How Does a Bug's Temperature Reading Change as It Moves Along a Curve?

[HerpMcDerp]

Homework Statement

[12] 3.Thetemperature T(x, y) at a point of the xy-plane is given by
T(x,y)= ye^(x^2).
A bug travels from left to right along the curve y = x^2
at a speed of 0.01m/sec. The bug
monitors T(x, y) continuously. What is the rate of change of T as the bug passes through
the point (1, 1)?

Homework Equations

Parameterizing x and y in terms of t, taking into account the velocity given (and assuming x and y are in meters, t is in seconds):

x = 0.01t
y = (0.01t)^2
t = 100 s

Chain rule
dT/dt = T_x dx/dt + T_y dy/dt

The Attempt at a Solution

dT/dt = 2xye^(x^2) * 0.01 + e^(x^2) * 0.02(0.01t) = 2*1*1*e * 0.01 + 0.02*e * 1 =
0.04*e degrees/sec

Right? Lol.

 

[SammyS]

For that parametrization, the bug's speed is 0.01m/sec only at the origin.

So the answer is likely incorrect.

 

[HallsofIvy]

You have the horizontal component of the bug's speed equal to 0.01 m/s, not its speed. If the bugs horizontal position is given by u(t), and its vertical position by v(t), then its speed is given by [itex]\sqrt{(u')^2+ (v')^2}= 0.01[/tex]

 

[HerpMcDerp]

Hey thanks for the replies... the horizontal component eh, so if it's traveling at 0.01m/s from left to right then x = u(t) = 0.01t, and then if y = x^2, so wouldn't y = (0.01t)^2 then? How do you arrive at sqrt{(u')^2+ (v')^2}= 0.01? Wouldn't the bug be at (1,1) at 100 seconds given the parametrizations?