Let x and y be functions of t with y = x^2 + x^(1/2). If dx/dt = 1/8, what is dy/dt when x = 4? Write an exact, simplified answer. ______

Identify Relationship: Identify the relationship and differentiate. Given y = x^2 + x^(1/2), we need to find dy/dt using the chain rule. dy/dx = 2x + (1/2)x^(-1/2) Substitute dx/dt: Substitute dx/dt into the equation. Given dx/dt = 1/8, substitute and find dy/dt. dy/dt = dy/dx * dx/dt = (2x + (1/2)x^(-1/2)) * (1/8) Evaluate at x=4: Evaluate dy/dt at x = 4. Substitute x = 4 into dy/dt. dy/dt = (2*4 + (1/2)*4^(-1/2)) * (1/8) = (8 + (1/2)*1/2) * (1/8) = (8 + 1/4) * (1/8) = (33/4) * (1/8) = 33/32

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Let $x$ and $y$ be functions of $t$ with $y = x^2 + x^{\frac{1}{2}}$ . If $\frac{dx}{dt} = \frac{1}{8}$ , 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 = x^2 + x^{\frac{1}{2}}

. If

\frac{dx}{dt} = \frac{1}{8}

, what is

\frac{dy}{dt}

when

x = 4

\newline

Write an exact, simplified answer.

Identify Relationship: Identify the relationship and differentiate. $\newline$ Given $y = x^2 + x^{1/2}$ , we need to find $\frac{dy}{dt}$ using the chain rule. $\newline$ $\frac{dy}{dx} = 2x + \left(\frac{1}{2}\right)x^{-1/2}$
Substitute $\frac{dx}{dt}$ : Substitute $\frac{dx}{dt}$ into the equation. $\newline$ Given $\frac{dx}{dt} = \frac{1}{8}$ , substitute and find $\frac{dy}{dt}$ . $\newline$ $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = (2x + (\frac{1}{2})x^{(-\frac{1}{2})}) \cdot (\frac{1}{8})$
Evaluate at $x=4$ : Evaluate $\frac{dy}{dt}$ at $x = 4$ . $\newline$ Substitute $x = 4$ into $\frac{dy}{dt}$ . $\newline$ \frac{dy}{dt} = (\(2\cdot $4$ + (\frac{ $1$ }{ $2$ })\cdot $4$ ^{(-\frac{ $1$ }{ $2$ })}) \cdot (\frac{ $1$ }{ $8$ }) $\newline$ = ( $8$ + (\frac{ $1$ }{ $2$ })\cdot\frac{ $1$ }{ $2$ }) \cdot (\frac{ $1$ }{ $8$ }) $\newline$ = ( $8$ + \frac{ $1$ }{ $4$ }) \cdot (\frac{ $1$ }{ $8$ }) $\newline$ = (\frac{ $33$ }{ $4$ }) \cdot (\frac{ $1$ }{ $8$ }) $\newline$ = \frac{ $33$ }{ $32$ }