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

. If

\frac{dx}{dt} = 3

, what is

\frac{dy}{dt}

when

x = \pi

?

\newline

Write an exact, simplified answer.

Identify Relationship: Identify the relationship and differentiate using the chain rule. $\newline$ Given $y = \sin(x)$ and $\frac{dx}{dt} = 3$ , use the chain rule to find $\frac{dy}{dt}$ . $\newline$ $\frac{dy}{dt} = \left(\frac{dy}{dx}\right) \cdot \left(\frac{dx}{dt}\right)$ $\newline$ $\frac{dy}{dx} = \cos(x)$ because the derivative of $\sin(x)$ is $\cos(x)$ . $\newline$ Substitute $\frac{dx}{dt} = 3$ into the equation. $\newline$ $\frac{dy}{dt} = \cos(x) \cdot 3$
Use Chain Rule: Substitute the value of $x$ to find $\frac{dy}{dt}$ when $x = \pi$ . At $x = \pi$ , $\cos(\pi) = -1$ . So, $\frac{dy}{dt} = -1 \times 3 = -3$

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