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A curve is defined by the parametric equations
x(t)=-6t^(3)-t-4 and
y(t)=7t^(3)-5t^(2). Find
(dy)/(dx).
Answer:

A curve is defined by the parametric equations $x(t)=-6 t^{3}-t-4$ and $y(t)=7 t^{3}-5 t^{2}$ . 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)=-6 t^{3}-t-4

and

y(t)=7 t^{3}-5 t^{2}

. Find

\frac{d y}{d x}

.

\newline

Answer:

Find $x'(t)$ : To find the derivative of $y$ with respect to $x$ , $\frac{dy}{dx}$ , we need to find the derivatives of $y(t)$ and $x(t)$ with respect to $t$ , and then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . First, let's find the derivative of $x(t)$ with respect to $t$ . $y$ $1$ $y$ $2$ $y$ $3$
Find $y'(t)$ : Now, let's find the derivative of $y(t)$ with respect to $t$ .
$y(t) = 7t^3 - 5t^2$
$\frac{dy}{dt} = \frac{d}{dt}(7t^3 - 5t^2)$
$\frac{dy}{dt} = 21t^2 - 10t$
Find $\frac{dy}{dx}$ : With both derivatives found, we can now find $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$ .
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{21t^2 - 10t}{-18t^2 - 1}$

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