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A curve is defined by the parametric equations
x(t)=-3cos(-t) and
y(t)=5e^(-9t). Find
(dy)/(dx).
Answer:

A curve is defined by the parametric equations $x(t)=-3 \cos (-t)$ and $y(t)=5 e^{-9 t}$ . 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)=-3 \cos (-t)

and

y(t)=5 e^{-9 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, let's find $\frac{dx}{dt}$ . The derivative of $x(t) = -3\cos(-t)$ with respect to $t$ is $\frac{dx}{dt} = -3 \cdot \frac{d}{dt}[\cos(-t)]$ . Since the derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$ , and using the chain rule with $u = -t$ , we get $\frac{dx}{dt} = -3 \cdot (-\sin(-t)) \cdot (-1) = 3\sin(-t)$ .
Calculate $\frac{dy}{dx}$ : Now, let's find $\frac{dy}{dt}$ . The derivative of $y(t) = 5e^{-9t}$ with respect to $t$ is $\frac{dy}{dt} = 5 \cdot \frac{d}{dt}[e^{-9t}]$ . Since the derivative of $e^{u}$ with respect to $u$ is $e^{u}$ , and using the chain rule with $u = -9t$ , we get $\frac{dy}{dt} = 5 \cdot e^{-9t} \cdot (-9) = -45e^{-9t}$ .
Simplify $\frac{dy}{dx}$ : To find $\frac{dy}{dx}$ , we divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . So, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-45e^{-9t}}{3\sin(-t)}$ .
Simplify $\frac{dy}{dx}$ : To find $\frac{dy}{dx}$ , we divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . So, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-45e^{-9t}}{3\sin(-t)}$ .We can simplify the expression for $\frac{dy}{dx}$ by dividing both the numerator and the denominator by $3$ . This gives us $\frac{dy}{dx} = \frac{-45/3e^{-9t}}{\sin(-t)} = -15e^{-9t} / \sin(-t)$ .

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