Multivariable calculus: find the rate of change

[reminiscent]

The problem is:
The temperature (in degrees Celsius) of a metal plate, located in the xy -plane, at any point (x, y ) is given by the function of two variables T(x, y ) = x sin y + y² sin x.
(a) Find the rate of change in temperature in the direction of the positive x-axis at the point (π, π).
(b) Find the rate of change in temperature in the direction of the positive y -axis at the point (π, π).

I am thinking that you would have to find the gradient vector of T(x,y), then plug the point (π, π). But the phrase "direction of the positive x-axis/y-axis at the point" is throwing me off. Does that mean you would have to find the directional derivative instead? So find the gradient vector then multiply it by a unit vector, but what is the unit vector?

Thanks.

 

[reminiscent]

Homework Statement

The temperature (in degrees Celsius) of a metal plate, located in the xy -plane, at any point (x, y ) is given by the function of two variables T(x, y ) = xsiny + y²sin x.
(a) Find the rate of change in temperature in the direction of the positive x-axis at the point (π, π).
(b) Find the rate of change in temperature in the direction of the positive y -axis at the point (π, π).

Homework Equations

∇T = <t_x, t_y>

The Attempt at a Solution

I am thinking that you would have to find the gradient vector of T(x,y), then plug the point (π, π). But the phrase "direction of the positive x-axis/y-axis at the point" is throwing me off. Does that mean you would have to find the directional derivative instead? So find the gradient vector then multiply it by a unit vector, but what is the unit vector?

Thanks.

 

[Aakash Gupta]

The gradient of the scalar T(x,y) is given by
∇T = <tx, ty> = <(∂T/∂x), (∂T/∂y)>
Here (∂T/∂x) represents the directional derivative of T w.r.t x
and (∂T/∂y) represents the directional derivative of T w.r.t y
The Directional derivative means rate of change of a quantity in a particular direction, so you can find gradient of T and need not to multiply by any unit vector, because it won't represent the true rate of change then(as its magnitude will be 1 then)
Hope this helps

 

[beamie564]

Yes, you're right, you need to find the directional derivative. You want the change in the x-direction, so what would be the unit vector in this direction?

 

[reminiscent]

Yes, you're right, you need to find the directional derivative. You want the change in the x-direction, so what would be the unit vector in this direction?

Would it be <1,0>? Do I even need a unit vector for this?

 

[reminiscent]

The gradient of the scalar T(x,y) is given by
∇T = <tx, ty> = <(∂T/∂x), (∂T/∂y)>
Here (∂T/∂x) represents the directional derivative of T w.r.t x
and (∂T/∂y) represents the directional derivative of T w.r.t y
The Directional derivative means rate of change of a quantity in a particular direction, so you can find gradient of T and need not to multiply by any unit vector, because it won't represent the true rate of change then(as its magnitude will be 1 then)
Hope this helps

So a) and b) are pretty much asking for the directional derivative of T w.r.t the variable, so a) is w.r.t x and b) is w.r.t y? It's that simple?

 

[beamie564]

Would it be <1,0>? Do I even need a unit vector for this?

Yes that's the unit vector you want. Here you don't actually NEED a unit vector, since the required direction being along the x-axis you can simply take the x component of the gradient. ( but can you see that you are taking the dot product of the unit vector and the gradient vector?)

 

[Aakash Gupta]

So a) and b) are pretty much asking for the directional derivative of T w.r.t the variable, so a) is w.r.t x and b) is w.r.t y? It's that simple?

yes!

 

[reminiscent]

Thanks guys! :D