating each derivative.
d
dt (u·v) = (ex)(0)(sin(z)) + (ex)(cos(z))dz
dt + 0(cos(t)) + 1
y
dy
dt (cos(t))
Step 5: Express the final answer.
d
dt (u·v)=(ex)(cos(z))dz
dt +cos(t)
y
dy
dt
Question 35
Question
Let v= (x2+y2)i+ 2xyj. Find dv
dx .
Solution
Step 1: To differentiate a vector with respect to x, we differentiate each com-
ponent of the vector separately. Let v= (f(x, y))i+ (g(x, y))j, where f(x, y) =
x2+y2and g(x, y)=2xy.
Step 2: Differentiating the first component of the vector:
∂f
∂x =∂
∂x (x2+y2)=2x
Step 3: Differentiating the second component of the vector:
∂g
∂x =∂
∂x (2xy)=2y
Step 4: Putting it all together, we get:
dv
dx =d
dx (x2+y2)i+ 2xyj=d
dx (x2+y2)i+d
dx (2xy)j= 2xi+ 2yj
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