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A curve is defined by the parametric equations
x(t)=-10 sin(10 t) and
y(t)=3t^(2). Find
(dy)/(dx).
Answer:

A curve is defined by the parametric equations $x(t)=-10 \sin (10 t)$ and $y(t)=3 t^{2}$ . Find $\frac{d y}{d x}$ . $\newline$ Answer:

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Transformations of quadratic functions

Full solution

Q. A curve is defined by the parametric equations

x(t)=-10 \sin (10 t)

and

y(t)=3 t^{2}

. 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 $\frac{dx}{dt}$ separately and then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ .
Find $\frac{dy}{dt}$ : First, let's find $\frac{dx}{dt}$ . The derivative of $x(t)=-10 \sin(10 t)$ with respect to $t$ is found using the chain rule. $\newline$ $\frac{dx}{dt} = \frac{d}{dt} [-10 \sin(10 t)] = -10 \cdot \cos(10 t) \cdot 10 = -100 \cos(10 t)$
Divide to find $\frac{dy}{dx}$ : Next, we find $\frac{dy}{dt}$ . The derivative of $y(t)=3t^{2}$ with respect to $t$ is found using the power rule. $\frac{dy}{dt} = \frac{d}{dt} [3t^{2}] = 2 \cdot 3 \cdot t^{2-1} = 6t$
Simplify $\frac{dy}{dx}$ : Now we divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ to find $\frac{dy}{dx}$ . $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6t}{-100 \cos(10 t)}$
Simplify $\frac{dy}{dx}$ : Now we divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ to find $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6t}{-100 \cos(10 t)}$ We simplify the expression for $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{-6t}{100 \cos(10 t)} = \frac{-3t}{50 \cos(10 t)}$

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