Chain rule: y = f(t, t^2, t^3) and y = g(t, h(t), k(t^2))

[939]

Homework Statement

I am confused because for each problem there is no equation and for one no intermediate variables.

Compute dy/dt when

a) y = f(t, t^2, t^3)
b) y = g(t, h(t), k(t^2))

Homework Equations

a) y = f(t, t^2, t^3)
b) y = g(t, h(t), k(t^2))

The Attempt at a Solution

a)

dy/dt = ∂f/∂t * ∂f/∂t^2 * ∂f/∂t^3


2)

dy/dt = ∂f/∂t + (∂g/∂h * dh/dt) + (∂g/∂k * dk/dt^2)

 

[LCKurtz]

Homework Statement

I am confused because for each problem there is no equation and for one no intermediate variables.

Compute dy/dt when

a) y = f(t, t^2, t^3)
b) y = g(t, h(t), k(t^2))

Homework Equations

a) y = f(t, t^2, t^3)
b) y = g(t, h(t), k(t^2))

The Attempt at a Solution

a)
View attachment 68161
dy/dt = ∂f/∂t * ∂f/∂t^2 * ∂f/∂t^3


2)View attachment 68162

dy/dt = ∂f/∂t + (∂g/∂h * dh/dt) + (∂g/∂k * dk/dt^2)

There are intermediate variables, they are just implied so you don't see them. For a) think of ##u = t,~v = t^2,~w=t^3## and ##y = f(u,v,w)##. Now do you see how to calculate ##\frac{dy}{dt}## using the chain rule?

 

[939]

There are intermediate variables, they are just implied so you don't see them. For a) think of ##u = t,~v = t^2,~w=t^3## and ##y = f(u,v,w)##. Now do you see how to calculate ##\frac{dy}{dt}## using the chain rule?

Does this work?

dy/dt = (∂y/∂u * du/dt) + (∂y/∂v * du/dt^2) + (∂y/∂w * dw/dt^3)

 

[LCKurtz]

There are intermediate variables, they are just implied so you don't see them. For a) think of ##u = t,~v = t^2,~w=t^3## and ##y = f(u,v,w)##. Now do you see how to calculate ##\frac{dy}{dt}## using the chain rule?

Does this work?

dy/dt = (∂y/∂u * du/dt) + (∂y/∂v * du/dt^2) + (∂y/∂w * dw/dt^3)

Look up that chain rule in your calculus book. Is that what it says?

 

[939]

Look up that chain rule in your calculus book. Is that what it says?

Sorry... If it was i.e. y = (x, y), x = (t), y = (t^2) I would get it and there my book says you do the same steps as I listed above.

I'm just not sure here... y = f(t, t^2, t^3). Thus, y = dependent. t, t^2, t^3 are independent. Intermediate variables must connect the two... If that is not it, maybe just find derivatives f t, t^2 and t^3 in the equation?

 

[Mark44]

Sorry... If it was i.e. y = (x, y), x = (t), y = (t^2) I would get it and there my book says you do the same steps as I listed above.

I'm just not sure here... y = f(t, t^2, t^3). Thus, y = dependent. t, t^2, t^3 are independent. Intermediate variables must connect the two... If that is not it, maybe just find derivatives f t, t^2 and t^3 in the equation?

Yes. Using LCKurtz's suggestion these would be du/dt, dv/dt, and dw/dt.

 

[LCKurtz]

Does this work?
View attachment 68164
dy/dt = (∂y/∂u * du/dt) + (∂y/∂v * du/dt^2) + (∂y/∂w * dw/dt^3)

Sorry... If it was i.e. y = (x, y), x = (t), y = (t^2) I would get it and there my book says you do the same steps as I listed above.

I'm just not sure here... y = f(t, t^2, t^3). Thus, y = dependent. t, t^2, t^3 are independent. Intermediate variables must connect the two... If that is not it, maybe just find derivatives f t, t^2 and t^3 in the equation?

You likely calculated them correctly but I was pointing out your notation was bad. Those two red factors should be dv/dt and dw/dt. The u was probably a typo but you shouldn't have t^2 and t^3 in the formulas.