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

A curve is defined by the parametric equations $x(t)=6 t^{3}+6 t+8$ and $y(t)=-4 \cos (-2 t)$ . 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)=6 t^{3}+6 t+8

and

y(t)=-4 \cos (-2 t)

. Find

\frac{d y}{d x}

.

\newline

Answer:

Find $\frac{dx}{dt}$ : To find the derivative of $y$ with respect to $x$ ( $\frac{dy}{dx}$ ) for parametric equations, we need to find $\frac{dy}{dt}$ and $\frac{dx}{dt}$ and then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . First, let's find $\frac{dx}{dt}$ , which is the derivative of $x$ with respect to $y$ $0$ . $y$ $1$ $y$ $2$
Find $\frac{dy}{dt}$ : Now, let's find $\frac{dy}{dt}$ , which is the derivative of $y$ with respect to $t$ .
$\frac{dy}{dt} = \frac{d}{dt} (-4\cos(-2t))$
Using the chain rule, we get:
$\frac{dy}{dt} = -4 \cdot (-\sin(-2t)) \cdot (-2)$
$\frac{dy}{dt} = 8\sin(-2t)$
Calculate dy/dx: Now that we have both derivatives, we can find $dy/dx$ by dividing $dy/dt$ by $dx/dt$ . $\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}$ $\frac{dy}{dx} = \frac{8\sin(-2t)}{18t^2 + 6}$
Simplify $\frac{dy}{dx}$ : We can simplify the expression for $\frac{dy}{dx}$ if possible. $\newline$ However, in this case, the expression is already in its simplest form. $\newline$ So, the final answer is: $\newline$ $\frac{dy}{dx} = \frac{8\sin(-2t)}{18t^2 + 6}$

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