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

. If

\frac{dx}{dt} = 6

, what is

\frac{dy}{dt}

when

x = \frac{1}{3}

?

\newline

Write an exact, simplified answer.

Identify Given Information: Identify the given information and the formula to use. $\newline$ Given $y = 4\pi x^3$ and $\frac{dx}{dt} = 6$ . We need to find $\frac{dy}{dt}$ using the chain rule.
Apply Chain Rule: Apply the chain rule to find $\frac{dy}{dt}$ . $\newline$ $\frac{dy}{dt} = \frac{d(4\pi x^3)}{dx} \cdot \frac{dx}{dt}$ $\newline$ $= 12\pi x^2 \cdot \frac{dx}{dt}$
Substitute Values: Substitute $\frac{dx}{dt} = 6$ and $x = \frac{1}{3}$ into the equation. $\newline$ $\frac{dy}{dt} = 12\pi\left(\frac{1}{3}\right)^2 \times 6$ $\newline$ $\quad\quad\quad = 12\pi\left(\frac{1}{9}\right) \times 6$ $\newline$ $\quad\quad\quad = \frac{12\pi}{9} \times 6$ $\newline$ $\quad\quad\quad = 8\pi$

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