There was a review of sigma (summation) notation.
The scores on Quiz 12 are 28,27,26,24,23,21,20,18,17,16,14,14,13,13,10,6,1 with μ=17.1 and σ=6.9. Individual scores are on Blackboard.
Check your numbers on Blackboard to make sure you and I agree.
Today’s topics were antiderivatives and initial value problems (§4.9).
Here is a solution for Extra Credit 7.
First, find the equation of the tangent line at \((a,1/a^2)\) for \(a>0\). To do this, differentiate to see its slope is \(y'(a)=-2/a^3\). Since it passes through \((a,1/a^2)\), the point slope formula and a little algebra gives its equation as \[ y = -\frac{2}{a^3}x+\frac{3}{a^2}. \] The \(y\)-intercept is \(B=3/a^2\). Setting \(y=0\) and solving for \(x\) gives \(A=3a/2\). The distance between \(A\) and \(B\) is \[ D = \sqrt{\left(\frac{3a}{2}\right)^2+\left(\frac{3}{a^2}\right)^2} = \sqrt{\frac{9a^2}{4}+\frac{9}{a^4}},\quad a>0.\tag*{🏀} \] To find the shortest possible distance, find the critical numbers for the function \(D\). \[ \frac{dD}{da} = \frac{\frac{9 a}{2}-\frac{36}{a^5}}{2 \sqrt{\frac{9}{a^4}+\frac{9 a^2}{4}}} \] Set the numerator to \(0\) and solve to see the only critical number is \(a=\sqrt{2}\). Since there is clearly no longest such segment, this must be where the minimum occurs. Plugging this into 🏀 gives the minimum distance as \[ D(\sqrt{2})=\frac{3\sqrt{3}}{2} \] when \(A=3/\sqrt{2}\) and \(B=3/2\). The tangent line is \(y=-\frac{1}{\sqrt{2}}x+\frac{3}{2}\).
Quiz 12 may include topics through §4.8.
Today’s topics included Newton’s method (§4.8) and a short introduction to antidifferentiation (§4.9).
The scores on Quiz 11 are 30,28,27,26,25,24,23,23,20,18,18,17,17,16,15,12,10,7,7 with μ=19.1 and σ=6.9. Individual scores are on Blackboard.
After Quiz 11, the final example from the sheet of optimization examples was presented.
More optimization examples from the handout were solved.
Extra Credit 7. (Due November 13.) A tangent line to \(y=1/x^2\) intersects the \(x\)-axis at the point \(A\) and the \(y\)-axis at the point \(B\). What is the length of the shortest such line segment \(AB\)?
Quiz 11 may include topics through §4.6.
The first four optimization problems from a handout were solved (§4.7).
The scores on Quiz 10 are 30,30,30,30,30,30,30,29,29,28,28,27,27,24,24,23,15,7 with μ=26.2 and σ=6.1. Individual scores are on Blackboard.
More graphing was done (§4.5).
Here are two solutions for Extra Credit 6.
Complete the square in \(x\) to see the equation becomes
\[
(x-2)^2+y^2=1.\tag{1}
\]
This is a circle of radius \(1\) centered at \((2,0)\). Its graph and the two tangent lines are shown below.
Solution 1. Suppose a line passing through the origin is tangent to the circle at \(T=(a,b)\). It has an equation of the form \(y=mx\). To determine \(m\), implicitly differentiate (1) to see
\[
y'=-\frac{(x-2)}{y}.\tag{2}
\]
At \((a,b)\), the tangent line has slope \(m=-(a-2)/b\). Since the tangent line passes through \((a,b)\), which is on the line,
\[
b=-\frac{a-2}{b}a
\implies
b^2=2a-a^2.
\]
Substituting into (1), it follows that
\[
(a-2)^2+2a-a^2=1
\implies a=\frac{3}{2}.
\]
Putting this back into (1), we see \(b=\pm\sqrt{3}/2\). From (2), \(m=\pm1/\sqrt{3}\), and the two lines are
\[
y=\pm\frac{x}{\sqrt{3}}.
\]
Quiz 10 may include material through §4.4.
l’Hôspital’s Rule was introduced (§4.4).
The scores on Quiz 9 are 30,30,30,28,27,27,27,26,26,25,24,22,21,19,19,18,16,15,5 with μ=22.9 and σ=6.4. Individual scores are on Blackboard.
