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A curve is defined by the parametric equations
x(t)=3sin(3t) and
y(t)=-9t^(3)-5t^(2)+9t. Find
(dy)/(dx).
Answer:

A curve is defined by the parametric equations $x(t)=3 \sin (3 t)$ and $y(t)=-9 t^{3}-5 t^{2}+9 t$ . 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)=3 \sin (3 t)

and

y(t)=-9 t^{3}-5 t^{2}+9 t

. Find

\frac{d y}{d x}

.

\newline

Answer:

Find $\frac{dx}{dt}$ : To find $\frac{dy}{dx}$ for parametric equations, 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 $\frac{dx}{dt}$ . $\frac{dx}{dt} = \frac{d}{dt} [3\sin(3t)]$ Using the chain rule, we get $\frac{dx}{dt} = 3 \cdot \cos(3t) \cdot \frac{d}{dt} [3t]$ $\frac{dx}{dt} = 3 \cdot \cos(3t) \cdot 3$ $\frac{dy}{dx}$ $0$
Find $\frac{dy}{dt}$ : Now, let's find $\frac{dy}{dt}$ .
$\frac{dy}{dt} = \frac{d}{dt} [-9t^{3} - 5t^{2} + 9t]$
Using the power rule, we get $\frac{dy}{dt} = -27t^{2} - 10t + 9$
Calculate $\frac{dy}{dx}:$ Now we have $\frac{dx}{dt}$ and $\frac{dy}{dt},$ we can find $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}.$ $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ $\frac{dy}{dx} = \frac{-27t^{2} - 10t + 9}{9\cos(3t)}$

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