Right circular cone The lateral surface area S of a right circular cone is related to the base radius r and height h by the equation
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S = πr √ r² + h².


c. How is dS/dt related to dr/dt and dh/dt if neither r nor h is constant?

Verified step by step guidance

1. First, identify the given formula for the lateral surface area of the cone: \( S = \pi r \sqrt{r^2 + h^2} \).
2. To find \( \frac{dS}{dt} \), apply the chain rule for differentiation, as both \( r \) and \( h \) are functions of time \( t \).
3. Differentiate \( S \) with respect to \( t \): \( \frac{dS}{dt} = \frac{d}{dt}(\pi r \sqrt{r^2 + h^2}) \). Use the product rule: \( \frac{d}{dt}(uv) = u \frac{dv}{dt} + v \frac{du}{dt} \).
4. Let \( u = \pi r \) and \( v = \sqrt{r^2 + h^2} \). Differentiate \( u \) and \( v \) with respect to \( t \): \( \frac{du}{dt} = \pi \frac{dr}{dt} \) and \( \frac{dv}{dt} = \frac{1}{2\sqrt{r^2 + h^2}}(2r\frac{dr}{dt} + 2h\frac{dh}{dt}) \).
5. Substitute these derivatives back into the product rule formula: \( \frac{dS}{dt} = \pi r \frac{1}{2\sqrt{r^2 + h^2}}(2r\frac{dr}{dt} + 2h\frac{dh}{dt}) + \sqrt{r^2 + h^2} \pi \frac{dr}{dt} \). Simplify to express \( \frac{dS}{dt} \) in terms of \( \frac{dr}{dt} \) and \( \frac{dh}{dt} \).