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Let $x$ and $y$ be functions of $t$ with $y = \frac{4}{3} \pi x^3$ . If $\frac{dx}{dt} = \frac{1}{16}$ , what is $\frac{dy}{dt}$ when $x = 6$ ? $\newline$ Write an exact, simplified answer.

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Write and solve direct variation equations

Full solution

Q. Let

x

and

y

be functions of

t

with

y = \frac{4}{3} \pi x^3

. If

\frac{dx}{dt} = \frac{1}{16}

, what is

\frac{dy}{dt}

when

x = 6

?

\newline

Write an exact, simplified answer.

Identify Relationship: Step $1$ : Identify the relationship between $y$ and $x$ . Given $y = \frac{4}{3} \pi x^3$ , this equation shows how $y$ changes with $x$ .
Differentiate with Chain Rule: Step $2$ : Differentiate $y$ with respect to $t$ using the chain rule. $\newline$ $\frac{dy}{dt} = \frac{d}{dt} \left(\frac{4}{3} \pi x^3\right) = 4\pi x^2 \cdot \frac{dx}{dt}$
Substitute Values: Step $3$ : Substitute the given values of $\frac{dx}{dt}$ and $x$ . $\newline$ $\frac{dx}{dt} = \frac{1}{16}$ and $x = 6$ . $\newline$ $\frac{dy}{dt} = 4\pi (6)^2 \times \frac{1}{16}$
Calculate dy/dt: Step $4$ : Calculate dy/dt. $\newline$ $\frac{dy}{dt} = 4\pi \times 36 \times \frac{1}{16} = 9\pi$

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