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?

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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?

Mathematics
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this looks like a related rates question... \[\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\]
Yeah it is :) hmm... so what are you doing there?
ok... so you need to find \[\frac{dy}{dx} ~~~given~~~~ y = (1 + x^3)^{\frac{1}{2}}\]

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ok... thats \[\frac{ 3x^2 }{ 2\sqrt(1+x^3) }\]
right?
looks good to me... now the point is (2, 3) so substitute x = 2 and get a value
That's all i gotta do? :O
i think we need to find \(\dfrac{dx}{dt}\), the rate of change of x coordinate
well that is just a step
The answer is correct @campbell_st, it's 2
:D
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
hmm... how do you get that? Is it just something i have to memorize?
well I'm using the chain rule I got \[\frac{dy}{dx} = \frac{6}{\sqrt{9}}\] so then if I use the chain rule \[\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} ~~~~then~~~~\frac{6}{\sqrt{9}} = 4 \times \frac{dt}{dx}\]
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...
that would be my approach to this question... other more learned people may have different views...
That makes a lot of sense now! Thanks so much @campbell_st :) I appreciate it!

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