Chapter 8-9 Test Review
Welcome to the Chapter 8-9 test review! This material covers key concepts, including arc length, surface area of revolution, probability using calculus, and solving separable equations. Let's get started!
Arc Length
The arc length formula allows us to calculate the length of a curve. If we have a function $y = f(x)$ on the interval $[a, b]$, the arc length $L$ is given by:
$$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$Example: Find the exact length of the curve $y = \ln(\sec(x))$, $0 \le x \le \frac{\pi}{6}$.
First, find the derivative:
$$\frac{dy}{dx} = \frac{\sec(x)\tan(x)}{\sec(x)} = \tan(x)$$Then, apply the arc length formula:
$$L = \int_0^{\pi/6} \sqrt{1 + \tan^2(x)} dx = \int_0^{\pi/6} \sqrt{\sec^2(x)} dx = \int_0^{\pi/6} \sec(x) dx$$ $$L = \ln|\sec(x) + \tan(x)| \Big|_0^{\pi/6} = \ln\left(\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{3}\right) - \ln(1 + 0) = \ln(\sqrt{3})$$Surface Area of Revolution
To find the surface area of a solid generated by revolving a curve $y = f(x)$ about the x-axis on the interval $[a, b]$, we use the following formula:
$$S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$Example: Find the exact area of the surface obtained by rotating the curve $y = \frac{x^3}{6} + \frac{1}{2x}$, $\frac{1}{2} \le x \le 1$ about the x-axis.
First, find the derivative:
$$\frac{dy}{dx} = \frac{x^2}{2} - \frac{1}{2x^2}$$Then, apply the surface area formula:
$$S = \int_{1/2}^1 2\pi \left(\frac{x^3}{6} + \frac{1}{2x}\right) \sqrt{1 + \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2} dx$$After simplification you find:
$$S = \frac{263\pi}{256}$$Probability
Calculus can be used to determine probabilities. The probability density function $f(x)$ describes the probability of a continuous random variable falling within a certain range. For $a \le X \le b$:
$$P(a \le X \le b) = \int_a^b f(x) dx$$A common example is the exponential density function: $f(t) = ce^{-ct}$ for $t \ge 0$.
Separable Equations
A separable equation is a differential equation that can be written in the form:
$$\frac{dy}{dx} = g(x)f(y)$$To solve, separate the variables and integrate both sides:
$$\int \frac{dy}{f(y)} = \int g(x) dx$$Example: Solve the differential equation $\frac{dP}{dt} = 7\sqrt{P}t$, given $P(1) = 6$.
Separate variables:
$$\frac{dP}{\sqrt{P}} = 7t dt$$Integrate both sides:
$$\int \frac{dP}{\sqrt{P}} = \int 7t dt$$ $$2\sqrt{P} = \frac{7}{2}t^2 + C$$Use the initial condition $P(1) = 6$ to find $C$:
$$2\sqrt{6} = \frac{7}{2}(1)^2 + C \implies C = 2\sqrt{6} - \frac{7}{2}$$Solve for P:
$$P = \left(\frac{7}{4}t^2 + \sqrt{6} - \frac{7}{4}\right)^2$$Good luck with your test! Remember to review these concepts and practice solving problems. You've got this!