A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Find the sum direction and magnitude of the greatest rate of change of the function z+ln(x+y^2) at (1,1)
anonymous
 5 years ago
Find the sum direction and magnitude of the greatest rate of change of the function z+ln(x+y^2) at (1,1)

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it should not say sum... it should just say find the direction and magnitude my bad

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know you need to take the directional derivative and turn (1,1) into a unit vector and I did that, but i'm stuck at this point. I have \[DuF(x,y) = (1+2y)\div(\sqrt{x}(x+y ^{2}))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I’m not very knowledgeable in vector calc, but this comes from http://tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx#Gradient_Defn There’s a theorem in vector calculus that says that the gradient of a function (grad(f(x)) points in the direction of maximal change, and that the magnitude of this change is grad(f(x)) at that point x. If f(x, y, z) = z + ln(x + y^2), then grad(f) = <1/(x + y^2), 2y/(x + y^2), 1>. Plugging in our point (1, 1), we see that the direction of maximal change is <1/2, 1, 1>, and the magnitude of that change is <1/2, 1, 1> = sqrt(9/4) = 3/2. Again, if you want, you can check out that link above for more explanation and the proof for the theorem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you thank you thank you
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.