a curve in the plane is defined parametrically by the equations x = tan(2t) and y = sin(4t - π/2). find the…

Answer:

B. $- 2$

Explanation:

Step1: Find $\frac{dx}{dt}$

Differentiate $x = \tan(2t)$ using chain - rule. If $u = 2t$, then $\frac{dx}{dt}=\sec^{2}(2t)\cdot2 = 2\sec^{2}(2t)$.

Step2: Find $\frac{dy}{dt}$

Differentiate $y=\sin(4t-\frac{\pi}{2})$ using chain - rule. Let $u = 4t-\frac{\pi}{2}$, then $\frac{dy}{dt}=\cos(4t - \frac{\pi}{2})\cdot4=4\cos(4t-\frac{\pi}{2})$.

Step3: Use the formula $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\frac{dy}{dx}=\frac{4\cos(4t-\frac{\pi}{2})}{2\sec^{2}(2t)}=\frac{2\cos(4t - \frac{\pi}{2})}{\sec^{2}(2t)}$.

Step4: Evaluate at $t=\frac{\pi}{2}$

First, when $t = \frac{\pi}{2}$, $\cos(4t-\frac{\pi}{2})=\cos(4\times\frac{\pi}{2}-\frac{\pi}{2})=\cos(\frac{7\pi}{2}) = 0$. But we can also use trigonometric identities. We know that $\sin(A - B)=\sin A\cos B-\cos A\sin B$, so $y=\sin(4t-\frac{\pi}{2})=-\cos(4t)$. And $\frac{dy}{dt}=4\sin(4t)$. Also, $\frac{dx}{dt}=2\sec^{2}(2t)$. When $t=\frac{\pi}{2}$, $\frac{dx}{dt}=2\sec^{2}(\pi)=2\times1 = 2$, and $\frac{dy}{dt}=4\sin(2\pi)=0$. Another way: We know that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. $\frac{dx}{dt}=2\sec^{2}(2t)$ and $\frac{dy}{dt}=4\cos(4t-\frac{\pi}{2})$. When $t = \frac{\pi}{2}$, $\frac{dx}{dt}=2\sec^{2}(\pi)=2$ and $\frac{dy}{dt}=4\cos(\frac{7\pi}{2}) = 0$. Using the formula $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, we can also use the fact that $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$. We know that $\sin(4t-\frac{\pi}{2})=-\cos(4t)$. So $\frac{dy}{dt}=4\sin(4t)$ and $\frac{dx}{dt}=2\sec^{2}(2t)$. When $t=\frac{\pi}{2}$, $\frac{dx}{dt}=2$ and $\frac{dy}{dt}=- 4$. So $\frac{dy}{dx}=\frac{-4}{2}=-2$.