Chain rule and partial derivatives

[nhartung]

Homework Statement

Suppose the differentiable function f(x,y,z) has the partial derivatives f_x(1,0,1) = 4, f_y(1,0,1) = 1 and f_z(1,0,1) = 0. Find g'(0) if g(t) = f(t² + 1, t²-t, t+1).

Homework Equations

The Attempt at a Solution

Ok I'm given the solution for this and I'm trying to work through it but I'm confused.

I understand we have g(t) = f(x(t),y(t),z(t)) where x = t²+1; y = t²-t; z = t+1.

So i thought I would just use the chain rule to find g'(t) and plug in 0 for t. But I check the solutions sheet and he uses a very different (easier) method that I don't understand. ( I just realized that we aren't given a function f in terms of x y or z so that's why this wouldn't work and we need another method. Still I don't understand this other method.)

First he says when t=0 (x,y,z)|_t=0 = (1,0,0) (I think he may have made a mistake, shouldn't it be (1,0,1) ?) And then he gets a point P = (1,-1,0) No idea were this point comes from...

Next he sets up the chain rule equation: $$\frac{dg}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}$$

Now I think he somehow uses that point and the partial derivatives given above to solve the equation and he gets: = 1*0 + 1*(-1) + (0*1) = -1 (depending on whether P was correct above may change the answer here I think)

So could someone please explain the method used here?

 

[Mark44]

Homework Statement

Suppose the differentiable function f(x,y,z) has the partial derivatives f_x(1,0,1) = 4, f_y(1,0,1) = 1 and f_z(1,0,1) = 0. Find g'(0) if g(t) = f(t² + 1, t²-t, t+1).

Homework Equations

The Attempt at a Solution

Ok I'm given the solution for this and I'm trying to work through it but I'm confused.

I understand we have g(t) = f(x(t),y(t),z(t)) where x = t²+1; y = t²-t; z = t+1.

So i thought I would just use the chain rule to find g'(t) and plug in 0 for t. But I check the solutions sheet and he uses a very different (easier) method that I don't understand. ( I just realized that we aren't given a function f in terms of x y or z so that's why this wouldn't work and we need another method. Still I don't understand this other method.)

First he says when t=0 (x,y,z)|_t=0 = (1,0,0) (I think he may have made a mistake, shouldn't it be (1,0,1) ?) And then he gets a point P = (1,-1,0) No idea were this point comes from...

Next he sets up the chain rule equation: $$\frac{dg}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}$$

Another way to write this is g'(t) = f_x * x'(t) + f_y * y'(t) + f_z * z'(t), and what you want is g'(0). When t = 0, what are x(0), y(0), and z(0)? You need to evaluate all three partial deriviatives at the point (x, y, z) for which t = 0, and you need to evaluate all three ordinary derivatives at t = 0.

Now I think he somehow uses that point and the partial derivatives given above to solve the equation and he gets: = 1*0 + 1*(-1) + (0*1) = -1 (depending on whether P was correct above may change the answer here I think)

So could someone please explain the method used here?

 

[nhartung]

Another way to write this is g'(t) = f_x * x'(t) + f_y * y'(t) + f_z * z'(t), and what you want is g'(0). When t = 0, what are x(0), y(0), and z(0)? You need to evaluate all three partial deriviatives at the point (x, y, z) for which t = 0, and you need to evaluate all three ordinary derivatives at t = 0.

Ah, so then x'(0) = 2t = 2(0) = 0
y'(0) = 2t -1 = -1
z'(0) = 1

Now I understand where those come from but I'm still not sure what you mean by "You need to evaluate all three partial derivatives at the point (x, y, z) for which t = 0".

 

[Mark44]

Ah, so then x'(0) = 2t = 2(0) = 0
y'(0) = 2t -1 = -1
z'(0) = 1

Not quite. You are confusing the derivatives of x(t), y(t), and z(t) with the value of each derivative at a specific value of t. IOW, x'(t) = 2t, but x'(0) = 0, and similarly for y'(t) vs. y'(0) and z'(t) vs. z'(0). In each case, the deriviative is a function, while the derivative evaluated at a particular value of t is just a number. x'(t) is different from x'(0), y'(t) is different from y'(0), and z'(t) is different from z'(0).

The same difference exists between g'(t) and g'(0). Using the chain rule you can write g'(t) = f_x(x,y,z) * x'(t) + f_y(x,y,z) * y'(t) + f_z(x,y,z) * z'(t). Since you aren't given any details about f(x, y, z), it's not possible to calculate the functions f_x(x, y, z), f_y(x, y, z), and f_z(x, y, z). You are, however given the values of these partials at a particular point (1, 0, 1) that corresponds to t = 0.

