Samantha Rodriguez Brewer MAT 267 ONLINE B Fall 2020
Assignment Section 11.4 due 11/06/2020 at 11:59pm MST
1. (1 point) Find an equation of the tangent plane to the sur-
face
z=5x5+8y5+4xy at the point (−2,2,80).
Solution:
SOLUTION:
fx(x,y) = 5·5x4+4y⇒fx(−2,2) = 408
fy(x,y) = 8·5y4+4x⇒fy(−2,2) = 632
Thus an equation of the tangent plane is
z=80 +fx(−2,2)(x+2) + fy(−2,2)(y−2)
⇒z=80 +408(x+2) + 632(y−2)
⇒z=408x−368 +632y
Correct Answers:
•z-(408*x+632*y) = -368
2. (1 point) Find the equation of the tangent plane to the
surface z=e4x/17 ln(y)at the point (−2,3,0.686231).
Solution:
SOLUTION
fx(x,y) = 4
17 e4x/17 ln(y)⇒fx(−2,3) = 4
17 e−8/17 ln(3)≈0.161466
fy(x,y) = e4x/17
y⇒fy(−2,3) = e−8/17
3≈0.208212
Thus an equation of the tangent plane is
z=0.686231 +fx(−2,3)(x+2) + fy(−2,3)(y−3)
⇒z=0.686231 +0.161466(x+2) + 0.208212(y−3)
Correct Answers:
•0.161466*x+0.208212*y-z = -0.384529
3. (1 point) Find the linearization of the function f(x,y) =
x√yat the point (9,25).
L(x,y) =
Solution:
SOLUTION
f(x,y) = x√y⇒f(9,25) = 9√25 =45
fx(x,y) = √y⇒fx(9,25) = √25 =5
fy(x,y) = x
2√y⇒fy(9,25) = 9
2√25 =9
10
Both fxand fyare continuous functions for y≥0, so fis differ-
entiable at (9,25)and the linearization of fat (9,25)is
L(x,y) = 45 +5(x−9) + 9
10 (y−25)
Correct Answers:
•45 + 5*(x - 9) + 0.9 * (y - 25)
4. (1 point) Find the linearization of the function f(x,y) =
p27 −x2−2y2at the point (−4,1).
L(x,y) =
Use the linear approximation to estimate the value of
f(−4.1,1.1)
f(−4.1,1.1)≈
Solution:
SOLUTION
f(x,y) = p27 −x2−2y2⇒f(−4,1) = 3
fx(x,y) = −1
2p27 −x2−2y2·2x⇒fx(−4,1) = 1.33333
fy(x,y) = −1
2p27 −x2−2y2·2·2y⇒fy(−4,1) = −0.666667
Both fxand fyare continuous functions for 27−x2−2y2>0, so
fis differentiable at (−4,1)and the linearization of fat (−4,1)
is
L(x,y) = 3+1.33333(x+4)−0.666667(y−1)
We use the linear approximation to estimate the value of
f(−4.1,1.1):
f(−4.1,1.1)≈L(−4.1,1.1) = 3+1.33333(−4.1+4)−0.666667(1.1−1) = 2.8
Correct Answers:
•3+1.33333*(x+4)-0.666667*(y-1)
•2.8
5. (1 point) Suppose that f(x,y)is a smooth function and
that its partial derivatives have the values, fx(−7,0) = −3 and
fy(−7,0) = −1. Given that f(−7,0) = −6, use this information
to estimate the following values:
Estimate of (integer value) f(−7,1):
Estimate of (integer value) f(−6,0):
Estimate of (integer value) f(−6,1):
Solution:
SOLUTION
We can estimate the required values by using the linear ap-
proximation of fat (−7,0), given by
f(x,y)≈f(−7,0) + fx(−7,0)(x+7) + fy(−7,0)(y−0)
=−6−3(x+7)−1(y−0).
Thus
f(−7,1)≈ −6−3(−7+7)−1(1−0) = −7
f(−6,0)≈ −6−3(−6+7)−1(0−0) = −9
f(−6,1)≈ −6−3(−6+7)−1(1−0) = −10
Correct Answers:
•-7
1
•-9
•-10
6. (1 point) Find the differential of the function w=
x4siny7z2
dw =dx+dy+dz
Solution:
SOLUTION
dw =∂w
∂xdx +∂w
∂ydy +∂w
∂zdz
=4x3siny7z2dx +x4·7y6z2cosy7z2dy +x4y7·2zcosy7z2dz
Correct Answers:
•4*xˆ3*sin(yˆ7*zˆ2)
•xˆ4*7*yˆ6*zˆ2*cos(yˆ7*zˆ2)
•xˆ4*yˆ7*2*z*cos(yˆ7*zˆ2)
7. (1 point) Use differentials to estimate the amount of mate-
rial in a closed cylindrical can that is 80 cm high and 32 cm in
diameter if the metal in the top and bottom is 0.1 cm thick, and
the metal in the sides is 0.1 cm thick.
Note, you are approximating the volume of metal which makes
up the can (i.e. melt the can into a blob and measure its volume),
not the volume it encloses.
The differential for the volume is
dV =dr+dh (enter your answer in
therms of rand h)
dr =and dh =(be careful)
The approximate volume of material is cm3.
Solution:
SOLUTION
Let rand hbe the radius and height of the can, respectively.
The volume of a can is V=πr2hand ∆V≈dV is an estimate of
the amount of metal.
dV =∂V
∂rdr +∂V
∂hdh = (2πrh)dr + (πr2)dh
Letting dr =0.1 and dh =0.2 (0.1 on top, 0.1 on bottom), we
get
dV =2π(16)(80) (0.1) + π(16)2(0.2) = 307.2π
Thus the amount of metal is about 307.2πcm3.
Correct Answers:
•2*3.14159265358979*r*h
•3.14159265358979*r**2
•0.1
•0.2
•965.097263182784
8. (1 point) The dimensions of a closed rectangular box are
measured as 80 centimeters, 70 centimeters, and 70 centime-
ters, respectively, with the error in each measurement at most .2
centimeters. Use differentials to estimate the maximum error in
calculating the surface area of the box.
square centimeters
Solution:
SOLUTION
Let x,yand zbe the dimensions of the box.
The surface area is S(x,y,z) = 2(xy +yz +xz)and
dS(x,y,z) = 2(y+z)dx +2(x+z)dy +2(y+x)dz.
The error in measurements are at most .2, so |dx| ≤.2,|dy| ≤.2
and |dz| ≤ .2.
An estimate of the error in the calculated area is then
dS(80,70,70) = 2(70+70)(.2)+2(80+70)(.2)+2(70+80)(.2) = 176
Correct Answers:
•176
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