# Related rates

1. Feb 22, 2007

### Rasine

a partical moves along the curve y=(17+x^3)^1/2. as it reaches the point (4,9) the y coordinate is increasing at a rate of 4 cm/s. how fast is the x-coordinate of the point changing at that instant?

so dy/dt=4 cm/s

i diffeientate the curve equation with respect to time to get
dy/dt=(3x^2/2(17+x^3)^1/2)dx/dt

i want to find dx/dt so i put 4 cm/s in for dy/dt and 4 for x (given by the point (4,9)) and then i solve for dx/dt, but i am not coming up with the right answer. please tell me what i am doing wrong.

2. Feb 22, 2007

### arildno

What did you get?

3. Feb 22, 2007

### f(x)

dy/dt=d((17+x^3)^1/2)/d(17+x^3) . d(17+x^3)/d(x) . d(x)/dt
4=1/2.(17+64)^1/2 . 48. dx/dt
72/48=dx/dt
or dx/dt=1.5 cm/s is what i expect you should have got.....
I dont understand how to use y=9 ?

4. Feb 22, 2007

### arildno

y=9 is a necessarily given value, due to the function relation.

5. Feb 22, 2007

### f(x)

yeah, but how does one use it? Or is it just given...

6. Feb 22, 2007

### arildno

It was a nice gesture of the exercise maker to provide the actual y-coordinate.
besides, remember that the derivative of y wrt. to x can be written as:
$$\frac{dy}{dx}=\frac{3x^{2}}{2y}$$

7. Feb 22, 2007

### Rasine

i think 9 was just given as to give the point with the x-coordinate .

i got the right answer by the way.