DOC PREVIEW
UIUC MATH 241 - Worksheet_091812_sol

This preview shows page 1 out of 4 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 4 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 4 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Tuesday, September 18 ∗ Solutions ∗ Partial derivatives and differentiability.1. Consider f (x, y) =xy2ex2. Compute the following:∂ f∂x∂ f∂y∂2f∂x∂y∂2f∂y∂xWhat is the relationship between your answers for the last two? This is an instance of Clairaut’sTheorem and holds for most functions.Solution.∂ f∂x=2x2y2ex2+y2ex2;∂ f∂y=2x yex2∂2f∂x∂y=∂∂xµ∂ f∂y¶=∂∂x(2x yex2) =4x2yex2+2yex2∂2f∂y∂x=∂∂yµ∂ f∂x¶=∂∂y(2x2y2ex2+y2ex2) =4x2yex2+2yex2So we see that∂2f∂x∂y=∂2f∂y∂x.2. Shown are some level curves for the function f : R2→ R. Determine whether the followingpartial derivatives are positive, negative, or zero at the point P:fxfyfxxfy yfx y= fyxSolution.(a) If we fix y and allow x to vary, the level curves indicate that the value of f increases as wemove through P in the positive x-direction, so fxis positive at P.(b) If we fix x and allow y to vary, the level curves indicate that the value of f is neither in-creasing nor decreasing as we move through P in the positive y-direction, so fyis zero atP.(c) fxx=∂∂x(fx), so if we fix y and allow x to vary, fxxis the rate of change of fxas x increases.Note that at points to the right of P the level curves are closer together (in the x-direction)than at points to the left of P, demonstrating that f increases more quickly with respect tox to the right of P. So as we move through P in the positive x-direction the (positive) valueof fxincreases, hence fxxis positive.(d) fy y=∂∂y(fy), so if we fix x and allow y to vary, fy yis the rate of change of fyas y increases.The level curves are closer together (in the y-direction) at points above P than at thosebelow P, demonstrating that f increases more quickly with respect to y above P. So as wemove through P in the positive y-direction the (positive) value of fyincreases, hence fy yis positive.(e) fx y=∂∂y(fx), so if we fix x and allow y to vary, fx yis the rate of change of fxas y increases.The level curves are closer together (in the x-direction) at points above P than at thosebelow P, demonstrating that f increases more quickly with respect to x for y-values aboveP. So as we move through P in the positive y-direction, the (positive) value of fxincreases,hence fx yis positive.3. The wind-chill index W = f (T,v) is the perceived temperature when the actual temperature isT and the wind speed is v. Here is a table of values for W .(a) Use the table to estimate∂ f∂Tand∂ f∂vat (T, v) =(−20,40).Solution.fT(−20,40) = limh→0f (−20 +h, 40) − f (−20,40)hwhich can be approximated by considering h =5 and h =−5 as follows:fT(−20,40) ≈f (−15,40) − f (−20,40)5=−27 −(−34)5=75,fT(−20,40) ≈f (−25,40) − f (−20,40)−5=−41 −(−34)−5=75.Averaging these values, we estimate fT(−20,40) to be 7/5.Similarly,fv(−20,40) = limh→0f (−20,40 +h) − f (−20,40)hwhich can be approximated by considering h =10 and h =−10:fv(−20,40) ≈f (−20,50) − f (−20,40)10=−35 −(−34)10=−110,fv(−20,40) ≈f (−20,30) − f (−20,40)−10=−33 −(−34)−10=−110.Averaging these values, we estimate fv(−20,40) to be −1/10.(b) Use your answer in (a) to write down the linear approximation to f at (−20, 40).Solution. The linear approximation of f at (−20, 40) isf (T,v) ≈ f (−20,40) + fT(−20,40)(T −(−20)) + fv(−20,40)(v −40)=−34 +75(T +20)−110(v −40).(c) Use your answer in (b) to approximate f (−22,45).Solution.f (−22,45) ≈−34 +75(−22 +20) −110(45 −40) =−37.3.4. Consider f (x, y) =p1 −x2−y2.(a) What is the domain of f ? That is, for which (x, y) does the function make sense?Solution.D ={(x, y) ∈R2: x2+y2≤1}.(b) Describe geometrically the surface which is the graph of f .Solution. It is the upper half of a unit sphere.(c) Find the tangent plane to the graph at (x, y) =(1/2, 1/2).Solution. fx(x, y) =−xp1−x2−y2and fy(x, y) =−yp1−x2−y2. Then fx(12,12) =−p22and fy(12,12) =−p22. So the equation of the tangent plane at (12,12) isz − f (12,12) = fx(12,12)(x −12) + fy(12,12)(y −12),z −p22=−p22(x −12) −p22(y −12),or z =p22(2 −x −y).(d) Consider the vector v =¡1/2,1/2,1/p2¢which goes from 0 to the point on the graph wherewe just found the tangent plane. What is the angle between v and a normal vector to thetangent plane?Solution. From part (c), a normal vector to the tangent plane is n = (p22,p22,1). Let θbe the angle between v and n, thencosθ =v ·n|v||n|=(1/2,1/2,1/p2) ·(p2/2,p2/2,1)q(1/2)2+(1/2)2+(1/p2)2q(p2/2)2+(p2/2)2+12=1,so θ =cos−1(1) =0.5. Consider f (x, y) =3px3+y3.(a) Compute fx(0,0). Note: this partial derivative exists.Solution.fx(0,0) = limh→0f (0 +h,0) − f (0,0)h= limh→0(h3+0)1/3−0h= limh→0hh=1.(b) Is f differentiable at (0,0)?Solution. No, f is not differentiable at (0,0). Consider that if we look along the line y = xthe function is f (x,x) =3px3+x3=3p2x. Suppose on the contrary that f is differentiableat (0,0). Then the linear approximation would bef (x, x) = f (0, 0) + fx(0,0)x + fy(0,0)x +E(x, x)=3p03+03+x +x +E(x,x) =2x +E(x, x).Solim(x,x)→(0,0)E(x, x)|(x, x)|= lim(x,x)→(0,0)(3p2 −2)xp2x6=0.Hence f is not differentiable at


View Full Document

UIUC MATH 241 - Worksheet_091812_sol

Documents in this Course
Notes

Notes

9 pages

16_05

16_05

29 pages

16.6

16.6

43 pages

16_07

16_07

34 pages

16_08

16_08

12 pages

16_09

16_09

13 pages

exam1

exam1

10 pages

exam2

exam2

7 pages

exam3

exam3

9 pages

15_03

15_03

15 pages

15_04

15_04

13 pages

15_04 (1)

15_04 (1)

13 pages

15_05

15_05

31 pages

15_10

15_10

27 pages

15_07

15_07

25 pages

15_08

15_08

12 pages

15_09

15_09

24 pages

15_10_B

15_10_B

8 pages

16_04

16_04

17 pages

14_01

14_01

28 pages

12_06

12_06

12 pages

12_05

12_05

19 pages

12_04

12_04

26 pages

Lecture1

Lecture1

31 pages

Lecture 9

Lecture 9

41 pages

Lecture 8

Lecture 8

35 pages

Lecture 7

Lecture 7

40 pages

Lecture 6

Lecture 6

49 pages

Lecture 5

Lecture 5

26 pages

Lecture 4

Lecture 4

43 pages

Lecture 3

Lecture 3

29 pages

Lecture 2

Lecture 2

17 pages

m2-1

m2-1

6 pages

-

-

5 pages

Load more
Download Worksheet_091812_sol
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Worksheet_091812_sol and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Worksheet_091812_sol 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?