Implicit differentiation

[-TesseracT-]

w(t) = arctan(5/L(t)). I want to calcuate w'(0). So I use implicit differentiation to get to: w'(t) = -5 * L'(t) / ( L(t) * w(t) * (1+ 25/L(t)^2)) I know that L(0) = 15, L'(0) = 600 and w(0) = arctan(1/3). When i put this in, I get that w'(0) = -559. The right answer (according to book) is w'(0) = 12. Have I done the differentiation wrong or something else? Thanks in advance. 

[Dash Lambda]

Neither w(t) nor L(t) is a function of the other, they're both functions of t. You just need to take the t derivative.

Though the answer is negative, so it looks like the book did make an error.

[colonel_mortis]

The left you've got, d/dt(w(t)) = w'(t)

For the right, you made a slight mistake:

d/dt (arctan(5/L(t))) = d/dp (arctan(p)) * d/dq (5/q) * d/dt (L(t))  (by chain rule; p = 5/L(t); q = L(t))

 = 1/(1+p²) * -5q^-2 * L'(t)

 = 1/(1+(5/L(t))²) * -5L(t)^-2 * L'(t)  (You got to here)

 = -5 * L'(t) / ((1 + 25 * L(t)^-2) * L(t)²) (You forgot that last term)

 = -5 * L'(t) / (25 + L(t)²)

This gives w(0) = -12, which is almost what the book says, so either I made a small mistake somewhere, you made a slight mistake transcribing into the post, or the book missed out the -.

[-TesseracT-]

41 minutes ago, Dash Lambda said:

Neither w(t) nor L(t) is a function of the other, they're both functions of t. You just need to take the t derivative.

Though the answer is negative, so it looks like the book did make an error.

 

9 minutes ago, colonel_mortis said:

The left you've got, d/dt(w(t)) = w'(t)

For the right, you made a slight mistake:

d/dt (arctan(5/L(t))) = d/dp (arctan(p)) * d/dq (5/q) * d/dt (L(t))  (by chain rule; p = 5/L(t); q = L(t))

 = 1/(1+p²) * -5q^-2 * L'(t)

 = 1/(1+(5/L(t))²) * -5L(t)^-2 * L'(t)  (You got to here)

 = -5 * L'(t) / ((1 + 25 * L(t)^-2) * L(t)²) (You forgot that last term)

 = -5 * L'(t) / (25 + L(t)²)

This gives w(0) = -12, which is almost what the book says, so either I made a small mistake somewhere, you made a slight mistake transcribing into the post, or the book missed out the -.

Roger that, thanks a lot for replies. I wrote it incorrectly, the book does say -12.

[-TesseracT-]

1 hour ago, Dash Lambda said:

Neither w(t) nor L(t) is a function of the other, they're both functions of t. You just need to take the t derivative.

Though the answer is negative, so it looks like the book did make an error.

When exactly do I need to put that for example f'(x) = f'(x)*f(x) instead of "just" f'(x)?

[Crossbred]

10 minutes ago, -TesseracT- said:

When exactly do I need to put that for example f'(x) = f'(x)*f(x) instead of "just" f'(x)?

Usually it’s formatted as dy/dx instead of f-prime, and then solved for dy/dx. (At least I learned it that way...)

[Dash Lambda]

1 hour ago, -TesseracT- said:

When exactly do I need to put that for example f'(x) = f'(x)*f(x) instead of "just" f'(x)?

It's the chain rule. Let's say you have two functions, g(t) and f(g(t)). If you want to take the t derivative of f, it goes like this: df/dt = (df/dg)(dg/dt) = g'(t)f'(g(t))

 

In your example, you have g(L(t))=5/L(t) and f(g((L(t)))=tan^-1(g(L(t))). Thus the derivative goes like this: fd/ft = (df/dg)(dg/dt) = (df/dg)(dg/dL)(dL/dt) = L'g'f'

 

50 minutes ago, Crossbred said:

Usually it’s formatted as dy/dx instead of f-prime, and then solved for dy/dx. (At least I learned it that way...)

When convenient. You really can't stick to one notation because sometimes one works best and other times others work best.

As for the primed notation, I've used that way more than the du/dt form. The latter gets really tiring. When you get into partial differentials, you also start using a subscript notation -The derivative of u with respect to t is u_t.

[Crossbred]

4 hours ago, Dash Lambda said:

Snip

We’re just finishing the implicit/logarithmic differentiation unit in my class, I’ve no context for anything later.