Examples of graphing functions using the first and second derivatives were presented (§4.3).
Extra Credit 6. (Due October 30.) There are two lines tangent to the curve \(x^2-4x+y^2+3=0\) that pass through the origin. Find their equations.
Quiz 9 may include topics through §4.2.
It was shown how the derivative indicates where a function is increasing or decreasing (§4.2).
Here is a solution for Extra Credit 5.
Let \(V\) be the volume of oil in the tank, \(A\) be the area of the top of the oil, \(r\) be the radius of the top of the oil and \(h\) be the depth of the oil.
Here is what we know:
\begin{align*}
&h=2r \tag{➀}\\
&V=\frac{1}{3}\pi r^2 h \tag{➁}\\
&A=\pi r^2 \tag{➂}\\
&\frac{dV}{dt}=-4 \tag{➃}
\end{align*}
We want to determine:
\[
\left.\frac{dA}{dt}\right\vert_{h=6,r=3}
\]
This will be done in two steps. The first step is to determine \(dr/dt\) and the second is to use \(dr/dt\) to find \(dA/dt\).
Use (➀) to write \(V\) as a function of \(r\) and differentiate both sides. Then use (➃) and a little algebra.
\begin{align*}
V&=\frac{2\pi}{3}r^3\\
\frac{dV}{dt}&=2\pi r^2\frac{dr}{dt}\\
-4&=2\pi r^2\frac{dr}{dt}\\
\frac{dr}{dt}&=-\frac{2}{\pi r^2} \tag{➄}
\end{align*}
Differentiate (➂) and substitute (➄).
\begin{align*}
\frac{dA}{dt}&=2\pi r\frac{dr}{dt}\\
&=
2\pi r\left(-\frac{2}{\pi r^2}\right)\\
&=
-\frac{4}{r}
\end{align*}
Finally,
\[
\left.\frac{dA}{dt}\right\vert_{h=6,r=3}
=
-\frac{4}{3}\text{ ft\(^2\)/min}
\]
The scores on Quiz 8 are 30,30,30,30,30,30,30,30,30,28,27,24,23,23,20,17,15,11,10,10 with μ=23.9 and σ=7.4. Individual scores are on Blackboard.
Rolle’s Theorem and the Mean Value Theorem were introduced (§4.2).
Topics included the extreme values of functions, Extreme Value Theorem, and Fermat’s Theorem (§4.1).
Here is a solution for Extra Credit 4.
Using the function for the shape of the slide, we see \[ \frac{dy}{dt}=-3e^{-x/3}\frac{dx}{dt}. \] Substitute the known value for \(dy/dt\) to get \[ -\frac{\sqrt{9-y}}{4}=-3e^{-x/3}\frac{dx}{dt} \implies \frac{dx}{dt}=\frac{e^{x/3}\sqrt{9-y}}{12} \] When \(x=9\) and \(y=9e^{-3}\) \[ \left.\frac{dx}{dt}\right\vert_{x=9,y=9/e^3}=\frac{e^3}{4}\sqrt{1-e^{-3}}\approx5\text{ ft/s } \]
Extra Credit 5. (Due October 20.) Oil is being withdrawn from a tank at 4 ft^{3}/min. The tank is shaped like a cone with height 10 ft and radius at the top 5 feet. How fast is the surface area on top of the oil in the tank changing when the oil is 6 ft deep?
Quiz 8 may include topics through §3.10.
Differentials were introduced along with their use in error approximation (§3.10).
Several more related rates examples were done (§3.9), followed by an introduction to tangent line approximation (§3.10).
The scores on Quiz 7 are 30,30,30,30,29,27,26,25,25,23,22,21,20,20,18,15,14,6,0 with μ=21.6 and σ=8.3. Individual scores are on Blackboard.
Extra Credit 4. (Due October 16.) A girl is sliding down a slide shaped like the curve \(y=9e^{-x/3}\) for \(0\le x\le9\). Her \(y\)-coordinate is changing at the rate \(dy/dt=(-1/4)\sqrt{9-y}\) ft/s. At what rate is her \(x\)-coordinate changing when she reaches the bottom of the slide?
Here is a solution for Extra Credit 3.
The tangent line at \(x=a\ne0\) passes through \((a,1/a)\) and has slope \(-1/a^2\). From the point-slope formula, its equation is seen to be
\[
y=-\frac{x}{a^2}+\frac{2}{a}.
