Questions: Find dy/dt for y=cos^2(4πt+3).

Transcript text: Find $\frac{d y}{d t}$ for $y=\cos ^{2}(4 \pi t+3)$.

Solution

Solution Steps

To find $\frac{d y}{d t}$ for $y = \cos^2(4\pi t + 3)$, we will use the chain rule and the power rule. First, recognize that $y = (\cos(u))^2$ where $u = 4\pi t + 3$. The derivative of $(\cos(u))^2$ with respect to $u$ is $2\cos(u)(-\sin(u))$. Then, find the derivative of $u$ with respect to $t$, which is $4\pi$. Multiply these derivatives together to get $\frac{d y}{d t}$.

Step 1: Define the Function

We start with the function given by \[ y = \cos^2(4\pi t + 3). \]

Step 2: Apply the Chain Rule

To find $\frac{d y}{d t}$, we apply the chain rule. Let $u = 4\pi t + 3$. Then, we can express $y$ as \[ y = \cos^2(u). \] The derivative of $\cos^2(u)$ with respect to $u$ is \[ \frac{d y}{d u} = 2\cos(u)(-\sin(u)) = -2\cos(u)\sin(u). \]

Step 3: Differentiate $u$

Next, we differentiate $u$ with respect to $t$: \[ \frac{d u}{d t} = 4\pi. \]

Step 4: Combine Derivatives

Now, we combine the derivatives using the chain rule: \[ \frac{d y}{d t} = \frac{d y}{d u} \cdot \frac{d u}{d t} = -2\cos(u)\sin(u) \cdot 4\pi = -8\pi\sin(u)\cos(u). \]

Step 5: Substitute Back for $u$

Substituting back for $u$, we have: \[ \frac{d y}{d t} = -8\pi\sin(4\pi t + 3)\cos(4\pi t + 3). \]