Exam 2-raw score
| min | max | median grade | Section 1.), raw | 2.) | 3.) | 4.) | |
|---|---|---|---|---|---|---|---|
| 31 | 107 | 75.5 | 30 points | 40 | 20 | 20 |
Exam 2 resubmissions
- Total raw points possible: 110. But I will throw out the question you scored lowest on. E.g. If you scored 75, and your lowest question was a 2/10, your final score is 73/100.
- If you scored less than 60% on any section, I encourage you to submit a resubmission of that section:
- Your final score for each section will be the average of your in-class and resubmission score. Example: you scored 8/20 on the final section. You resubmit it, and score 18/20. Your final score for that section will be $\frac{8+18}{2}=13$ out of 20... +5!
- Resubmissions are "open book": You may consult any written / printed online resource, but no other person. And I will re-collect your "in-class" exam today.
- For the derivatives (particularly section 4) show the intermediate steps you use to get to your final answer. Name the derivative rules you use. No credit will be given for answers without work shown.
- Resubmissions are due Friday, at the beginning of class.
- Section 3: Very few people scored more than half of the points possible on this section. I don't think I gave you enough opportunity to practice the kind of material in that section. So, instead of a resubmission, we will have a...
- Re-do of section 3 in Thursday's lab: Brief instruction today on using the derivative rules to calculate derivatives from table values. Then a re-do of section 3 (different questions) for full credit under test conditions (closed book) the first 15 minutes of Lab.
- If you are not re-doing section 3, you may come to lab 15 minutes after the start.
Using table values
Given the following table of values...
If $b(x)=f(g(x))$, calculate $b'(4)$.
- First, recognize and write down any useful derivative rules. Here we know that $b'(x)=[f(g(x))]'=f'(g(x))g'(x)$
- If you have a table of the values of the function and its derivatives, look up the values for the $x$ value requested: $$b'(4)=f'(g(4))g'(4)$$ No clue yet what to do with $f'$! but we can look up $g(4)=?$ and $g'(4)=?$".
- Subbing these in: $$b'(4)=f'(0)*(-4)$$.
- Perhaps now it's clearer what to do next with $f'$...[look up $f'(0)$]
- $$b'(4)=3*(-4)=-12.$$
Other problems like this you should be able to do:
- Let $c(x)=f(x)/g(x)$. Find $c'(2)$.
- What would you do if you are given the problem above, and have a table that only has data for $f(x)$, $g(x)$, and $g'(x)$?
Estimate $g(2.03)$ using this table of values:
For this problem $x=2.03$. That's close to a value in the table, call it $x_0=2.0$. You know that the slope of a line tangent to the function at $(x_0,g(x_0)$ has a slope $m=g'(x_0)$.
The "point-slope" form of the equation of a line tells you the other points $(x,y)$ on the same tangent line satisfy: $$g'(x_0)=m=\frac{\Delta y}{\Delta x}=\frac{y-g(x_0)}{x-x_0}.$$ These are the points on a tangent line. But as long as $x$ and $x_0$ are close together, $y(x)\approx g(x)$, and we have: $$g'(x_0)\approx\frac{g(x)-g(x_0)}{x-x_0}.$$ Re-arranging for $g(x)$ we get: $$g(x)=g(x_0)+\Delta g = g(x_0)+g'(x_0)(x-x_0)$$ [This is the same as the formula that we saw on Monday for changes of a function of 2 variable, but simpler for just one independent variable.]
Now we can take $x_0=2$ and $x=2.03$, and we get...
$$g(2.03)=g(2)+g'(2)\left(2.03-2)\right).$$ and subbing in values... $$g(2.03)=1+(-3)*0.03=1-0.09=0.91$$
Another problem like this:
- Given the table of values above, estimate $g'(2.03)$.
What is the slope of the line that is tangent to the graph of $g'(x)$ at $x_0=2$? What is the function that gives the "rate of change of $g'(x)$"?
In the table above, let's say that Sally is a cross country runner, and $x$ is the number of days after Sept 30. She notes her pace (time per mile) which is 8 minutes + $f(x)\times 10 s$ is the number of seconds *more* than 8 minutes per mile for her training run on that day. What does $f(1)=2$ mean?
