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Let $x$ and $y$ be functions of $t$ with $y = \pi x^2$ . If $\frac{dx}{dt} = -\frac{1}{8}$ , what is $\frac{dy}{dt}$ when $x = 16$ ? $\newline$ 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 = \pi x^2

. If

\frac{dx}{dt} = -\frac{1}{8}

, what is

\frac{dy}{dt}

when

x = 16

?

\newline

Write an exact, simplified answer.

Identify Relationship: Identify the relationship and differentiate implicitly. $\newline$ Given $y = \pi x^2$ , differentiate both sides with respect to $t$ . $\newline$ $\frac{dy}{dt} = 2\pi x\left(\frac{dx}{dt}\right)$
Differentiate Implicitly: Substitute the values of $\frac{dx}{dt}$ and $x$ . $\newline$ $\frac{dx}{dt} = -\frac{1}{8}$ and $x = 16$ . $\newline$ $\frac{dy}{dt} = 2\pi(16)(-\frac{1}{8})$
Substitute Values: Simplify the expression to find $\frac{dy}{dt}$ . $\frac{dy}{dt} = 2\pi(16)(-\frac{1}{8}) = -4\pi$

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