Question

Suppose that the celsius temperature at the point (x,y) in the xy-plane is T(x,y)=xsin(2y) and that distance in the xy-plane is measured in meters. A particle is moving clockwise around the circle of radius 1 m centered at the origin at a constant rate of 2m/sec.

A) how fast is the temp experienced by the particle changing in degress cel per meter at the point: P(1/2, sqrt(3)/2)

* I don't need the answer just how to find the unit vector, which I know is 13 years ago

Answer 1

Are you studying parametric equations?

Answer 2

multivar calc

Answer 3

so, parametric eq are fair game

Answer 4

not sure what unit vector you're talking about, but it's pretty quick:
express the circle in terms of x(t) and y(t).
plug those into T(x,y)
take the derivative with respect to time
plug in the time that corresponds to (1/2, sqrt(3)/2)

Answer 5

I guess maybe you're talking about the direction of velocity...?

Answer 6

I'm trying to solve it using gradients, I apologize, this is probably very obvious, but my ability to visualize these types of problems is very limited

Answer 7

I have a solutions manual, and everything makes sense except for the very first step in which it says simply: "The unit tangent vector @ (1/2, Sqrt(3)/2) in the direction of motion is:
u= (sqrt(3)/2i-1/2j)"
I just can't see how they calculated that.

Answer 8

if you're asking about the direction of velocity at the point (1/2, sqrt(3)/2)
then plug those coord.s into the equations of motion I gave:
1/2 = cos(2*t)
sqrt(3) /2 = sin(-2*t)
find t
find the time derivative of
x(t), y(t)
(-2sin(2*t) , -2cos(2*t) )
plug in the t you found to find the velocity vector at that point :)

Answer 9

that's how they calculated it.

Answer 10

Thanks!

Answer 11

sure, don't forget that the vector you get will need to be divided by magnitude to get a unit vector... (the vector you get is just <Vx, Vy> )

Answer 12

Got it, thanks again.