Let x and y be functions of t with y = 15cos x. If dx/dt = -1/9, what is dy/dt when x = π /6? Write an exact, simplified answer. ______

Differentiation using chain rule: Step 1: Use the chain rule for differentiation to find dy/dt. Given y = 15cos(x), differentiate both sides with respect to t. dy/dt = 15 * -sin(x) * dx/dtSubstitute given values: Step 2: Substitute the given values into the differentiated equation. dx/dt = -1/9 and x = π/6 dy/dt = 15 * -sin(π/6) * (-1/9)Calculate and simplify: Step 3: Calculate sin(π/6) and simplify the expression. sin(π/6) = 1/2 dy/dt = 15 * -1/2 * (-1/9) dy/dt = 15 * 1/2 * 1/9 dy/dt = 15/18 dy/dt = 5/6

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Let $x$ and $y$ be functions of $t$ with $y = 15\cos x$ . If $\frac{dx}{dt} = -\frac{1}{9}$ , what is $\frac{dy}{dt}$ when $x = \frac{\pi}{6}$ ? $\newline$ Write an exact, simplified answer.

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Precalculus

Find trigonometric ratios using a Pythagorean or reciprocal identity

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Q. Let

x

and

y

be functions of

t

with

y = 15\cos x

. If

\frac{dx}{dt} = -\frac{1}{9}

, what is

\frac{dy}{dt}

when

x = \frac{\pi}{6}

\newline

Write an exact, simplified answer.

Differentiation using chain rule: Step $1$ : Use the chain rule for differentiation to find $\frac{dy}{dt}$ . Given $y = 15\cos(x)$ , differentiate both sides with respect to $t$ . $\frac{dy}{dt} = 15 \cdot (-\sin(x)) \cdot \frac{dx}{dt}$
Substitute given values: Step $2$ : Substitute the given values into the differentiated equation. $\newline$ $\frac{dx}{dt} = -\frac{1}{9}$ and $x = \frac{\pi}{6}$ $\newline$ $\frac{dy}{dt} = 15 \cdot -\sin\left(\frac{\pi}{6}\right) \cdot \left(-\frac{1}{9}\right)$
Calculate and simplify: Step $3$ : Calculate $\sin(\frac{\pi}{6})$ and simplify the expression. $\newline$ $\sin(\frac{\pi}{6}) = \frac{1}{2}$ $\newline$ $\frac{dy}{dt} = 15 \times -\frac{1}{2} \times (-\frac{1}{9})$ $\newline$ $\frac{dy}{dt} = 15 \times \frac{1}{2} \times \frac{1}{9}$ $\newline$ $\frac{dy}{dt} = \frac{15}{18}$ $\newline$ $\frac{dy}{dt} = \frac{5}{6}$