Multivariable calculus, extremes problem

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1. Homework Statement

[itex] \text{Let f be a differentiable function from ℝ^2 to ℝ that satisfies:}\ [/itex]

[itex]1) f(x,y)=0\ \text{for all}\ (x,y)\ \text{in the circumference}\ x^2+y^2=2[/itex]
[itex]2) \text{If we consider the function g from ℝ to ℝ^2 given by g(t)=(t+1,e^​t), then the function fog has a relative minimum at t=0}\ [/itex]

[itex] \text{Find the gradient of f at the point (1,1)}\ [/itex]

The attempt at a solution

By the chain rule, I know that 0= D(fog(0))= D(f(g(0))g'(0)=([itex]\frac{\partial f}{\partial x}[/itex](1,1),[itex]\frac{\partial f}{\partial y}[/itex](1,1))[itex]
\begin{pmatrix}
1\\
1
\end{pmatrix}
[/itex]

[itex] \text{Doing matrix multiplication, I get the equation}\ [/itex][itex]\frac{\partial f}{\partial x}[/itex][itex](1,1)[/itex] + [itex]\frac{\partial f}{\partial y}[/itex][itex](1,1)=0[/itex]

So, I could use the information of 2) but I don't know how to use the data on 1). The only thing I know with 2) is that the partial derivatives satisfy [itex]\frac{\partial f}{\partial x}[/itex][itex](1,1)[/itex]=-[itex]\frac{\partial f}{\partial y}[/itex][itex](1,1)[/itex]. Could anyone tell me how to use the information given in 1)?

I'm trying to use latex but I'm clumsy, sorry about that, I'll try to improve my latex skills in the near future.
 
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Answers and Replies

  • #2
verty
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This confused me at first.

Hint: What constraints can you place on the tangent plane to f at (1, 1)?
Hint: What information about f is that statement about g giving you?
 
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  • #3
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g(t)=(t+1,e^​x)

what is this supposed to mean?
 
  • #4
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g(t)=(t+1,e^​x)

what is this supposed to mean?
Oops, I corrected my mistake, it made no sense with the two variables.
 
  • #5
Ray Vickson
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1. Homework Statement

[itex] \text{Let f be a differentiable function from ℝ^2 to ℝ that satisfies:}\ [/itex]

[itex]1) f(x,y)=0\ \text{for all}\ (x,y)\ \text{in the circumference}\ x^2+y^2=2[/itex]
[itex]2) \text{If we consider the function g from ℝ to ℝ^2 given by g(t)=(t+1,e^​t), then the function fog has a relative minimum at t=0}\ [/itex]

[itex] \text{Find the gradient of f at the point (1,1)}\ [/itex]

The attempt at a solution

By the chain rule, I know that 0= D(fog(0))= D(f(g(0))g'(0)=([itex]\frac{\partial f}{\partial x}[/itex](1,1),[itex]\frac{\partial f}{\partial y}[/itex](1,1))[itex]
\begin{pmatrix}
1\\
1
\end{pmatrix}
[/itex]

[itex] \text{Doing matrix multiplication, I get the equation}\ [/itex][itex]\frac{\partial f}{\partial x}[/itex][itex](1,1)[/itex] + [itex]\frac{\partial f}{\partial y}[/itex][itex](1,1)=0[/itex]

So, I could use the information of 2) but I don't know how to use the data on 1). The only thing I know with 2) is that the partial derivatives satisfy [itex]\frac{\partial f}{\partial x}[/itex][itex](1,1)[/itex]=-[itex]\frac{\partial f}{\partial y}[/itex][itex](1,1)[/itex]. Could anyone tell me how to use the information given in 1)?

I'm trying to use latex but I'm clumsy, sorry about that, I'll try to improve my latex skills in the near future.
Assume, instead of simple differentiability, that f has *continuous* partial derivatives near the circle. The fact that f is constant on the circle introduces a relationship between the partial derivatives ##f_x## and ##f_y## at points on the circle. Rather than spelling out what the relationship IS, I would rather tell you how you could get it for yourself. So, say you start at a point (x,y) = (a,b) on the circle, and you go to another point point (x',y') = (a+u,b+v) on the circle, where u| and |v| are small. Using f(a,b) = 0 and f(a+u,b+v) = 0, expand up to terms linear in u and v. The fact that (a+u,v+b) is also on the circle introduces a relationship between u and v; find this relationship to first order in u and v. Finally, see what first-order condition in u (or v) you must have in order to keep f = 0 at the new point. Then apply all this to points given in terms of t as in the question.
 
  • #6
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This confused me at first.

Hint: What constraints can you place on the tangent plane to f at (1, 1)?
Hint: What information about f is that statement about g giving you?
I'm still very confused. By 2) I know that f has a relative minimum at the point (1,1) and by 1) I also know that f(x,y)=0 at (1,1) because this point belongs to the circumference. Could I say that ∇f(1,1) is perpendicular to the tangent plane at the point (1,1)?. If this is true, then for all (x,y) belonging to the level set f(x,y)=0 (all the points of the circumference), the tangent plane is given by the equation ∇f(1,1).(x-1,y-1)=0. From here I don't know what to do, because if I put in (x,y)=(1,1) in the equation above, I trivially get 0=0.
 
