A curve is defined by the parametric equations $x(t)=7 t^{3}$ and $y(t)=t^{2}-2 t-5$ . 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}

and

y(t)=t^{2}-2 t-5

. Find

\frac{d y}{d x}

\newline

Answer:

Identify Derivatives: To find $(\frac{dy}{dx})$ , we first need to find the derivatives of $x(t)$ and $y(t)$ with respect to $t$ , which are denoted as $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , respectively.
Calculate $\frac{dx}{dt}$ : Calculate $\frac{dx}{dt}$ , which is the derivative of $x(t)$ with respect to $t$ .
$\frac{dx}{dt} = \frac{d(7t^3)}{dt}$
$\frac{dx}{dt} = 21t^2$
Calculate $\frac{dy}{dt}$ : Calculate $\frac{dy}{dt}$ , which is the derivative of $y(t)$ with respect to $t$ .
$\frac{dy}{dt} = \frac{d(t^2 - 2t - 5)}{dt}$
$\frac{dy}{dt} = 2t - 2$
Find $(\frac{dy}{dx}):</b> Now, to find \$(\frac{dy}{dx}), we divide \$\frac{dy}{dt}$ by $\frac{dx}{dt}$ . $\newline$ (\frac{dy}{dx}) = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}$\newline(\frac{dy}{dx}) = \frac{(\(2$t - $2$)}{($21$t^$2$)}
Simplify the Expression: Simplify the expression for \((\frac{dy}{dx}). $(\frac{dy}{dx}) = \frac{2t - 2}{21t^2}$ $(\frac{dy}{dx}) = \frac{2(t - 1)}{21t^2}$

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A curve is defined by the parametric equations x(t)=7t3 x(t)=7 t^{3} x(t)=7t3 and y(t)=t2−2t−5 y(t)=t^{2}-2 t-5 y(t)=t2−2t−5. Find dydx \frac{d y}{d x} dxdy​.\newlineAnswer:

A curve is defined by the parametric equations $x(t)=7 t^{3}$ and $y(t)=t^{2}-2 t-5$ . Find $\frac{d y}{d x}$ . $\newline$ Answer: