Let x and y be functions of t with y = 16x^2 + 4x + 3. If dx/dt = 1/16, what is dy/dt when x = 4? Write an exact, simplified answer. ______

Identify Function: Identify the function and differentiate it with respect to x. Using the chain rule, differentiate y = 16x^2 + 4x + 3 with respect to x. dy/dx = 32x + 4 Differentiate with Chain Rule: Substitute the value of dx/dt into the equation. Given dx/dt = 1/16, use the chain rule to find dy/dt. dy/dt = dy/dx * dx/dt = (32x + 4) * (1/16) Substitute dx/dt: Substitute x = 4 into the equation to find dy/dt when x = 4. dy/dt = (32*4 + 4) * (1/16) dy/dt = (128 + 4) * (1/16) dy/dt = 132 * (1/16) dy/dt = 8.25

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Let $x$ and $y$ be functions of $t$ with $y = 16x^2 + 4x + 3$ . If $\frac{dx}{dt} = \frac{1}{16}$ , what is $\frac{dy}{dt}$ when $x = 4$ ? $\newline$ Write an exact, simplified answer.

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Algebra 2

Write and solve direct variation equations

Full solution

Q. Let

x

and

y

be functions of

t

with

y = 16x^2 + 4x + 3

. If

\frac{dx}{dt} = \frac{1}{16}

, what is

\frac{dy}{dt}

when

x = 4

\newline

Write an exact, simplified answer.

Identify Function: Identify the function and differentiate it with respect to $x$ . Using the chain rule, differentiate $y = 16x^2 + 4x + 3$ with respect to $x$ . $\frac{dy}{dx} = 32x + 4$
Differentiate with Chain Rule: Substitute the value of $\frac{dx}{dt}$ into the equation. $\newline$ Given $\frac{dx}{dt} = \frac{1}{16}$ , use the chain rule to find $\frac{dy}{dt}$ . $\newline$ $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = (32x + 4) \cdot \left(\frac{1}{16}\right)$
Substitute $\frac{dx}{dt}$ : Substitute $x = 4$ into the equation to find $\frac{dy}{dt}$ when $x = 4$ .
$\frac{dy}{dt} = (32\cdot4 + 4) \cdot (\frac{1}{16})$
$\frac{dy}{dt} = (128 + 4) \cdot (\frac{1}{16})$
$\frac{dy}{dt} = 132 \cdot (\frac{1}{16})$
$\frac{dy}{dt} = 8.25$