Question

The temperature at a point (x,y) is T(x,y), measured in degrees C. A bug crawls so that its position after t seconds is given by x = sqrt(1 + t), y = 2 + 1/3 t, where x and y are measured in centimeters. The temperature function satisfies Tx(2,3) = 4 and Ty(2,3) = 3. How fast is the temperature rising on the bugs path after 3 seconds.

Answer 1

I think im having a conceptual block on this one. If I take dx/dt this gives me the rate of change in position with respect to time, but im not sure how to realate that to the temp change.

Answer 2

T' (3) = Tx x' (3) + Ty y' (3)

Answer 3

Im still a little confused. Could you elaborate a little on that. (please :) )

Answer 4

Find the instantaneous rate at position (x,y) after 3 secs:
x = √(1 + t)
-> x' = 1/ 2 √(1 + t)
=>x' (3) = 1/4
Similarly for y:

Answer 5

ok thats where I was going and found 1/4 but what does it represent. does that mean the temp increases 1/4 degrees in three seconds in the x direction?

Answer 6

Did you find y' (3) ?

Answer 7

its just 1/3 right?

Answer 8

oh wait...bah

Answer 9

Then plug into the equation to obtain T' !

Answer 10

But I don't have the equation for T' Do you mean T' = Tx + Ty ?

Answer 11

Ok, I think I get it now. I was so confused trying to solve for Tx and Ty, but the values are given.

So its just T'(3) = 4 (1/4) + 3 (1/3)

T'(3) = 2. Read the temperature in the path increases 2 degrees after 3 seconds. Is this correct?

Answer 12

Bingo =)

Answer 13

Thank you. I need to get my brain duster out of the closet I think.

Answer 14

....Should probably bring a ShopVac.....

Answer 15

My honorable pleasure =D

Answer 16

This can only end with a black hole joke.... Thank you again.

Answer 17

Been a while since I've been back and things have changes...do I just mark this as correct and then close it?

Answer 18

Yes, if you want ask another question, then close this post!