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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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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 $x = 4$ .
$\frac{dy}{dt} = (32\cdot4 + 4) \cdot (\frac{1}{16})$
$\frac{dy}{dt} = (128 + 4) \cdot (\frac{1}{16})$
$\frac{dy}{dt} = 132 \cdot (\frac{1}{16})$
$\frac{dy}{dt} = 8.25$

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