\]
Since its \(y\) intercept is \(2/a\) and its \(x\) intercept is \(2a\), the area of the triangle is \[\frac{1}{2}(2a)\frac{2}{a}=2.\]
There will indeed be a Quiz 7 next Wednesday. It may include topics through §3.8.
Continuous interest was defined (§3.8). A few related rates examples were done (§3.9).
The scores on Quiz 6 are 30,30,30,30,30,29,29,28,27,22,20,20,20,19,19,19,16,9,5 with μ=22.7 and σ=7.4. Individual scores are on Blackboard.
We’re skipping §3.7. In class, we moved right on to §3.8 concerning exponential growth and decay.
Extra Credit 3. (Due October 6.) A triangle is formed by the coordinate axes and the tangent line to the graph of \(y=1/x\) at the point \((a,1/a)\), \(a\ne0\). What is the area of the triangle?
Quiz 6 may include topics through §3.5.
The topic of the day was differentiation of logarithms and exponentials (§3.6).
Here is a solution for Extra Credit 2.
If they are tangent to each other, then they have the same tangent line at \((1,0)\). This means they both pass through \((1,0)\) with the same derivative when \(x=1\). The two curves have derivatives \(y'=2x+a\) and \(y'=c-2x\). Using this information gives three equations with three unknowns. \begin{align*} &1+a+b=0\\ &c-1=0\\ &2+a=c-2\\ \end{align*} This system is easily solved to see \(a=-3,b=2,c=1\).
Implicit differentiation was used to find the derivatives of the inverse trigonometric functions.
The scores on Quiz 5 are 30 30,30,30,29,29,29,29,28,28,28,27,24,22,20,20,20,19,15,10,8,2 with μ=23.0 and σ=8.0. Individual scores are on Blackboard.
More examples of the chain rule were considered and implicit differentiation was introduced.
Extra Credit 2. (Due September 29.) Find constants \(a\), \(b\) and \(c\) so that the two curves \(y=x^{2}+ax+b\) and \(y=cx-x^{2}\) are tangent to each other at the point \((1,0)\).
Quiz 5 may include topicsthrough §3.3.
Derivatives of the rest of the trigonometric functions were done today, and then we moved on to the chain rule.
The scores on Quiz 4 are 30,30,30,30,30,30,25,25,24,23,23,20,20,20,20,18,16,16,15,13,12,10 with μ=21.8 and σ=6.5. Individual scores are on Blackboard.
Quiz 4 may include topics through §3.1.
The constant multiple and addition rules for differentiation were introduced. At this point you should be able to differentiate any polynomial without any trouble. After this, the exponential function and its derivative were introduced (§3.1).
Here is a solution for Extra Credit 1.
Any solution of the equation \(f(x)=x\) will also satisfy \(f(x)-x=0\), so the search for solutions to the original equation is the same as determining where \(f(x)-x=0\).
If \(g(x)=f(x)-x\), then \(g\) is continuous on \([0,1]\) because it is the sum of two continuous functions. If \(f(0)=0\) or \(f(1)=1\), then \(f(x)=x\) does have a solution in \([0,1]\). Otherwise, it must be the case that \(f(0)>0\) and \(f(1)\lt1\); i.e., \(g(0)=f(0)-0>0\) and \(g(1)=f(1)-1\lt0\). Since \(g\) is continuous, the Intermediate Value Theorem guarantees there is a \(c\in(0,1)\) with \(g(c)=0\). From the definition of \(g\), we see \(f(c)-c=0\), or \(f(c)=c\).
After the quiz, we discussed higher order derivatives, velocity and acceleration (§2.8).
The scores on Quiz 3 are 30,30,30,27,27,26,25,25,24,24,24,23,22,22,21,17,17,16,13,11,11,10 with μ=21.6 and σ=6.3. Individual scores are on Blackboard.
Today’s topic was derivatives as functions (§2.8). It was shown that if \(f'(x)\) exists, then \(f\) is continuous at \(x\). The function \(f(x)=|x|\) shows the converse is not true because \(f\) is continuous everywhere, but \(f'(0)\) does not exist.
Quiz 3 may include topics through §2.7.
Beginning on Monday, the class will meet in NS 128.
Today the definition of the derivative was given (§2.7). Remember that if \(f'(a)\) exists, then either \[ f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a} \text{ or } f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h} \] may be used to compute its value. We’ll use whichever is most convenient for the problem at hand.
