A curve is defined by the parametric equations $x(t)=9 t^{3}+2 t^{2}-5 t-10$ and $y(t)=-8 t^{2}$ . Find $\frac{d y}{d x}$ . $\newline$ Answer:

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Find derivatives of using multiple formulae

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Q. A curve is defined by the parametric equations

x(t)=9 t^{3}+2 t^{2}-5 t-10

and

y(t)=-8 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 $\frac{dx}{dt}$ for $x(t)=9t^{3}+2t^{2}-5t-10$ . Using the power rule for derivatives, we get: $\frac{dx}{dt} = \frac{d}{dt}(9t^{3}) + \frac{d}{dt}(2t^{2}) - \frac{d}{dt}(5t) - \frac{d}{dt}(10)$ $\frac{dx}{dt} = 3\cdot9t^{2} + 2\cdot2t^{1} - 5 - 0$ $\frac{dx}{dt} = 27t^{2} + 4t - 5$
Calculate $(\frac{dy}{dx}):</b> Next, we find \$\frac{dy}{dt}$ for $y(t)=-8t^{2}$ . Using the power rule for derivatives, we get: $\frac{dy}{dt} = \frac{d}{dt}(-8t^{2})$ $\frac{dy}{dt} = 2*(-8)t^{1}$ $\frac{dy}{dt} = -16t$
Simplify the expression: Now we have $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , so we can find $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$ .
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{-16t}{27t^{2} + 4t - 5}$
Simplify the expression: Now we have $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , so we can find $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . $\newline$ $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ $\newline$ $\frac{dy}{dx} = \frac{-16t}{27t^{2} + 4t - 5}$ We can simplify the expression by factoring out $t$ from the numerator and denominator if possible. $\newline$ However, in this case, we cannot factor out $t$ from the denominator because not all terms in the denominator are multiples of $t$ . $\newline$ So, the final simplified form of $\frac{dy}{dx}$ is: $\newline$ $\frac{dy}{dx} = \frac{-16t}{27t^{2} + 4t - 5}$

A curve is defined by the parametric equations $x(t)=9 t^{3}+2 t^{2}-5 t-10$ and $y(t)=-8 t^{2}$ . Find $\frac{d y}{d x}$ . $\newline$ Answer:

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