Question

a point is moving along the graph of $x^{3}y^{2}=500$. when the point is at $(5,2)$, its $x$-coordinate is changing at the rate of $-7$ units per minute. how fast is the $y$-coordinate changing at that moment?
the $y$-coordinate is changing by
(type an integer or a fraction. simplif

Explanation:

Step1: Differentiate both sides with respect to \(t\)

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\). Let \(u = x^{3}\) and \(v = y^{2}\). Then \(\frac{d}{dt}(x^{3}y^{2})=\frac{d}{dt}(500)\).

The derivative of \(x^{3}\) with respect to \(t\) is \(3x^{2}\frac{dx}{dt}\) (by the chain rule \(\frac{d}{dt}(x^{n})=nx^{n - 1}\frac{dx}{dt}\)), and the derivative of \(y^{2}\) with respect to \(t\) is \(2y\frac{dy}{dt}\).

So, \(3x^{2}y^{2}\frac{dx}{dt}+2x^{3}y\frac{dy}{dt}=0\).

Step2: Substitute \(x = 5\), \(y = 2\), and \(\frac{dx}{dt}=-7\)

Substitute into \(3x^{2}y^{2}\frac{dx}{dt}+2x^{3}y\frac{dy}{dt}=0\):

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Step3: Solve for \(\frac{dy}{dt}\)

Add \(2100\) to both sides: \(500\times\frac{dy}{dt}=2100\).

Then \(\frac{dy}{dt}=\frac{2100}{500}=\frac{21}{5} = 4.2\).

Answer:

\(\frac{21}{5}\) units per minute