  • #7
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Assume, instead of simple differentiability, that f has *continuous* partial derivatives near the circle. The fact that f is constant on the circle introduces a relationship between the partial derivatives ##f_x## and ##f_y## at points on the circle. Rather than spelling out what the relationship IS, I would rather tell you how you could get it for yourself. So, say you start at a point (x,y) = (a,b) on the circle, and you go to another point point (x',y') = (a+u,b+v) on the circle, where u| and |v| are small. Using f(a,b) = 0 and f(a+u,b+v) = 0, expand up to terms linear in u and v. The fact that (a+u,v+b) is also on the circle introduces a relationship between u and v; find this relationship to first order in u and v. Finally, see what first-order condition in u (or v) you must have in order to keep f = 0 at the new point. Then apply all this to points given in terms of t as in the question.
I'm not so sure if this was what you were trying to say to me: I know that f is differentiable, in particular, is differentiable at (1,1), so I can make a linear approximation of f(x,y) at any point "close enough" to (1,1) by L(x,y)= f(1,1) + ∇f(1,1).(x-1,y-1)= ∇f(1,1).(x-1,y-1).

'Finally, see what first-order condition in u (or v) you must have in order to keep f = 0 at the new point. Then apply all this to points given in terms of t as in the question.' I don't know how to go on there. Last thing: Why can I assume that f is C^1? What if f doesn't have continuous partial derivatives for points near the circumference? Thanks for the help.
 
  • #8
verty
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I'm still very confused. By 2) I know that f has a relative minimum at the point (1,1) and by 1) I also know that f(x,y)=0 at (1,1) because this point belongs to the circumference. Could I say that ∇f(1,1) is perpendicular to the tangent plane at the point (1,1)?. If this is true, then for all (x,y) belonging to the level set f(x,y)=0 (all the points of the circumference), the tangent plane is given by the equation ∇f(1,1).(x-1,y-1)=0. From here I don't know what to do, because if I put in (x,y)=(1,1) in the equation above, I trivially get 0=0.
The gradient is not normal to the tangent plane. Conceptually, this is wrong.
 
  • #9
Ray Vickson
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The gradient is not normal to the tangent plane. Conceptually, this is wrong.
NO, that is not correct. For a C^1 function, if the gradient is non-zero it is, indeed, perpendicular to the tangent plane. If the gradient is zero the whole concept of tangent plane may be problematical. I am not sure exactly what happens if the function is just differentiable but not C^1.
 
  • #10
verty
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NO, that is not correct. For a C^1 function, if the gradient is non-zero it is, indeed, perpendicular to the tangent plane. If the gradient is zero the whole concept of tangent plane may be problematical. I am not sure exactly what happens if the function is just differentiable but not C^1.
Sorry, I was trying to avoid a lengthy explanation but what I ended up saying was probably unhelpful.

Let f(x,y) = x^2 + y^2, then ∇f = <2x, 2y>. The level curves of this function are circles and the gradient at a point is perpendicular to the level curve through that point. But the gradient at that point is not perpendicular to the tangent plane at that point. The difference is that the gradient is a vector in the xy-plane, while the tangent plane is in 3-space. Geometrically speaking, the tangent to the level curve is the intersection of the tangent plane and a horizontal plane through the point in question.

By saying it was conceptually wrong, I meant that the tangent plane has an extra dimension, an extra degree of freedom.

Ray, I suspect you were thinking of the higher dimensional case of level surfaces. A level surface would of course have a tangent plane to which the gradient would be perpendicular.
 
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  • #11
Ray Vickson
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Sorry, I was trying to avoid a lengthy explanation but what I ended up saying was probably unhelpful.

Let f(x,y) = x^2 + y^2, then ∇f = <2x, 2y>. The level curves of this function are circles and the gradient at a point is perpendicular to the level curve through that point. But the gradient at that point is not perpendicular to the tangent plane at that point. The difference is that the gradient is a vector in the xy-plane, while the tangent plane is in 3-space. Geometrically speaking, the tangent to the level curve is the intersection of the tangent plane and a horizontal plane through the point in question.

By saying it was conceptually wrong, I meant that the tangent plane has an extra dimension, an extra degree of freedom.

Ray, I suspect you were thinking of the higher dimensional case of level surfaces. A level surface would of course have a tangent plane to which the gradient would be perpendicular.
I was thinking of "tangent line" in 2D. In general, the gradient of ##f(x_1,x_2,\ldots,x_n)## is orthogonal to the tangent hyperplane of the level surface of f. Note that the surface f = c and the tangent hyperplane all live in ##R^n.##
 
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