Calculus problems, where to begin?

  • Thread starter Jeebus
  • Start date
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Hello! I have a math problem (mostly proofs) that im stuck
on, partially because I do not know where to begin and partially
because I believe I dont even fully understand the problem. I
was wondering if any of you would be kind enough to show me what to
do? Thank you.

1. Suppose f(x,y) is differentiable for all (x,y), f(x,y)=17 on the
unit circle x^2+y^2=1 and grad f is never zero on the unit circle. For
any real number K, find a unit vector parallel to grad
f(cos(k),sin(k))....grad f stands for the gradient of f. But isnt it contradicting what its saying? It says f(x,y)=17 on the unit circle x^2+y^2. How the...?

I'm just supposing f(x,y) is differentiable for all (x,y), f(x,y)=17 on the unit circle x^2+y^2=1 and grad(f) is never zero on the unit circle(?) So you just find a unit vector parallel to grad f(cos(k),sin(k)), for k real, right?

PS- Do level curves apply to this problem?
 
80
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Hello Jeebus,

You're being asked to find the direction of ∇f on the unit circle (k is just an angle). I think it's easier if we use polar coordinates (r,θ), and their corresponding unit vectors r and θ (don't know how to make the little hats yet). If we look at the gradient on the circle, and dot it with θ: θ dot ∇f, we get the rate of change of f in the θ direction. But what is the rate of change of f in the θ-direction on the unit circle? And what can you conclude about the direction of ∇f from this?

Hope this helps,
dhris
 

HallsofIvy

Science Advisor
Homework Helper
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dhris was giving hints that should help you but I got the impression that you really had no idea what was going on (and so need more than hints).

You ask "Do level curves apply to this problem?" Well, yes, of course. You are GIVEN that f(x,y)= 17 on the unit on the unit circle. The point is that f(x,y) is CONSTANT on that circle. The unit circle IS a level curve. Now, what is the relationship between level curves of a function and the gradient of the function at points on a level curve?
 

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