A curve is defined by the parametric equations $x(t)=7 t^{3}-t$ and $y(t)=-9 \sin (-2 t)$ . Find $\frac{d y}{d x}$ . $\newline$ Answer:

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Find the vertex of the transformed function

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Q. A curve is defined by the parametric equations

x(t)=7 t^{3}-t

and

y(t)=-9 \sin (-2 t)

. Find

\frac{d y}{d x}

\newline

Answer:

Calculate Derivative of $x(t)$ : To find $\frac{dy}{dx}$ , we need to calculate the derivatives of $y(t)$ with respect to $t$ , denoted as $\frac{dy}{dt}$ , and the derivative of $x(t)$ with respect to $t$ , denoted as $\frac{dx}{dt}$ , and then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ .
Calculate Derivative of $y(t)$ : First, let's find $\frac{dx}{dt}$ . The derivative of $x(t) = 7t^3 - t$ with respect to $t$ is $\frac{dx}{dt} = 21t^2 - 1$ .
Find $(dy)/(dx)$ : Next, we calculate $(dy)/(dt)$ . The derivative of $y(t) = -9\sin(-2t)$ with respect to $t$ is $(dy)/(dt) = -9 \times (-2) \times \cos(-2t) = 18\cos(-2t)$ , using the chain rule and the fact that the derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$ .
Find $(\frac{dy}{dx}):</b> Next, we calculate \$(\frac{dy}{dt}). The derivative of \$y(t) = -9\sin(-2t)$ with respect to $t$ is $(\frac{dy}{dt}) = -9 \times (-2) \times \cos(-2t) = 18\cos(-2t)$ , using the chain rule and the fact that the derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$ .Now we have $(\frac{dx}{dt}) = 21t^2 - 1$ and $(\frac{dy}{dt}) = 18\cos(-2t)$ . To find $(\frac{dy}{dx})$ , we divide $(\frac{dy}{dt})$ by $t$ $0$ : $t$ $1$ .

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A curve is defined by the parametric equations x(t)=7t3−t x(t)=7 t^{3}-t x(t)=7t3−t and y(t)=−9sin⁡(−2t) y(t)=-9 \sin (-2 t) y(t)=−9sin(−2t). Find dydx \frac{d y}{d x} dxdy​.\newlineAnswer:

A curve is defined by the parametric equations $x(t)=7 t^{3}-t$ and $y(t)=-9 \sin (-2 t)$ . Find $\frac{d y}{d x}$ . $\newline$ Answer: