AP Calculus AB Exam Question:

Consider the curve given by the equation:

(b) Determine whether the length of the curve from $x = 0$ to $x = \dfrac{\pi}{2}$ is greater than, equal to, or less than $\dfrac{\pi}{2}$.

Show all your work.

Answer:

(a) To find the length of the curve from $x=0$ to $x=\dfrac{\pi}{2}$, we will use the formula for arc length denoted by $L$: [L = \int_{a}^{b} \sqrt{1 + \left(\dfrac{dy}{dx}\right)^2} \ dx]

First, let's find the derivative $\dfrac{dy}{dx}$ of the equation $y = e^x\cos(x)$. Using the product rule and chain rule, we have:

Now we can substitute this derivative into the formula for arc length:

Simplifying the expression inside the square root, we have:

Now let's evaluate the integral. We denote $u = 1 + e^{2x}$, so $\dfrac{du}{dx} = 2e^{2x}$, which implies $dx = \dfrac{du}{2e^{2x}}$. Substituting this into the integral, we have:

Next, we need to express the limits of integration in terms of $u$ instead of $x$. We know that $u = 1 + e^{2x}$. Substituting $x = 0$, we get $u(0) = 1 + e^0 = 2$. Substituting $x = \dfrac{\pi}{2}$, we have $u(\frac{\pi}{2}) = 1 + e^{\pi} \approx 24.27$. Therefore, the integral becomes:

Applying the substitution rule, $\int f(g(x))g'(x) \ dx = \int f(u) \ du$, we have:

Simplifying this expression, we obtain the arc length of the curve from $x=0$ to $x=\dfrac{\pi}{2}$ as:

Since $L$ is less than $\dfrac{\pi}{2}$, we can conclude that the length of the curve is less than $\dfrac{\pi}{2}$.