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A curve is defined by the parametric equations
x(t)=4t^(3) and
y(t)=7sin(-8t). Find
(dy)/(dx).
Answer:

A curve is defined by the parametric equations $x(t)=4 t^{3}$ and $y(t)=7 \sin (-8 t)$ . Find $\frac{d y}{d x}$ . $\newline$ Answer:

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Write a quadratic function from its x-intercepts and another point

Full solution

Q. A curve is defined by the parametric equations

x(t)=4 t^{3}

and

y(t)=7 \sin (-8 t)

. Find

\frac{d y}{d x}

.

\newline

Answer:

Find $\frac{dx}{dt}$ : To find the derivative of $y$ with respect to $x$ , we need to find $\frac{dy}{dt}$ and $\frac{dx}{dt}$ and then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . First, let's find $\frac{dx}{dt}$ . $\frac{dx}{dt} = \frac{d(4t^3)}{dt}$ $\frac{dx}{dt} = 12t^2$
Find $\frac{dy}{dt}$ : Now, let's find $\frac{dy}{dt}$ .
$\frac{dy}{dt} = \frac{d(7\sin(-8t))}{dt}$
$\frac{dy}{dt} = 7 \cdot \cos(-8t) \cdot (-8)$
$\frac{dy}{dt} = -56\cos(-8t)$
Calculate $\frac{dy}{dx}$ : Now we divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ to find $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{-56\cos(-8t)}{12t^2}$
$\frac{dy}{dx} = \frac{-56\cos(-8t)}{12t^2}$
$\frac{dy}{dx} = \frac{-14\cos(-8t)}{3t^2}$

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