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A curve is defined by the parametric equations
x(t)=-4sin(-5t) and
y(t)=-3e^(-8t). Find
(dy)/(dx).
Answer:

A curve is defined by the parametric equations $x(t)=-4 \sin (-5 t)$ and $y(t)=-3 e^{-8 t}$ . Find $\frac{d y}{d x}$ . $\newline$ Answer:

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Write equations of cosine functions using properties

Full solution

Q. A curve is defined by the parametric equations

x(t)=-4 \sin (-5 t)

and

y(t)=-3 e^{-8 t}

. 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, we find $\frac{dx}{dt}$ . The derivative of $x(t) = -4\sin(-5t)$ with respect to $t$ is $\frac{dx}{dt} = -4 \times -5 \times \cos(-5t) = 20\cos(-5t)$ , since the derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$ and we apply the chain rule for the inner function $-5t$ .
Calculate $\frac{dy}{dx}$ : Next, we find $\frac{dy}{dt}$ . The derivative of $y(t) = -3e^{-8t}$ with respect to $t$ is $\frac{dy}{dt} = -3 \times -8 \times e^{-8t} = 24e^{-8t}$ , since the derivative of $e^u$ with respect to $u$ is $e^u$ and we apply the chain rule for the inner function $-8t$ .
Simplify $\frac{dy}{dx}$ : Now we have $\frac{dx}{dt} = 20\cos(-5t)$ and $\frac{dy}{dt} = 24e^{-8t}$ . To find $\frac{dy}{dx}$ , we divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ : $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{24e^{-8t}}{20\cos(-5t)}$ .
Simplify $\frac{dy}{dx}$ : Now we have $\frac{dx}{dt} = 20\cos(-5t)$ and $\frac{dy}{dt} = 24e^{-8t}$ . To find $\frac{dy}{dx}$ , we divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ : $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{24e^{-8t}}{20\cos(-5t)}$ .Simplify the expression for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{24}{20} \cdot \frac{e^{-8t}}{\cos(-5t)} = \frac{6}{5} \cdot \frac{e^{-8t}}{\cos(-5t)}$ .

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