Using the table of values
Estimate $\int_1^4 f(x)\,dx$.
Start by plotting the points you know:
We want the area "underneath" the graph, between $1\lt x\lt 4$:
=
The area above is equal to the sum of rectangles, where the height of the rectangle is the average of the value on the left and on the right of each sub interval. What is the width of each rectangle drawn?
$$\int_1^4 f(x)\,dx\approx 1*(+1)+1*(-1)+1*(-2.5)=-2.5$$
The width of the interval of integration is 3 units. What *are* the units?
The interval runs from $x$=1 to $x$=4. The units of $x$ are days.
The formula for average value of a function in an interval is $$f_\text{average}=\frac{1}{(b-a)}\int_a^b f(x)\,dx$$ In our case, what do you expect that the units of $f_\text{average}$ should be? What are the units and value of our interval from 1 to 4? Do the units work out right?
- The units of the integral are [units of $f$]$\cdot$[units of $x$]=sec$\times 10\cdot$days.
- The units of 1/interval are 1/[days].
- So, the units on the right are: $$\frac{1}{\text{days}}\cdot\text{sec}\times 10\cdot \text{days}$$
- The days cancel, and we're left with "sec$\times$10" which is the same as the units of $f_\text{average}$.
Calculate and interpret: $$f_\text{average}=\frac{1}{4-1}\int_1^4 f(x)\,dx$$
$$f_text{average}=\frac13\cdot(-2.5)=-2.5/3=-0.83\times\text{10 sec}=-8.3\text{ sec}$$ Sally's average pace for training runs from Oct 1 to Oct 4 was 8.3 sec *less* than 8 min/mile, that is, about 7 minutes and 52 seconds / mile.
Partial derivatives - algebraically
Meaning of partial derivatives
Assuming that $z=f(x,y)$,
$$f_x\equiv \frac{\del z}{\del x}$$
This is the rate of change of $f$ as $x$ changes (and $y$ remains constant).
Example: #1, $f=x^2+5xy+y^2$
The partial derivative w.r.t $\color{blue}{x}$ only has eyes for functions of $\color{blue}{x}$. Everything else is constants: $$\begineq f_{\color{blue} x}=&\frac{\del}{\color{blue}\del x}f(\color{blue}x\color{gray},y) =\color{gray} \frac{\del}{\color{blue}\del x} (\color{blue}{x^2} \color{black}+5y\color{blue}x\color{gray}+y^2)\\ =&\color{gray}{ \frac{\del}{\color{blue}{\del x}} \color{blue}{x^2} \color{black}+5y \frac{\del}{\color{blue}{\del x}}\color{blue}{x} \color{black}+\frac{\del}{\color{blue}{\del x}}y^2 }\\ =&\color{gray}{ \color{blue}{2x} \color{gray}+5y \color{blue}{1} +0 }\\ =& 2x+5y \endeq $$
The partial derivative w.r.t $\color{red}{y}$ only has eyes for functions of $\color{red}{y}$. Everything else is constants: $$\begineq f_{\color{red}y}=&\color{gray}\frac{\del}{\color{red}\del y}\color{gray}f(x,\color{red}y\color{black}) =\color{gray} \frac{\del}{\color{red}\del y} \color{gray}(x^2 +5x\color{red}y+y^2\color{black})\\ =&\color{gray} \frac{\del}{\color{red}\del y} x^2 +5x \frac{\del}{\color{red}\del y}\color{red}{y} +\frac{\del}{\color{red}\del y} \color{red}y^2 \\ =&\color{gray}{ 0+ +5x \color{red}{1} + \color{red}{2y} }\\ =& 5x+2y \endeq $$
Example: #2, $f=x^2\ln(x^2y)$
