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.