# Tough derivative problem

1. Feb 19, 2006

### jesuslovesu

whoops

k = 3/2*w
y=1/w^2 <- w is the x component of y = 1/x^2

w is increasing at a constant rate of 7 units per second. When w = 5, what is the rate of change of k with respect to time?
dw/dt = 7

I'm not sure but can I do this? It seems like this is wrong or that I'm skipping a step.
dk/dt = 3/2 * dw/dt
dk/dt = 10.5 ??

Last edited: Feb 19, 2006
2. Feb 19, 2006

### arildno

w=1/w^2 means that the only real number w can be is 1.

3. Feb 19, 2006

### benorin

Your missing k now: Post the problem as given if need be, or the source if it is a commom text.

4. Feb 19, 2006

### HallsofIvy

Staff Emeritus
"w is the x component of y= 1/x^2" doesn't make much sense. I think you just mean y= 1/w^2. In fact, I don't see reason to mention y or x. I also assume you mean k= (3/2)w rather than k= 3/(2w)

dk/dt= (dw/dk)(dk/dt)

dw/dk= (3/2) and you are told that dw/dt= 7 so dk/dt= (3/2)(7)= 10.5 just as you say.

But are you sure you've copied the problem correctly? There is no need to know that w= 5 if dk/dw is a constant.

5. Feb 19, 2006

### jesuslovesu

Thanks for all your help... I'm pretty sure that it must be 10.5 now, I don't see how any other relationship with time could be formed, although the w = 5 is a bit suspicious.

Here's the full picture, sorry I should have posted it first but I thought I could condense it.

6. Feb 19, 2006

### benorin

it seems that $$\ell$$ is the tangent line to the curve y=1/w^2 at the point w, and k is the x-intercept thereof. Hence

$$\ell : y-\frac{1}{w^2}=-\frac{2}{w^3}\frac{dw}{dt}(x-w)$$

is the equation of the tangent line.

7. Feb 19, 2006

### benorin

the x-intercept of the above line is k.

Since w=5 and $$\frac{dw}{dt}=7\frac{units}{sec}$$ we have...

Can you get it from here?

8. Feb 19, 2006

### jesuslovesu

Hmm, I see where you got the equation for the line but now I'm a bit confused...

Would I use the equation of the line to find k and then take the derivative of k?( i get 7.5 that way as opposed to 10.5)

Last edited: Feb 19, 2006
9. Feb 20, 2006

### VietDao29

If you are a little bit confused, then you should check it up in your textbook, there should be a part that says:
The equation of the tangent line of the function f(x) at the point (x0, f(x0)) is:
y - f(x0) = f'(x0) (x - x0)
Yes, you need to express k in terms of w, then use the chain rule:
$$\frac{dk}{dt} = \frac{dk}{dw} \times \frac{dw}{dt}$$
You were told that:
$$\frac{dw}{dt} = 5$$, so what you have to do now is to find k in terms of w, then differentiate k with respect to w to get:
$$\frac{dk}{dw}$$
Do you know how to find k in terms of w?
Hint: k is the x-intercept of the tangent line at the point P(w, 1 / w2). And the tangent line is:
$$y - \frac{1}{w ^ 2} = \frac{-2}{w ^ 3} (x - w)$$
--------------
$$k'_t(5) = k'_w(5) \times w'_t(5) = 7 k'_w(5) = ?$$
Can you go from here? :)
--------------
@benorin:
I don't really understand this.
$$\ell : y-\frac{1}{w^2}=-\frac{2}{w^3}\frac{dw}{dt}(x-w)$$
Why is there dw / dt in the tangent line equation???
Am I missing something?

Last edited: Feb 20, 2006
10. Feb 21, 2006

### benorin

No, you're not missing anything, I just goofed. It should be:

$$\ell : y-\frac{1}{w^2}=-\frac{2}{w^3}(x-w)$$

as you pointed out.

11. Feb 21, 2006

### benorin

related rates

To finish setting-up the related rates problem:

note that since k is the x-intercept of $$\ell$$, one may put y=0 and x=k into the above equation of the tangent line (a.k.a. $$\ell$$,) to get the following implicit relationship between w and k:

$$0-\frac{1}{w^2}=-\frac{2}{w^3}(k-w)$$

multiply by $-w^3$ to get

$$w=2(k-w)$$

which simplifies to... eh, you'll get it from here.