$$\begineq f_{\color{blue}x}=\color{gray} \frac{\del}{\color{blue}\del x}\color{blue}f =&\color{gray} \frac{\del}{\color{blue}\del x} [\color{blue}x^2\cdot\ln(x^2\color{gray}y)] \\ =&\color{gray} \left(\frac{\del}{\color{blue}\del x} \color{blue}x^2\color{black}\right) \cdot\ln(x^2\color{gray}y) +x^2\cdot \color{blue}\frac{\del}{\color{blue}\del x}\ln(\color{blue}x^2\color{gray}y) \color{black}\ \ \ \text{Product rule}\\ =&2x\cdot\ln(x^2y) + x^2\cdot\frac{1}{x^2y}\cdot \frac{\del}{\color{blue}\del x}(\color{blue}x^2\color{gray}y) \ \ \ \text{Power + Chain rule}\\ =& 2x\cdot\ln(x^2y) + x^2\cdot\frac{1}{x^2y}\cdot 2xy \\ f_x=& 2x\ln(x^2y) + 2x \endeq$$
$$\begineq f_{\color{red}y}\color{gray}= \frac{\del}{\color{red}\del y}f =&\color{gray} \frac{\del}{\color{red}\del y} [x^2\ln(x^2\color{red}y\color{gray})] \\ =&\color{gray}x^2 \frac{\del}{\color{red}\del y} \ln(x^2\color{red}y\color{gray}) \\ =&\color{gray}x^2\frac{1}{x^2y}\frac{\del}{\color{red}\del y} (x^2\color{red}{y}\color{black})\ \ \ \text{Chain rule} \\ =&x^2\cdot\frac{1}{x^2y} \cdot x^2\cdot \color{red}1 \\ =&\frac{x^2}{y} \endeq$$
Example: the ideal gas law
The ideal gas law (chemistry) links the pressure, volume, and temperature of a quantity $n$ (number of moles) of a gas. $$PV=nRT$$ where $R$ is a positive constant (called "the gas constant"!). Let's say that we have $n=1$, one mole of gas. Solve the gas law for pressure: $$P=\frac{RT}{V}$$ So, we can think of pressure, $P(T,V)$, as a function of two variables, temperature and volume. Now we calculate the partial derivatives: $$P_T=\frac{\del P}{\del T}=\frac{\del }{\del T}\left( \frac{RT}{V}\right)$$ Since we're taking the derivative with respect to $T$, the other variable, $V$ is constant. So we can use the constant multiple rule to write this as $$\frac{\del P}{\del T}=\frac{R}{V}\frac{\del }{\del T}\left(T\right)=\frac{R}{V}$$ $\Rightarrow$ The rate of change of pressure with respect to changes in temperature is a positive number. That is, heating a gas up while keeping the volume the same will increase the pressure of a gas!
Now, calculate the partial derivative wrt volume (while holding the temperature constant). $$\begineq P_V=&\frac{\del P}{\del V}=\frac{\del }{\del V}\left(\frac{RT}{V}\right) =RT\frac{\del }{\del V}\left(\frac{1}{V}\right)\\ =&RT*(-1)V^{-2}=-\frac{RT}{V^2}\\ \endeq $$
The rate of change, $P_V$ is negative. That means that if you hold the temperature constant but increase (expand) the volume, the pressure of the gas will go down.
Second order derivatives
$$f_{xx}\equiv (f_x)_x\approx \frac{\Delta f_x}{\Delta x}$$
This is the rate of change of $f_x$ as you change $x$.
(Is the surface concave up or down as you move in the $x$ direction?)
$$f_{xy}\equiv (f_x)_y\approx \frac{\Delta f_x}{\Delta y}$$
This is the rate of change of $f_x$ as you change $y$!
(In multi-variable calculus, we'll learn that this is the "twist" of the surface...)
- Graphically: $f_{xx}$ involves holding $y$ constant. At the origin, $y=0$ and so $f(x,0)=??$
- Graph $f(x,0)$.
- Now, using that function, evaluate $f_{xx} at $x=0$.
- Graphically: $f_{yy}$ involves holding $y$ constant. At the origin, $x=0$ and so $f(0,y)=??$
- Graph $f(0,y)$.
- Now, using that function, evaluate $f_{yy}$ at $y=0$.