Determine the highest and lowest elevation on a path

[Miley4567]

Homework Statement

Determine the highest and lowest elevations given by the height
z = f(x,y) = 1 - (1/16)x^2 - (1/9)y^2
on the path r(t) = <2cos(t), 3sin(t)>. The xy position on the path at time t is given by r(t).

Homework Equations

Lagrange Multipliers
Partial derivatives

The Attempt at a Solution

I was going to try using the Lagrange Multiplier method to find the extreme values of the function f(x,y), but I would need the xy position to be given as an equation and in x and y terms. I am confused as to how to do this because we are given a position vector instead, in terms of t.

Could anyone help me out and guide me with how to do this?

 

[Mark44]

Homework Statement

Determine the highest and lowest elevations given by the height
z = f(x,y) = 1 - (1/16)x^2 - (1/9)y^2
on the path r(t) = <2cos(t), 3sin(t)>. The xy position on the path at time t is given by r(t).

Homework Equations

Lagrange Multipliers
Partial derivatives

The Attempt at a Solution

I was going to try using the Lagrange Multiplier method to find the extreme values of the function f(x,y), but I would need the xy position to be given as an equation and in x and y terms. I am confused as to how to do this because we are given a position vector instead, in terms of t.

Could anyone help me out and guide me with how to do this?

On the path, x(t) = 2cos(t), and y(t) = 3sin(t). So you could write f(x, y) as a function of t alone, and then use the derivative to find the maximum and minimum values. That's how I would approach this problem.

 

[Simon Bridge]

Before choosing the method, understand the problem.
The path is being given to you as a position vector that varies with t. What shape is the path?
As you travel along that path, how does z vary?

[edit] mark44 beat me to it.

 

[Miley4567]

On the path, x(t) = 2cos(t), and y(t) = 3sin(t). So you could write f(x, y) as a function of t alone, and then use the derivative to find the maximum and minimum values. That's how I would approach this problem.

That sounds like a good idea, I should have thought of that. But then I am confused as to how to turn the position vector into a function that I can use to find extreme values (min/max).

 

[LCKurtz]

That sounds like a good idea, I should have thought of that. But then I am confused as to how to turn the position vector into a function that I can use to find extreme values (min/max).

Here's what Mark44 said above:

On the path, x(t) = 2cos(t), and y(t) = 3sin(t). So you could write f(x, y) as a function of t alone, and then use the derivative to find the maximum and minimum values. That's how I would approach this problem.

Try that.

 

[Simon Bridge]

That sounds like a good idea, I should have thought of that. But then I am confused as to how to turn the position vector into a function that I can use to find extreme values (min/max).

If you are having trouble understanding mark44's suggestion, try the focussing questions in post #3.
It is very difficult to tell you more detail without effectively telling you the answer too.

 

[Miley4567]

That sounds like a good idea, I should have thought of that. But then I am confused as to how to turn the position vector into a function that I can use to find extreme values (min/max).

Here's what Mark44 said above:Try that.

Ya I get that, but to use Lagrange Multipliers, I need two sets of equations, and plugging in the 2cost and 3sint into the f(x,y) equation still only leaves me with one equation to use to find a max and min.

 

[Simon Bridge]

So don't use Lagrange multipliers... have you heard of the 1st and second derivative tests?

 

[Ray Vickson]

Ya I get that, but to use Lagrange Multipliers, I need two sets of equations, and plugging in the 2cost and 3sint into the f(x,y) equation still only leaves me with one equation to use to find a max and min.

I don't see why you would want or need to use Lagrange multipliers in this problem. Lagrange multipliers are designed to deal with certain types of problems, and this is not one of them.