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A curve is defined by the parametric equations
x(t)=-10t^(2)+6t and
y(t)=-4t^(3)-8t^(2). Find
(dy)/(dx).
Answer:

A curve is defined by the parametric equations $x(t)=-10 t^{2}+6 t$ and $y(t)=-4 t^{3}-8 t^{2}$ . Find $\frac{d y}{d x}$ . $\newline$ Answer:

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Quadratic equation with complex roots

Full solution

Q. A curve is defined by the parametric equations

x(t)=-10 t^{2}+6 t

and

y(t)=-4 t^{3}-8 t^{2}

. Find

\frac{d y}{d x}

.

\newline

Answer:

Find Derivative of $x(t)$ : Find the derivative of $x(t)$ with respect to $t$ , denoted as $\frac{dx}{dt}$ . The derivative of $x(t) = -10t^2 + 6t$ is found using the power rule for derivatives. $\frac{dx}{dt} = \frac{d}{dt}(-10t^2 + 6t) = -20t + 6$ .
Find Derivative of $y(t)$ : Find the derivative of $y(t)$ with respect to $t$ , denoted as $\frac{dy}{dt}$ . The derivative of $y(t) = -4t^3 - 8t^2$ is found using the power rule for derivatives. $\frac{dy}{dt} = \frac{d}{dt}(-4t^3 - 8t^2) = -12t^2 - 16t$ .
Find Derivative $\frac{dy}{dx}$ : Find the derivative of $y$ with respect to $x$ , denoted as $\frac{dy}{dx}$ . The derivative $\frac{dy}{dx}$ is the ratio of $\frac{dy}{dt}$ to $\frac{dx}{dt}$ . $\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}$ . Substitute the derivatives from Step $1$ and Step $2$ . $\frac{dy}{dx} = \frac{-12t^2 - 16t}{-20t + 6}$ .
Simplify $\frac{dy}{dx}$ : Simplify the expression for $\frac{dy}{dx}$ if possible. $\frac{dy}{dx} = \frac{-12t^2 - 16t}{-20t + 6}$ cannot be simplified further without specific values of $t$ .

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