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

A curve is defined by the parametric equations $x(t)=-3 \sin (4 t)$ and $y(t)=-2 e^{10 t}$ . 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)=-3 \sin (4 t)

and

y(t)=-2 e^{10 t}

. Find

\frac{d y}{d x}

.

\newline

Answer:

Find dx/dt: To find the derivative of $y$ with respect to $x$ ( $\frac{dy}{dx}$ ) for parametric equations, we need to find $\frac{dy}{dt}$ and $\frac{dx}{dt}$ separately and then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . First, let's find $\frac{dx}{dt}$ . $\frac{dx}{dt} = \frac{d}{dt} [-3\sin(4t)]$ Using the chain rule, we get: $\frac{dx}{dt} = -3 \cdot \frac{d}{dt} [\sin(4t)]$ $x$ $0$ $x$ $1$
Find $\frac{dy}{dt}$ : Now, let's find $\frac{dy}{dt}$ .
$\frac{dy}{dt} = \frac{d}{dt} [-2e^{10t}]$
Using the exponential rule, we get:
$\frac{dy}{dt} = -2 \cdot \frac{d}{dt} [e^{10t}]$
$\frac{dy}{dt} = -2 \cdot e^{10t} \cdot 10$
$\frac{dy}{dt} = -20e^{10t}$
Find $\frac{dy}{dx}$ : With $\frac{dx}{dt}$ and $\frac{dy}{dt}$ calculated, we can now find $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
Substitute the values we found for $\frac{dy}{dt}$ and $\frac{dx}{dt}$ :
$\frac{dy}{dx} = \frac{-20e^{10t}}{-12\cos(4t)}$
Simplify the expression:
$\frac{dy}{dx} = \frac{20}{12} \cdot \frac{e^{10t}}{\cos(4t)}$
$\frac{dy}{dx} = \frac{5}{3} \cdot \frac{e^{10t}}{\cos(4t)}$

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