Samantha Rodriguez Brewer MAT 267 ONLINE B Fall 2020
Assignment Section 11.5 due 11/08/2020 at 11:59pm MST
1. (1 point) Suppose w=x
y+y
z, where
x=e4t,y=2+sin(5t), and z=2+cos(7t).
(1) Use the chain rule to find dw
dt as a function of x,y,z,
and t. Do not rewrite x,y,and zin terms of t, and do not
rewrite e4tas x.
dw
dt =
Note: Your answer should be an expression in x,y,z,
and t; e.g. “3x - 4y + 2t”
(2) Use part 1. to evaluate dw
dt when t=0.
Solution:
SOLUTION:
(1)
∂w
∂t=∂w
∂x
∂x
∂t+∂w
∂y
∂y
∂t+∂w
∂z
∂z
∂t
=1
y4e4t+1
z−x
y2(5cos(5t)) + −y
z2(−7sin(7t))
(2) When t=0,x=1,y=2 and z=3. Thus
∂w
∂t=1
2(4) + 1
3−1
4(5) + −2
9(0) = 29
12
Correct Answers:
•1/y*4*eˆ(4*t)+[1/z-x/(yˆ2)]*5*cos(5*t)+-[y/(zˆ2)]*-7*sin(7*t)
•2.41667
2. (1 point) Suppose z=x2sin(y),x=−2s2−2t2,y=8st.
A. Use the chain rule to find ∂z
∂sand ∂z
∂tas functions of x,y,s
and t.
∂z
∂s=
∂z
∂t=
B. Find the numerical values of ∂z
∂sand ∂z
∂twhen (s,t) = (2,2).
∂z
∂s(2,2) =
∂z
∂t(2,2) =
Solution:
SOLUTION
∂z
∂s=∂z
∂x
∂x
∂s+∂w
∂y
∂y
∂s= (2xsin(y))(−(2·2s)) + x2cos(y)(8t)
∂z
∂t=∂z
∂x
∂x
∂t+∂w
∂y
∂y
∂t= (2xsin(y))(−(2·2t)) + x2cos(y)(8s)
When s=2 and t=2, x=−16 and y=32. Thus
∂z
∂s=256sin(32) + 4096cos(32)≈3558.14
∂z
∂t=256sin(32) + 4096 cos(32)≈3558.14
Correct Answers:
•2*x*sin(y)*-(2*2*s)+xˆ2*cos(y)*8*t
•2*x*sin(y)*-(2*2*t)+xˆ2*cos(y)*8*s
•3558.14
•3558.14
3. (1 point) Use the chain rule to find ∂z
∂sand ∂z
∂t, where
z=exy tan(y),x=4s+5t,y=3s
3t
First the pieces:
∂z
∂x=
∂z
∂y=
∂x
∂s=
∂x
∂t=
∂y
∂s=
∂y
∂t=
And putting it all together:
∂z
∂s=∂z
∂x
∂x
∂s+∂z
∂y
∂y
∂sand ∂z
∂t=∂z
∂x
∂x
∂t+∂z
∂y
∂y
∂t
Correct Answers:
•eˆ(x*y)*y*tan(y)
•eˆ(x*y)*x*tan(y)+eˆ(x*y)*[sec(y)]ˆ2
•4
•5
•3*3*t/[(3*t)ˆ2]
•-(3*3*s/[(3*t)ˆ2])
4. (1 point) Let
w=−3xy −3yz −5xz,x=st,y=est ,z=t2
Compute
∂w
∂s(3,5) =
∂w
∂t(3,5) =
Solution:
SOLUTION:
∂w
∂s=∂w
∂x
∂x
∂s+∂w
∂y
∂y
∂s+∂w
∂z
∂z
∂s
= (−3y−5z)(t)+(−3x−3z)(test )+(−3y−5x)(0)
∂w
∂t=∂w
∂x
∂x
∂t+∂w
∂y
∂y
∂t+∂w
∂z
∂z
∂t
= (−3y−5z)(s)+(−3x−3z)(sest )+(−3y−5x)(2t)
When s=3 and t=5, x=15,y=e15 and z=25. Thus
1
∂w
∂s(3,5) = −3e15 −125(5)−1205e15
=−615e15 −625
∂w
∂t(3,5) = −3e15 −125(3)−1203e15+−3e15 −75(10)
=−399e15 −1125
Correct Answers:
•-2010446309.07035
•-1304339056.61637
5. (1 point) Let W(s,t) = F(u(s,t),v(s,t)) where
u(1,0) = 8,us(1,0) = −7,ut(1,0) = 4
v(1,0) = −1,vs(1,0) = 1,vt(1,0) = 7
Fu(8,−1) = −2,Fv(8,−1) = 9
Ws(1,0) = Wt(1,0) =
