# What is the difference between these two problems?

## Homework Statement

I have a problem from thomas' calculus. the topic is Tangent planes and diffrentials. The problem is as follows: the celcus temparature in a region in space is given by T(x,y,z)= 2*x^2-xyz; a particle is moving in the region and it's position at time t is given by x=2*t^t ; y=3*t; z=-t^2, where time is measured in seconds and distance in meters. then how fast is the temparature experienced by the particle changing in degrees per meter when the particle is at the point P(8,6,-4)?

ther was another problem just above this problem. there says ; suppose that the celcius temparature at the point (x,y) in the xy-plane is T(x,y)=x-sin(2y) and the distance is in meters. a particle is moving clockwise aroun the circle of radius 1m centered at te origin at the constant rate of 2m/sec.
and it asks : how fast is the temparature experienced by the particle changing in degrees per meter when the particle is at the point
P(1/2,sqrt(3)/2) ?
I have sold the second queston as folows. I find the direction of the particle when iti s at the poin P. then I took he directional derivative at this point. However, in the first question how can ı find the direction o the particle?

Dick
Homework Helper
You could take the directional derivative along the tangent vector to the curve (x',y',z'). But why bother? If you put x, y and z into the expression for T, you can get T as a function of t. Then just take the derivative.

" You could take the directional derivative along the tangent vector to the curve (x',y',z'). "
how can I do that? I mean how can ı find a directon ,from the given equations, that is like xi+yj+zk ?

Dick
Homework Helper
The tangent vector is i*dx/dt+j*dy/dt+k*dz/dt. The length of the tangent vector is the speed. A direction vector pointing in that direction is tangent/speed. But as I said, you could also do it as dT/dt/speed. Your choice. Does that help?

Last edited:
The tangent direction is i*dx/dt+j*dy/dt+k*dz/dt. Its length is the speed. A direction vector pointing in that direction is tangent/speed. But as I said, you could also do it as dT/dt/speed. Your choice. Does that help?

yes, it helped a lot. thank you very much.

HallsofIvy
Homework Helper
This is more a "chain rule" problem than "tangents and differentials".

$$\frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}$$

Dick
Homework Helper
This is more a "chain rule" problem than "tangents and differentials".

$$\frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}$$

He's not actually looking for dT/dt. He's looking for dT/ds, where s is arclength along the curve.

This is more a "chain rule" problem than "tangents and differentials".

$$\frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}$$

yes you make a good point. And this is actually my question? If you look to the two problems, which I wrote on the top of the page, they are same(at least i tink so). However, in the first part we have to take a directional derivative. then why do we here have to take a derivative with respect to t? I mean what is te diffference between these two problems_

Dick
Homework Helper
There is no real difference between the problems. If they had given you a particle moving along x=cos(2*t), y=sin(2*t) and asked for the rate at t=pi/6 in the second problem, they would look even more alike. But you would have gotten the same answer.

There is no real difference between the problems. If they had given you a particle moving along x=cos(2*t), y=sin(2*t) and asked for the rate at t=pi/6 in the second problem, they would look even more alike. But you would have gotten the same answer.

last question. In the first question i find the direction vector in a geometric way (I draw a circle with radius 1...) . Is there a algebratic way to obtain the position vector which is there sqrt(3)i - j .

Dick
Homework Helper
last question. In the first question i find the direction vector in a geometric way (I draw a circle with radius 1...) . Is there a algebratic way to obtain the position vector which is there sqrt(3)i - j .

Well, that was the idea of what I just sent you. If I write a parametric equation for the circle (like x=cos(2*t), y=sin(2*t)) then you can find the direction vector by differentiating in the same way as you did the first problem.

Well, that was the idea of what I just sent you. If I write a parametric equation for the circle (like x=cos(2*t), y=sin(2*t)) then you can find the direction vector by differentiating in the same way as you did the first problem.

ok, i think ı understood you. when we take the dot product of the gradient f and the directional vector (which is 4ti+3j-2tk in this case) we obtain the change per meter. However in the second part of the question it asks the changing in degrees celcius per second. Since I do not know the velocity of the particle (or do ı know it?) how can ı solve this part?

ok i understood now. As you stated before "the length of the tangent vector is the speed" then in our case the speed is sqrt(89). And then the answer is velocity(Celcius per meter)* sqrt(89).
Soory for bothering you. you have helped me to understand the concept of the partial derivatives. thanks again...