Which of the following is the derivative (dy)/(dx) for the plane curve defined by the equations x(t)=-sin pi t, y(t)=cos pi t, and 0 <= t <= 2 ? Select the correct answer below: (A) -cot pi t (B) cot pi t (C) pi tan pi t (D) tan pi t

Find dx/dt: To find the derivative (dy)/(dx), we need to find dy/dt and dx/dt first and then divide dy/dt by dx/dt.Find dy/dt: Let's find dx/dt. Given x(t) = -sin(pi*t), we differentiate with respect to t to get dx/dt. dx/dt = d/dt [-sin(pi*t)] = -pi*cos(pi*t)Calculate (dy)/(dx): Now, let's find dy/dt. Given y(t) = cos(pi*t), we differentiate with respect to t to get dy/dt. dy/dt = d/dt [cos(pi*t)] = -pi*sin(pi*t)Simplify the expression: Now we have dx/dt and dy/dt. To find (dy)/(dx), we divide dy/dt by dx/dt. (dy)/(dx) = (dy/dt) / (dx/dt) = (-pi*sin(pi*t)) / (-pi*cos(pi*t))Final Result: We can simplify the expression by canceling out the -pi terms in the numerator and the denominator. (dy)/(dx) = sin(pi*t) / cos(pi*t)The expression sin(pi*t) / cos(pi*t) is the definition of tan(pi*t). (dy)/(dx) = tan(pi*t)

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Which of the following is the derivative (dy)/(dx) for the plane curve defined by the equations x(t)=-sin pi t, y(t)=cos pi t, and 0 <= t <= 2 ?
Select the correct answer below:
(A) -cot pi t
(B) cot pi t
(C) pi tan pi t
(D) tan pi t

Which of the following is the derivative $\frac{d y}{d x}$ for the plane curve defined by the equations $x(t)=-\sin \pi t$ , $y(t)=\cos \pi t$ , and $0 \leq t \leq 2$ ? $\newline$ Select the correct answer below: $\newline$ (A) $-\cot \pi t$ $\newline$ (B) $\cot \pi t$ $\newline$ (C) $\pi \tan \pi t$ $\newline$ (D) $\tan \pi t$

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Algebra 2

Inverses of sin, cos, and tan: degrees

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Q. Which of the following is the derivative

\frac{d y}{d x}

for the plane curve defined by the equations

x(t)=-\sin \pi t

y(t)=\cos \pi t

, and

0 \leq t \leq 2

\newline

Select the correct answer below:

\newline

(A)

-\cot \pi t

\newline

(B)

\cot \pi t

\newline

(C)

\pi \tan \pi t

\newline

(D)

\tan \pi t

Find $\frac{dx}{dt}$ : To find the derivative $\frac{dy}{dx}$ , we need to find $\frac{dy}{dt}$ and $\frac{dx}{dt}$ first and then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ .
Find $\frac{dy}{dt}$ : Let's find $\frac{dx}{dt}$ . Given $x(t) = -\sin(\pi*t)$ , we differentiate with respect to $t$ to get $\frac{dx}{dt}$ . $\newline$ $\frac{dx}{dt} = \frac{d}{dt} [-\sin(\pi*t)] = -\pi*\cos(\pi*t)$
Calculate $(\frac{dy}{dx}):</b> Now, let's find \$\frac{dy}{dt}$ . Given $y(t) = \cos(\pi*t)$ , we differentiate with respect to $t$ to get $\frac{dy}{dt}$ . $\newline$ $\frac{dy}{dt} = \frac{d}{dt} [\cos(\pi*t)] = -\pi*\sin(\pi*t)$
Simplify the expression: Now we have $\frac{dx}{dt}$ and $\frac{dy}{dt}$ . To find $\frac{dy}{dx}$ , we divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . $\newline$ $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\pi\sin(\pi t)}{-\pi\cos(\pi t)}$
Final Result: We can simplify the expression by canceling out the $-\pi$ terms in the numerator and the denominator. $\newline$ $\frac{dy}{dx} = \frac{\sin(\pi t)}{\cos(\pi t)}$
Final Result: We can simplify the expression by canceling out the $-\pi$ terms in the numerator and the denominator. $\newline$ $(dy)/(dx) = \sin(\pi*t) / \cos(\pi*t)$ The expression $\sin(\pi*t) / \cos(\pi*t)$ is the definition of $\tan(\pi*t)$ . $\newline$ $(dy)/(dx) = \tan(\pi*t)$

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