## anonymous one year ago A particle moves along the curve y=√(1+x^3). As it reaches the point (2,3), the y-coordinate is increasing at the rate of 4cm/s. How fast is the x-coordinate of the point changing at this instant?

1. campbell_st

this looks like a related rates question... $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$

2. anonymous

Yeah it is :) hmm... so what are you doing there?

3. campbell_st

ok... so you need to find $\frac{dy}{dx} ~~~given~~~~ y = (1 + x^3)^{\frac{1}{2}}$

4. anonymous

ok... thats $\frac{ 3x^2 }{ 2\sqrt(1+x^3) }$

5. anonymous

right?

6. campbell_st

looks good to me... now the point is (2, 3) so substitute x = 2 and get a value

7. anonymous

That's all i gotta do? :O

8. ganeshie8

i think we need to find $$\dfrac{dx}{dt}$$, the rate of change of x coordinate

9. campbell_st

well that is just a step

10. anonymous

The answer is correct @campbell_st, it's 2

11. anonymous

:D

12. campbell_st

so remember its $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$ so you need a numeric value for dy/dx so use the x value at the point

13. anonymous

hmm... how do you get that? Is it just something i have to memorize?

14. campbell_st

well I'm using the chain rule I got $\frac{dy}{dx} = \frac{6}{\sqrt{9}}$ so then if I use the chain rule $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} ~~~~then~~~~\frac{6}{\sqrt{9}} = 4 \times \frac{dt}{dx}$

15. campbell_st

then I'd say $\frac{6}{4\sqrt{9}} = \frac{dt}{dx}$ take the reciprocal of both sides $\frac{dx}{dt} = \frac{4\sqrt{9}}{6}$ you can simplify it...

16. campbell_st

that would be my approach to this question... other more learned people may have different views...

17. anonymous

That makes a lot of sense now! Thanks so much @campbell_st :) I appreciate it!