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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 dydx \frac{d y}{d x} for the plane curve defined by the equations x(t)=sinπt x(t)=-\sin \pi t , y(t)=cosπt y(t)=\cos \pi t , and 0t2 0 \leq t \leq 2 ?\newlineSelect the correct answer below:\newline(A) cotπt -\cot \pi t \newline(B) cotπt \cot \pi t \newline(C) πtanπt \pi \tan \pi t \newline(D) tanπt \tan \pi t

Full solution

Q. Which of the following is the derivative dydx \frac{d y}{d x} for the plane curve defined by the equations x(t)=sinπt x(t)=-\sin \pi t , y(t)=cosπt y(t)=\cos \pi t , and 0t2 0 \leq t \leq 2 ?\newlineSelect the correct answer below:\newline(A) cotπt -\cot \pi t \newline(B) cotπt \cot \pi t \newline(C) πtanπt \pi \tan \pi t \newline(D) tanπt \tan \pi t
  1. Find dxdt\frac{dx}{dt}: To find the derivative dydx\frac{dy}{dx}, we need to find dydt\frac{dy}{dt} and dxdt\frac{dx}{dt} first and then divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt}.
  2. Find dydt\frac{dy}{dt}: Let's find dxdt\frac{dx}{dt}. Given x(t)=sin(πt)x(t) = -\sin(\pi*t), we differentiate with respect to tt to get dxdt\frac{dx}{dt}.\newlinedxdt=ddt[sin(πt)]=πcos(πt)\frac{dx}{dt} = \frac{d}{dt} [-\sin(\pi*t)] = -\pi*\cos(\pi*t)
  3. Calculate (dydx):</b>Now,letsfind$dydt(\frac{dy}{dx}):</b> Now, let's find \$\frac{dy}{dt}. Given y(t)=cos(πt)y(t) = \cos(\pi*t), we differentiate with respect to tt to get dydt\frac{dy}{dt}.\newlinedydt=ddt[cos(πt)]=πsin(πt)\frac{dy}{dt} = \frac{d}{dt} [\cos(\pi*t)] = -\pi*\sin(\pi*t)
  4. Simplify the expression: Now we have dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}. To find dydx\frac{dy}{dx}, we divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt}. \newlinedydx=dydtdxdt=πsin(πt)πcos(πt)\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\pi\sin(\pi t)}{-\pi\cos(\pi t)}
  5. Final Result: We can simplify the expression by canceling out the π-\pi terms in the numerator and the denominator.\newlinedydx=sin(πt)cos(πt)\frac{dy}{dx} = \frac{\sin(\pi t)}{\cos(\pi t)}
  6. Final Result: We can simplify the expression by canceling out the π-\pi terms in the numerator and the denominator.\newline(dy)/(dx)=sin(πt)/cos(πt)(dy)/(dx) = \sin(\pi*t) / \cos(\pi*t) The expression sin(πt)/cos(πt)\sin(\pi*t) / \cos(\pi*t) is the definition of tan(πt)\tan(\pi*t).\newline(dy)/(dx)=tan(πt)(dy)/(dx) = \tan(\pi*t)

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