The following material should be read prior to attending lab. You are responsible for preparing for lab so that you don't slow down your group.
In calculus, we often need to solve an equation for a specific variable. You have solved equations such as $x^2 - x - 6 = 0$ many times in your mathematics courses. In calculus sometimes we encounter more complicated equations involving logarithms, exponentials, and trigonometric functions. There will also be situations where you may want to solve an equation with multiple variables for a specific variable and you may want to solve a system of equations to find where two curves intersect. The following examples will illustrate some of these situations that you may encounter in first semester calculus.
Using the properties of logarithms, we have that $\log[(x - 3) \cdot (x)] = 1$. Since the base on log is 10, we will rewrite the logarithmic equation as an exponential equation and solve. This gives us the following solution.
| $\log[(x - 3) \cdot (x)]$ | $=$ | $1$ | We use the laws of logarithms to combine the terms into one logarithm. | |
| Then we rewrite the equation as exponential form rather than logarithmic form (see Lab 3). | $10^1$ | $=$ | $(x - 3)(x)$ | |
| $10$ | $=$ | $x^2 - 3x$ | ||
| $0$ | $=$ | $x^2 - 3x - 10$ | Once we recognize it is a quadratic equation, we know that we must solve by having everything on one side and zero on the other. | |
| $0$ | $=$ | $(x - 5)(x + 2)$ | ||
| $ \begin{array}{rcl} x - 5 & = & 0 \\ x & = & 5 \end{array}$ | or or | $\begin{array}{rcl} x + 2& = & 0 \\ x &=& -2 \end{array}$ |
In calculus, we usually work with natural logarithms, which can be thought of as $\log_e(x)$ but is always written as $\ln(x)$.
| $\ln(x)$ | $=$ | $-5$ | ||
| Again, we recognize that we need to use exponentials to solve so we raise both sides to the power of e. | $e^{\ln(x)}$ | $=$ | $e^{-5}$ | |
| $x$ | $=$ | $e^{-5}$ | Then we remember Property 5 from Lab 3, $e^{\ln(x)} = x$. | |
| $x$ | $=$ | $\frac{1}{e^5}$ |
This is an exponential equation and we will need to use a property of logarithms (Property 4 from Lab 3) to solve the equation. To do this, we will take the natural logarithm of both sides, which will adhere to the algebra rule that "whatever you do to one side, you must do to the other side." We can apply the natural log to both sides and get a unique answer because the natural logarithmic function is one-to-one and onto. This gives us the equation $\ln(e^{5730k}) = \ln\!\left(\frac{1}{2}\right)$. Continuing the solving process gives us the following result.
Example 7.3 is the type of equation that you would need to solve when dealing with a radioactive decay problem. A problem involving population growth would have a similar solution process. Those are the best real world examples of exponential functions.
Another exponential equation that you will often encounter in Calculus I can be seen in Example 7.4.
First, we may want to rewrite the equation, $2te^{2t} + e^{2t} = 0$. Now in order to solve for t, we need to apply one of the rules of factoring, factor out all common factors first. Then solving we will get the following.
| $2te^{2t} + e^{2t}$ | $=$ | $0$ | ||
| $e^{2t}(2t + 1)$ | $=$ | $0$ | ||
| Looking at this equation and remembering the graph of the exponential function and its range, we should realize that $e^{2t} \neq 0$ for any values of t. | $e^{2t} = 0$ | or | $ \begin{array}{rcl} 2t + 1 & = & 0 \\ 2t &= & -1 \\ t & = & \frac{-1}{2} \\ \end{array} $ |
Other equations that are often solved in calculus involve trigonometric functions as seen in the next example. These examples are particularly important when finding relative maxima and minima.
(a) Solve the following equation for x.
This problem involves factoring (see Lab 1) and remembering your unit circle (see the Prerequisite Material).
| $-2\, \sin(x) + 2\, \sin(x) \cos(x)$ | $=$ | $0$ | ||
| We factor out a common factor of $-2\, \sin(x)$. | $-2\, \sin(x)(1 - \cos(x))$ | $=$ | $0$ | |
| $\begin{array}{rcl}-2\, \sin(x) & = & 0 \\ \sin(x) &=& 0\end{array}$ | or or | $ \begin{array}{rcl}1 - \cos(x) & = & 0 \\1 &=& \cos(x) \end{array} $ | ||
| $\sin(x) = 0$ when $x = k\pi, k \in \mathbb{Z}$ $\cos(x) = 1$ when $x = 2k\pi, k \in \mathbb{Z}$ | ||||
Therefore, our equation has infinitely many solutions and since the solution set of the second equation is a subset of the solution set of the first equation, we can write our solution set as follows.
| $\{x | x = k\pi, k \in \mathbb{Z}\}$ | The notation $k \in \mathbb{Z}$ means k is an element of the integers. Therefore, the solutions set is the set of all integer multiples of $\pi$. |
(b) Find all solutions to the following equation on the interval $\left[\frac{\pi}{4}, \frac{7\pi}{4}\right]$.
| $1 - \frac{1}{2}\, \csc^2\!\left(\frac{x}{2}\right)$ | $=$ | $0$ | ||||
| $1$ | $=$ | $\frac{1}{2}\, \csc^2\!\left(\frac{x}{2}\right)$ | ||||
| $2 \cdot 1$ | $=$ | $2 \cdot \frac{1}{2}\, \csc^2\!\left(\frac{x}{2}\right)$ | ||||
| We take the square root of both sides of the equation. | $2$ | $=$ | $\csc^2\!\left(\frac{x}{2}\right)$ | |||
| $\pm\sqrt{2}$ | $=$ | $\csc\!\left(\frac{x}{2}\right)$ | The cosecant function is the reciprocal of the sine function. | |||
| $\pm\sqrt{2}$ | $=$ | $\frac{1}{\sin\left(\frac{x}{2}\right)}$ | ||||
| $\sin\!\left(\frac{x}{2}\right)$ | $=$ | $\pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$ | ||||
This gives us that $\frac{x}{2} = \frac{\pi}{4}$, $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, $\frac{7\pi}{4}$, $\ldots $
Then solving for x yields $x = \frac{2\pi}{4}$, $\frac{6\pi}{4}$, $\frac{10\pi}{4}$, $\frac{14\pi}{4}$, $\ldots$ and simplifying we have that $x = \frac{\pi}{2}$, $\frac{3\pi}{2}$, $\frac{5\pi}{2}$, $\frac{7\pi}{2}$, $\ldots$ but the only solutions that are in the interval are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.