RaySix

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at November 3rd 2023, 5:25:57 pm.

AP Calculus AB Exam Question:

A curve is defined by the equation:

y = 3x^2 - 4x + 1

a) Determine the length of the curve between $x = 0$ and $x = 2$ . Show all calculations and provide the exact answer.

b) Find the length of the curve between $x = -1$ and $x = 3$ . Give the answer to three decimal places.

Answer:

a) To find the length of the curve given by the equation, we will use the formula for the length of a curve:

L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx

where $\frac{dy}{dx}$ is the derivative of $y$ with respect to $x$ , and $a$ and $b$ are the two endpoints of the curve.

First, let's find $\frac{dy}{dx}$ by taking the derivative of $y$ with respect to $x$ :

\frac{dy}{dx} = \frac{d}{dx} \left(3x^2 - 4x + 1\right) = 6x - 4

Now, we can substitute this expression into the formula for the length of the curve:

L = \int_0^2 \sqrt{1 + \left(6x - 4\right)^2} dx

Next, we simplify the integrand by expanding the square:

L = \int_0^2 \sqrt{1 + 36x^2 - 48x + 16} dx = \int_0^2 \sqrt{36x^2 - 48x + 17} dx

To evaluate this integral, we can use a trigonometric substitution.

Let $u = 6x - 4$ , so $du = 6dx$ and $dx = \frac{du}{6}$ .

When $x = 0$ , $u = -4$ , and when $x = 2$ , $u = 8$ .

Substituting into the integral, we have:

L = \int_{-4}^8 \sqrt{u^2 + 17} \frac{du}{6}

Next, we simplify the integral:

L = \frac{1}{6} \int_{-4}^8 \sqrt{u^2 + 17} du

This integral can be evaluated by using the trigonometric substitution $u = 5\sqrt{2} \tan(\theta)$ .

Applying this substitution, we get:

du = 5\sqrt{2} \sec^2(\theta) d\theta

and

\sqrt{u^2 + 17} = \sqrt{50 \tan^2(\theta) + 17}

Simplifying further:

L = \frac{1}{6} \int_{-4}^8 \sqrt{50 \tan^2(\theta) + 17} \cdot 5\sqrt{2} \sec^2(\theta) d\theta

L = \frac{5\sqrt{2}}{6}\int_{-4}^8 \sec^3(\theta) d\theta

Using the reduction formula for integrating $\sec^n(\theta)$ , where $n$ is an odd number:

\int \sec^n(\theta) d\theta = \frac{1}{n-1}\sec^{n-2}(\theta)\tan(\theta) + \frac{n-2}{n-1}\int \sec^{n-2}(\theta) d\theta

Using this reduction formula with $n=3$ :

\int \sec^3(\theta) d\theta = \frac{1}{2}\sec(\theta)\tan(\theta) + \frac{1}{2}\int \sec(\theta) d\theta

Now, we can substitute back and evaluate the definite integral:

L = \frac{5\sqrt{2}}{6}\left[ \frac{1}{2}\sec(\theta)\tan(\theta) + \frac{1}{2}\ln|\sec(\theta) + \tan(\theta)| \right]_{-4}^8

Using the limits of integration, we find:

\sec(\theta) = \frac{5\sqrt{2}}{4}

\tan(\theta) = \frac{3\sqrt{2}}{4}

\ln|\sec(\theta) + \tan(\theta)| = \ln\left(\frac{7\sqrt{2}}{4}\right)

Substituting these values back into the expression:

L = \frac{5\sqrt{2}}{6}\left[ \frac{1}{2}\frac{5\sqrt{2}}{4}\frac{3\sqrt{2}}{4} + \frac{1}{2}\ln\left(\frac{7\sqrt{2}}{4}\right) \right]

L = \frac{25}{6} + \frac{5}{6}\ln\left(\frac{7\sqrt{2}}{4}\right) \approx 10.498

Therefore, the length of the curve between $x = 0$ and $x = 2$ is approximately 10.498 units.

b) We can use the same formula for the length of the curve to find the length between $x=-1$ and $x=3$ :

L = \int_{-1}^3 \sqrt{1 + \left(6x - 4\right)^2} dx

Following the same steps as in part a, we obtain:

L = \frac{5\sqrt{2}}{6}\left[ \frac{1}{2}\frac{3\sqrt{2}}{4}\frac{9\sqrt{2}}{4} + \frac{1}{2}\ln\left(\frac{11\sqrt{2}}{4}\right) \right]

L = \frac{135}{32} + \frac{5}{6}\ln\left(\frac{11\sqrt{2}}{4}\right) \approx 8.025

Therefore, the length of the curve between $x = -1$ and $x = 3$ is approximately 8.025 units.