# Homework Help: Change in Temperature of a 2-D Temperature Field

1. Jan 24, 2010

### ColdFusion85

1. The problem statement, all variables and given/known data
A two-dimensional temperature field is given by the expression $$T=4x^2y^2+3y^3$$
(a) What is the magnitude and direction of the temperature gradient at the point (2,3)?
(b) What is the change in temperature $$dT/dS$$ at the point (2,3) in a direction along the curve $$3x-4y^2=-30$$ passing through the point (2,3)?

2. Relevant equations
$$T=4x^2y^2+3y^3$$
$$3x-4y^2=-30$$

3. The attempt at a solution

So part (a) is pretty easy. I just found the gradient and evaluated it at the given point. However, I'm stuck on part (b). I think I have to take the dot product of the gradient of T with some unit vector of that curve. However, I don't know how to formulate this dot product. I figure that I first evaluate the gradient of T at (2,3), and then get a unit vector for that curve, evaluate that result at (2,3), then take the dot product to get dT/dS. Is this correct? If so, how do I handle the RHS of the curve? Would I move it over to the LHS and then find the unit vector? Would the 30 have to be squared when finding the magnitude of the curve too? I guess I'm a little confused with how to handle the nonzero RHS.

2. Jan 24, 2010

### Dick

Since you can find x=(4y^2-30)/3, you can write the points (x,y) on the curve as a function only of the parameter y, like ((4y^2-30)/3,y). Now take d/dy to find a tangent. The RHS of the curve disappears when you differentiate.

3. Jan 24, 2010

### ColdFusion85

Sorry, I'm still a little confused. I see how you got x=(), but then are you saying I differentiate that equation, x(y), or do I plug that into T for x, and take grad(T)?

4. Jan 24, 2010

### Dick

((4y^2-30)/3,y) parametrizes the curve in terms of the variable y. If I take d/dy of that vector I get a another vector that is tangent to the curve. You want to find the tangent vector at (2,3) to dot with your gradient at (2,3).

5. Jan 24, 2010

### ColdFusion85

But if I parametrize the equation in terms of only y, there will be no x to plug in the 2 of the (2,3) point. I would have curve=(4y^2-30)/3 i + 1 j. Differentiating that would give me (8/3)y i evaluated at (2,3).

6. Jan 24, 2010

### Dick

No, the curve is i*(4y^2-30)/3+j*y. You get (8/3)yi+j for the derivative. You don't need to plug x=2 into anything. If you know y, then you know x. Just plug in y=3. That gives you 2i+3j or (2,3) on the original curve.

7. Jan 24, 2010

### ColdFusion85

x=(4y^2-30)/3*i
y=y*j

So, S=(4y^2-30)/3*i + y*j

dS=(8/3)y*i+1*j

eval @ (2,3),

dS=8*i+1*j

So, I do (8xy^2*i + (8x^2y+9y^2)*j) DOT (8*i + 1*j) to get dT/dS

Is this correct?

8. Jan 24, 2010

### Dick

I thought you were planning on normalizing the tangent vector so it was a unit vector.

9. Jan 24, 2010

### ColdFusion85

Oh,right. So it would be 8/sqrt(65)*i + 1/sqrt(65)*j

Then dot grad(T) with this, and eval @ (2,3) to get dT/dS = 1152/sqrt(65)*i + 177/sqrt(65)*j?

Is this correct?

10. Jan 24, 2010

### Dick

dT/dS isn't a vector, is it? It's a real number. It's the dot product of two vectors.

11. Jan 24, 2010

### ColdFusion85

Oh yeah, that's what I meant to do. Thanks for your help.