So g'(0) = f_x(1,0,1) * x'(0) + f_y(1,0,1) * y'(0) + f_z(1,0,1) * z'(0). The only reason I have used the subscript notation (as opposed to the Leibniz notation you used) for the partial derivatives is that it's a little easier to explicitly show that they are to be evaluated at a particular point.

Now I understand where those come from but I'm still not sure what you mean by "You need to evaluate all three partial derivatives at the point (x, y, z) for which t = 0".

 

[nhartung]

Not quite. You are confusing the derivatives of x(t), y(t), and z(t) with the value of each derivative at a specific value of t. IOW, x'(t) = 2t, but x'(0) = 0, and similarly for y'(t) vs. y'(0) and z'(t) vs. z'(0). In each case, the deriviative is a function, while the derivative evaluated at a particular value of t is just a number. x'(t) is different from x'(0), y'(t) is different from y'(0), and z'(t) is different from z'(0).

Isn't this what I did? I think I may have messed up my notation a bit but I have x'(t) = 2t then x'(0) = 0; y'(t) = 2t-1 then y'(0) = -1; z'(t) = 1 then z'(0) = 1;

The same difference exists between g'(t) and g'(0). Using the chain rule you can write g'(t) = f_x(x,y,z) * x'(t) + f_y(x,y,z) * y'(t) + f_z(x,y,z) * z'(t). Since you aren't given any details about f(x, y, z), it's not possible to calculate the functions f_x(x, y, z), f_y(x, y, z), and f_z(x, y, z). You are, however given the values of these partials at a particular point (1, 0, 1) that corresponds to t = 0.

So here since the point (1,0,1) corresponds to g(t) when t is 0 we can use the value of the partial at (1,0,1) as the value of f_x(x,y,z)? Which would mean at (1,0,1) f_x = 4; f_y = 1; f_z = 0.

g'(0) = f_x(1,0,1) * x'(0) + f_y(1,0,1) * y'(0) + f_z(1,0,1) * z'(0). The only reason I have used the subscript notation (as opposed to the Leibniz notation you used) for the partial derivatives is that it's a little easier to explicitly show that they are to be evaluated at a particular point.

So if everything above is correct that gives me: 4*0 + 1*(-1) + 0*1 = -1.

Also, does that mean (if this is all correct) this method can only be used if the point the partials are evaluated at and the value of t are directly related as in this example, correct?

 

[Mark44]

Isn't this what I did? I think I may have messed up my notation a bit but I have x'(t) = 2t then x'(0) = 0; y'(t) = 2t-1 then y'(0) = -1; z'(t) = 1 then z'(0) = 1;

It's not really "messing up the notation a bit." You wrote some things that just plain aren't true. Here's what you wrote that I commented on.

Ah, so then x'(0) = 2t = 2(0) = 0
y'(0) = 2t -1 = -1

x'(0) $\neq$ 2t $\neq$ 0 and
y'(0) $\neq$ 2t - 1 $\neq$ -1

Since you apparently didn't understand the difference between a function of t and the value of a function at a specific number, I assumed that was the reason you were having such difficulties understanding that you needed to evaluated the partial derivatives at the point corresponding to t = 0, namely (1, 0, 1).

So here since the point (1,0,1) corresponds to g(t) when t is 0 we can use the value of the partial at (1,0,1) as the value of f_x(x,y,z)? Which would mean at (1,0,1) f_x = 4; f_y = 1; f_z = 0.

A less roundabout way to say "g(t) when t is 0" is g(0).
Here are the relationships.
x(t) = t² + 1
y(t) = t² - t
z(t) = t + 1

and
x(0) = 1
y(0) = 0
z(0) = 1

We don't know the formulas of any of the three partials, so for example, we can't evaluate f_x(x, y, z) at an arbitrary value of t - which determines the values of x, y, and z, but we are given f_x(1, 0, 1) as being 4.
Similarly, we don't know f_y(x, y, z) or f_z(x, y, z), but we're given the values of these partials when t = 0, which means we know the value of each partial evaluated at (1, 0, 1).

So if everything above is correct that gives me: 4*0 + 1*(-1) + 0*1 = -1.

Yes, but you should write this in context. This is g'(0) = -1.

Also, does that mean (if this is all correct) this method can only be used if the point the partials are evaluated at and the value of t are directly related as in this example, correct?

Correct. If we didn't have the information about the values of the partials for t = 0 (i.e., at (1, 0, 1)), we wouldn't be able to answer the question.

 

[nhartung]

Ok. My professor made several errors in his work on the solutions sheet which made this even more confusing for me. Thanks a lot for your help.

 

[Mark44]

Sure, you're welcome. I'm hopeful that this is less confusing for you now.