Let x and y be functions of t with y = 4/3 π x^3. If dx/dt = 1/16, what is dy/dt when x = 6? Write an exact, simplified answer. ______

Identify Relationship: Step 1: Identify the relationship between y and x. Given y = 4/3 π x^3, this equation shows how y changes with x.Differentiate with Chain Rule: Step 2: Differentiate y with respect to t using the chain rule. dy/dt = d/dt (4/3 π x^3) = 4π x^2 * dx/dtSubstitute Values: Step 3: Substitute the given values of dx/dt and x. dx/dt = 1/16 and x = 6. dy/dt = 4π (6)^2 * (1/16)Calculate dy/dt: Step 4: Calculate dy/dt. dy/dt = 4π * 36 * 1/16 = 9π

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Let $x$ and $y$ be functions of $t$ with $y = \frac{4}{3} \pi x^3$ . If $\frac{dx}{dt} = \frac{1}{16}$ , what is $\frac{dy}{dt}$ when $x = 6$ ? $\newline$ Write an exact, simplified answer.

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Math Problems

Grade 8

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}{16}

, what is

\frac{dy}{dt}

when

x = 6

\newline

Write an exact, simplified answer.

Identify Relationship: Step $1$ : Identify the relationship between $y$ and $x$ . Given $y = \frac{4}{3} \pi x^3$ , this equation shows how $y$ changes with $x$ .
Differentiate with Chain Rule: Step $2$ : Differentiate $y$ with respect to $t$ using the chain rule. $\newline$ $\frac{dy}{dt} = \frac{d}{dt} \left(\frac{4}{3} \pi x^3\right) = 4\pi x^2 \cdot \frac{dx}{dt}$
Substitute Values: Step $3$ : Substitute the given values of $\frac{dx}{dt}$ and $x$ . $\newline$ $\frac{dx}{dt} = \frac{1}{16}$ and $x = 6$ . $\newline$ $\frac{dy}{dt} = 4\pi (6)^2 \times \frac{1}{16}$
Calculate dy/dt: Step $4$ : Calculate dy/dt. $\newline$ $\frac{dy}{dt} = 4\pi \times 36 \times \frac{1}{16} = 9\pi$