Question

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?

Answer 1

this looks like a related rates question...

\[\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\]

Answer 2

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

Answer 3

ok... so you need to find

\[\frac{dy}{dx} ~~~given~~~~ y = (1 + x^3)^{\frac{1}{2}}\]

Answer 4

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

Answer 5

right?

Answer 6

looks good to me...

now the point is (2, 3) so substitute x = 2

and get a value

Answer 7

That's all i gotta do? :O

Answer 8

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

Answer 9

well that is just a step

Answer 10

The answer is correct @campbell_st, it's 2

Answer 11

:D

Answer 12

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

Answer 13

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

Answer 14

well I'm using the chain rule

I got

\[\frac{dy}{dx} = \frac{6}{\sqrt{9}}\]

so then

if I use the chain rule

\[\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} ~~~~then~~~~\frac{6}{\sqrt{9}} = 4 \times \frac{dt}{dx}\]

Answer 15

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...

Answer 16

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

Answer 17

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