Solution:
SOLUTION:
By the chain rule, ∂W
∂s=∂W
∂u
∂u
∂s+∂W
∂v
∂v
∂s. Then
Ws(1,0) = Fu(u(1,0),v(1,0))us(1,0) + Fv(u(1,0),v(1,0))vs(1,0)
=Fu(8,−1)us(1,0) + Fv(8,−1)vs(1,0)
=−2(−7) + 9(1)
=23
Similarly, ∂W
∂t=∂W
∂u
∂u
∂t+∂W
∂v
∂v
∂t. Then
Wt(1,0) = Fu(u(1,0),v(1,0))ut(1,0) + Fv(u(1,0),v(1,0))vt(1,0)
=Fu(8,−1)ut(1,0) + Fv(8,−1)vt(1,0)
=−2(4) + 9(7)
=55
Correct Answers:
•23
•55
6. (1 point) Consider the curve x5+3xy +y5=5
The equation of the tangent line to the curve at the point (1,1)
has the form y=mx +bwhere
m=and b=
Solution:
SOLUTION
Let F(x,y) = x5+3xy +y5−5=0. Then
dy
dx =−Fx
Fy
=−5x4+3y
3x+5y4
The slope of the tangent at (1,1)is then
m=−5+3
3+5=−1
The equation of the tangent line at (1,1)is y−1=−1(x−1)or
y=−1x+2.
Thus the y-intercept is b=2.
Correct Answers:
•-1
•2
7. (1 point) Consider the surface F(x,y,z) = x7z3+
siny8z3−4=0.
Find the following partial derivatives
∂z
∂x=
∂z
∂y=
Solution:
SOLUTION
∂z
∂x=−Fx
Fz
=−7x6z3
x7·3z2+y8·3z2cos(y8z3)
∂z
∂y=−Fy
Fz
=−8y7z3cosy8z3
x7·3z2+y8·3z2cos(y8z3)
Correct Answers:
•-(7*xˆ6*zˆ3)/[xˆ7*3*zˆ2+yˆ8*3*zˆ2*cos(yˆ8*zˆ3)]
•-[8*yˆ7*zˆ3*cos(yˆ8*zˆ3)]/[xˆ7*3*zˆ2+yˆ8*3*zˆ2*cos(yˆ8*zˆ3)]
8. (1 point)
The radius of a right circular cone is increasing at a rate of
2 inches per second and its height is decreasing at a rate of 4
inches per second. At what rate is the volume of the cone chang-
ing when the radius is 40 inches and the height is 40 inches?
NOTE: The volume of a cone with base radius rand height his
given by V=1
3πr2h.
cubic inches per second
Solution:
SOLUTION:
dV
dt =∂V
∂r
dr
dt +∂V
∂h
dh
dt
=2πrh
3(2) + πr2
3(−4)
=6400
3π−6400
3π
=0πin3/sec
Correct Answers:
•0
9. (1 point) In a simple electric circuit, Ohm’s law states
that V=IR, where Vis the voltage in Volts, Iis the current in
Amperes, and Ris the resistance in Ohms. Assume that, as the
battery wears out, the voltage decreases at 0.03 Volts per second
and, as the resistor heats up, the resistance is increasing at 0.04
Ohms per second. When the resistance is 200 Ohms and the
current is 0.03 Amperes, at what rate is the current changing?
Amperes per second
Solution:
SOLUTION
I=V
R⇒
dI
dt =∂I
∂V
dV
dt +∂I
∂R
dR
dt =1
R
dV
dt −V
R2
dR
dt
Since V=IR,
dI
dt =1
R
dV
dt −I
R
dR
dt =1
200 (−0.03)−0.03
200 (0.04) = −0.000156
Correct Answers:
•-0.000156
2
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