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

- Page ID
- 3350

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.

**Answer**-
In this video we're going to be talking about all of the basic rules of exponents. And remember, when we're talking about exponents we can have an exponent here like X to the fourth where x is the base what we call the base and four is the exponent this small number in the upper right hand corner. It means that we're going to multiply X by itself four times or it means we have four factors of X multiplied together. So, if we expand this out its x times X times X times X. if we collapse it its X to the fourth. So, what happens when we do addition, subtraction, multiplication, and division of exponents? Well, in all cases we have to be really careful about like terms. For example, when we add terms that have exponents in them together both the bases and the exponents have to be the same in order for us to add them together. So, if we look at this first example 3x squared plus 2x squared the base here is X and the base here is X so the bases are the same which is good because we need that. and the exponents we have 2 and 2 which is good because we also need the exponents to be the same in order to add these together. So basically we have 3x squared added to 2x squared is going to give us five of them, 5x squared. So, if you're going to do addition and subtraction the bases and the exponents have to be the same. In this case we have X to the third plus x squared our bases X are the same but our exponents are different we have three and two. These are not like terms, so we can't add these together we can't simplify this at all. What happens when we do subtraction well again we're looking for similar basis so we have X and X for our base and then we have exponents of four and four. So because the bases and the exponents of the scene we can combine these like terms. We have six of them were subtracting and applied one of them which is going to leave us with five of them. So 5 times X to the fourth, but in this problem despite having the same base they will have a base of X we have different exponents we have a 4 and a 3 and because we're doing subtraction we can't combine these. We can't simplify this at all. What happens when we multiply two values together where exponents are involved? Well, here in order to simplify all we care about is that the bases are the same. The exponents do not have to be the same. So here we have base X and base X and we know already that's all we need to multiply these together it doesn't matter that the exponents are also the same we just add them. So we have three times to these are coefficients on our x squared terms. We multiply those together. So three times two is six, so that's going to be the first part and then we have x squared times x squared. And if we look at that x squared times x squared what we're going to do is add the exponents together. And the reason is because if we expand these out we know that x squared is two factors of X multiplied together. We're multiplying that by another x squared, so we're multiplying that by two more factors of X multiplied together. All together this is X to the fourth. Which we know because this essentially becomes the rule x to the a x x to the B is X to the a plus B. We just add the exponents together. So two plus two is four we get X to the fourth. Here's another example we have X to the third times x squared remember there's an implied one coefficient in front of both of these when we multiply 1 x 1 we get one so there will be a implied one coefficient on our final answer. x cubed + x squared. We just care that the bases are the same and they both have a base X so we know will be able to multiply them together. We have X to the third times x squared and remember that is going to be X to the three plus two so when we simplify we get X to the fifth and that should make sense because we have 3 factors of x x 2 factors of X adding them all up we get five factors of X so X to the fifth. The quotient rule for exponents tells us that in the same way as when we multiplied we didn't have to have the same exponent. When we divide we also don't have to have the same exponent we only care about the bases so here we have like basis. We have base X for both of these the exponents happened to be the same but that doesn't matter we're just going to leave this six and our final answer, so we'll get six here. And then what we're going to do is subtract the exponent in the denominator from the exponent in the numerator so the result is going to be X to the 4 minus 4. This is the four from the numerator this is the four from the denominator. 4-4 is 0 so we get 6 x 20 x to the 0 is 1 so this is 6 times 1 or just six. Even if we have different numbers again we only care about the bases both of these have the same base of X so again we'll just keep our two and our final answer and then we'll have X to the 4-3 because we say numerator exponent minus denominator exponent. That's going to give us 2 times X to the 4 minus 3 is 1. so X to the first which is of course just equal to 2x. What about a power raised to another power or an exponent raised to another exponent? Well, just like before in this example here when we said X to the fourth means multiply X by itself four times here we're saying multiply x squared by itself three times. So this is going to be equal to x squared times x squared times x squared and now we're really just back at this right here for multiplying like bases together and we add the exponents. So, this is just the same as X to the two plus two plus two. Two plus two plus two is six so we get x to the sixth power. What we realize then is that we can expand this and then add the exponents together using this rule over here or we can just multiply these two exponents together. Two times three gives us six and so we can do it that way as well. We can even do this when we have a negative base. So this problem here is telling us multiply 3 factors of negative x squared together so this is going to be negative x squared times negative x squared times negative x squared. We can deal with the negatives separately. Remember we can cancel every two negatives and they become a positive so negative and negative become a positive we're just left with this single negative sign here. so our answer will be negative and then x squared times x squared times x squared we know is X to the sixth. You can also think about it this way when you have this negative sign inside the parentheses. It's the same thing as saying negative 1 times x squared all raised to the third power and then you can apply this exponent to the negative 1 negative 1 times negative 1 times negative 1 is going to give you negative 1 which is this part right here. And then x squared to the third is going to be X to the 60 you get this X to the sixth and when you multiply them together you get negative x to the sixth. So those are just some of the most basic exponent rules that you need to know.

**Credit:****Krista King**

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?

**Answer**-
\(x^{1 / b}=\left(a y^{b}\right)^{1 / b}=a^{1 / b}\left(y^{b}\right)^{1 / b}=a^{1 / b} y\)

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

#### Quiz 1-2: Solving integrals and differentials.

Now it's time to to take another quiz. Again, I * highly* recommend that you begin by taking the Practice Quiz before completing the graded Quiz, since it will make you more competent and confident to take the graded Quiz : ).

- Go to the Canvas and find
**Practice Quiz 1-2**. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz. - When you feel you are ready, take
**Quiz 1-2**. You will be allowed to take this quiz only**once**. This quiz is timed, so after you start, you will have a limited amount of time to complete it and submit it. Good luck!