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Let $x$ and $y$ be functions of $t$ with $y = \frac{4}{3} \pi x^3$ . If $\frac{dx}{dt} = \frac{1}{8}$ , what is $\frac{dy}{dt}$ when $x = 3$ ? $\newline$ 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}{8}

, what is

\frac{dy}{dt}

when

x = 3

?

\newline

Write an exact, simplified answer.

Identify Relationship: Identify the relationship and differentiate implicitly. $\newline$ Given $y = \frac{4}{3}\pi x^3$ , differentiate both sides with respect to $t$ . $\newline$ Using the chain rule, $\frac{dy}{dt} = \frac{4}{3}\pi \cdot 3x^2 \cdot \frac{dx}{dt}$ .
Differentiate Implicitly: Substitute the given values. $\newline$ Substitute $x = 3$ and $\frac{dx}{dt} = \frac{1}{8}$ into $\frac{dy}{dt} = 4\pi x^2 \times \frac{1}{8}$ . $\newline$ $\frac{dy}{dt} = 4\pi(3^2) \times \frac{1}{8}$ .
Apply Chain Rule: Simplify the expression. $\newline$ $\frac{dy}{dt} = 4\pi(9) \times \frac{1}{8} = 36\pi \times \frac{1}{8} = 4.5\pi$ .

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