# 1.3: If you thought practice makes perfect, you could be right

Calculus is an integral part of a meteorologist’s training. The ability to solve problems with calculus differentiates meteorologists from weather readers. You should know how to perform both indefinite and definite integrals. Brush up on the derivatives for variables raised to powers, logarithms, and exponentials. We will take many derivatives with respect to time and to distance.

#### Need Extra Practice?

Visit the Khan Academy website that explains calculus with lots of examples, practice problems, and videos. You can start with single variable calculus, but may find it useful for more complicated calculus problems.

Simple Integrals and Derivatives That are Frequently Used to Describe the Behavior of Atmospheric Phenomena

1. $$\frac{d a}{d t}=-k a$$

$$\frac{d a}{a}=-k d t$$

$$\int_{a_{o}}^{a_{1}} \frac{d a}{a}=-\int_{t_{o}}^{t_{1}} k d t$$

$$\ln \left(a_{1}\right)-\ln \left(a_{0}\right)=-k\left(t_{1}-t_{0}\right)$$

$$\ln \left(a_{1} / a_{0}\right)=-k\left(t_{1}-t_{0}\right)$$

$$a_{1} / a_{0}=e^{\left(-k\left(t_{1}-t_{0}\right)\right)}=\exp \left(-k\left(t_{1}-t_{0}\right)\right)$$

$$a_{1}=a_{0} e^{\left(-k\left(t_{1}-t_{0}\right)\right)}=a_{0} \exp \left(-k\left(t_{1}-t_{0}\right)\right)$$

2. $$p=p_{o} e^{(-z / H)} \quad ; \quad \int_{0}^{\infty} p d z=? \quad$$ (Do the definite integral.)

$$\int_{0}^{\infty} p d z=-\left.H p_{o} e^{-2 I H}\right|_{0} ^{\infty}=-H p_{o}(0-1)=p_{o} H$$

3. $$p=p_{0} e^{\left(-\frac{z}{H}\right)} ; \frac{1}{p} \frac{d p}{d z}=?$$

$$\frac{d p}{d z}=-\frac{1}{H} p_{0} e^{\frac{-z}{H}}=-\frac{1}{H} p ; \frac{1}{p} \frac{d p}{d z}=-\frac{1}{H}$$

4. $$\frac{d \ln (a x)}{d t}=? \quad \frac{d \ln (a x)}{d t}=\frac{1}{a x} \frac{d(a x)}{d t}=\frac{1}{a x} \frac{a d x}{d t}=\frac{1}{x} u,$$ where $$u=$$ velocity

5. $$d(\cos (x))=? \quad d(\cos (x))=-\sin (x) d x$$

### You have the power.

Often in meteorology and atmospheric science you will need to manipulate equations that have variables raised to powers. Sometimes, you will need to multiply variables at different powers together and then rearrange your answer to simplify it and make it more useful. In addition, it is very likely that you will need to invert an expression to solve for a variable. The following rules should remind you about powers of variables.

Laws of Exponents

\begin{aligned} a^{x} a^{y} &=a^{x+y} \\(a b)^{x} &=a^{x} b^{y} \\\left(a^{x}\right)^{y} &=a^{x y} \\ a^{-x} &=\frac{1}{a^{x}} \\ \frac{a^{x}}{a^{y}} &=a^{x-y} \\ a^{0} &=1 \end{aligned}

$$\left(\frac{a}{b}\right)^{x}=a^{x}\left(\frac{1}{b}\right)^{x}=\left(\frac{1}{a}\right)^{-x} b^{-x}=\left(\frac{b}{a}\right)^{-x}$$

\begin{aligned} \text { If } a=& b^{x}, \text { then raise both sides to the exponent } \frac{1}{x} \text { to move the } \\ & \text { exponent to the other side: } a^{\frac{1}{x}}=\left(b^{x}\right)^{\frac{1}{x}}=b^{\frac{x}{x}}=b \end{aligned}

If $$a^{x} b^{y}$$ , and you want to get an equation with a raised to no power,
then raise both sides to the exponent $$\frac{1}{x}$$ :


\left(a^{x} b^{y}\right)^{\frac{1}{x}}=\left(a^{x}\right)^{\frac{1}{x}}\left(b^{y}\right)^{\frac{1}{x}}=a b^{\frac{y}{x}}=\text { new constant }

This brief video (7:42) sums up these important rules:

Rules of Exponents

Click Answer for transcript of the Rules of Exponents.

Are you ready to give it a try? Solve the following problem on your own. After arriving at your own answer, click on the link to check your work. Here we go:

$$x=a y^{b}$$
What does y equal?

$$x^{1 / b}=\left(a y^{b}\right)^{1 / b}=a^{1 / b}\left(y^{b}\right)^{1 / b}=a^{1 / b} y$$
$$y=x^{1 / b} / a^{1 / b}=\left(\frac{x}{a}\right)^{1 / b}$$