Question

suppose that a dimension x and the area a = 4x² of a shape are differentiable functions of t. write an equation that relates (\frac{da}{dt}) to (\frac{dx}{dt}). (\frac{da}{dt}=square)

Explanation:

Step1: Differentiate \(A = 4x^{2}\) with respect to \(t\)

Use the chain - rule \(\frac{dA}{dt}=\frac{dA}{dx}\cdot\frac{dx}{dt}\). First, find \(\frac{dA}{dx}\) for \(A = 4x^{2}\). By the power rule \(\frac{d}{dx}(ax^{n})=nax^{n - 1}\), so \(\frac{dA}{dx}=\frac{d}{dx}(4x^{2})=8x\).

Step2: Apply the chain - rule formula

Since \(\frac{dA}{dt}=\frac{dA}{dx}\cdot\frac{dx}{dt}\) and \(\frac{dA}{dx}=8x\), then \(\frac{dA}{dt}=8x\frac{dx}{dt}\).

Answer:

\(8x\frac{dx}{dt}\)