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Feb/Mar 2020 p12 q4
1104
A curve has equation \(y = x^2 - 2x - 3\). A point is moving along the curve in such a way that at \(P\) the \(y\)-coordinate is increasing at 4 units per second and the \(x\)-coordinate is increasing at 6 units per second.
Find the \(x\)-coordinate of \(P\).
Solution
Given the curve equation \(y = x^2 - 2x - 3\), we need to find \(\frac{dy}{dx}\).
Differentiate \(y\) with respect to \(x\):
\(\frac{dy}{dx} = 2x - 2\).
We know \(\frac{dy}{dt} = 4\) and \(\frac{dx}{dt} = 6\).
Using the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).