Related rates
\[\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\]
Area of circle: dA/dt = 2πr dr/dt. Volume of sphere: dV/dt = 4πr² dr/dt. Cube: dV/dt = 3a² da/dt.
Applications of Derivatives
If \(y=f(x)\), then \(\frac{dy}{dx}\) (or \(f\,'(x_0)\) at a point) is the rate of change of \(y\) with respect to \(x\). When both quantities depend on time \(t\), the chain rule links their rates:
\[\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\]
Example. The radius of a circle grows at \(0.5\) cm/s. How fast is the area growing when \(r=4\) cm? \(A=\pi r^2\Rightarrow\frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi(4)(0.5)=4\pi\) cm²/s.
Economics. If \(C(x)\) is the total cost of producing \(x\) units, the marginal cost is \(C\,'(x)\), the approximate cost of one more unit. Likewise marginal revenue is \(R\,'(x)\).
\[\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\]
Area of circle: dA/dt = 2πr dr/dt. Volume of sphere: dV/dt = 4πr² dr/dt. Cube: dV/dt = 3a² da/dt.
Differentiate \(V=\frac43\pi r^3\) first, then put \(r=5\). If you substitute first, \(V\) becomes a constant with derivative 0.
"Decreasing at 2 cm/s" means \(\frac{dx}{dt}=-2\).
A fresh set of questions from this topic. Change answers freely – solutions appear after you submit.
Start PracticeWrite the relation, differentiate with respect to time, then substitute the instant’s values. Units of a rate: (units of y) per (unit of t).
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