The topics du jour were limits at infinity and asymptotes (§2.6).
Extra Credit 1. (Due September 15.) If \(f\) is a continuous function with \(f(0)\lt0\) and \(f(1)\gt1\), then why must the equation \(f(x)=x\) have a solution?
The scores on Quiz 2 are 30,30,30,30,30,30,30,28,27,26,24,23,23,23,20,20,20,18,16,10,10,0 with μ=22.6 and σ=8.0. Individual scores are on Blackboard.
On Wednesday, Quiz 2 may include topics through §2.5
Today’s topic was continuity and continuous functions (§2.5).
Here is a solution to problem 7 from §2.4.
§2.4, #7: For the limit \[ \lim_{x\to2}\left( x^3-3x+4 \right)=6 \] illustrate Definition 2 by finding values of \(\d\) that correspond to \(\e=0.2\) and \(\e=0.1\).
The solution given here is for \(\e=0.2\). The other case is similar.
Consider a graph of the function \(f(x)=x^3-3x+4\).
The horizontal lines are at \(6+\e=6.2\) and \(6-\e=5.8\). A graph at this scale isn’t very useful for determining \(\d\), so we zoom in to a small region near the point \((2,6)\).
From this second graph, it appears any positive \(\d\le0.02\) will suffice. To see this, note that if \(1.98\lt x\lt2.02\), then
\[
5.8\lt5.82\approx f(1.98)\lt f(x)\lt f(2.02)\approx 6.18
\lt 5.2.
\]
The scores on Quiz 1 are 30,30,30,25,25,25,25,23,22,20,20,20,20,20,20,18,18,18,17,13,12,12,11,6 with μ=20.0 and σ=6.2. Individual scores are on Blackboard.
After Quiz 1, the technical definition of the limit (§2.4) was introduced and discussed.
The topic today was calculating limits using the basic limit laws (§2.3). In particular, it was shown that when \(r(x)\) is a rational function and \(r(a)\) exists, then \(\lim_{x\to a}r(x)=r(a)\). (A special case of this is when \(r(x)\) is a polynomial.)
The due date for the homework assignment “Getting Started with WebAssign” has been extended to September 8 because some students didn’t have their WebAssign stuff squared away in time. The problems for the next several sections have also been posted.
Quiz 1 may include topics through §2.2.
Topics today included instantaneous velocity (§2.1) and the intuitive definition of limit (§2.2).
There are two categories of problems in the WebAssign assignment list, those labelled Homework and those labelled Practice. The ones with the Homework appelation are the ones that count in the WebAssign gradebook. The Practice ones don’t affect the gradebook. The Practice problems have a due date after the end of the semester as a workaround for some rigidity in the WebAssign environment.
If you had trouble with WebAssign yesterday, here’s why.
Dear Valued Customer,
Yesterday evening from approximately 4:45 PM ET to 6:00 PM ET and 8:30 PM ET to 11:49 PM ET students and instructors were unable to access the WebAssign platform. We sincerely apologize for this inconvenience.
We know that you and your students need reliable access to your course and learning tools and we take that responsibility seriously. Our team worked quickly to investigate and resolve the issue. As of 11:49 PM ET on August 23rd, WebAssign is now accessible and fully functional.
Our team will continue to work to prevent future interruptions from occurring to ensure that your students have the best learning experience with our platforms.
â€”The Cengage WebAssign Team
The tangent line and velocity problems were discussed.
If you’re working a problem on WebAssign, it might have a string of characters at the top of the window looking something like SCalcET8 5.2.002. You can translate this to a problem in the text.
SCalcET8 means it is from Stewart’s calculus text, 8th edition
5.2.002 means it is problem #2 at the end of §5.2
I’ve been getting so many emails from students telling me they won't be in class on Eclipse Day, that I've decided to move the first class period to Wednesday.
Here are some things to do in lieu of Monday’s class.
1. Read the syllabus. It is on Blackboard and can be found at the class Web site.
2. Chapter 1 in the text is prerequisite material. Look through that chapter to make sure you are familiar with it. In particular, trigonometry with radian measure is important; degree measure is almost never used in calculus. Understanding the inverse trigonometric functions and logarithms will also be necessary. There are some sugggested homework problems from the text on the class Web site. If you find there is a lot of material in Chapter 1 with which you are unfamiliar, then you’re in the wrong course!
3. Get squared away with WebAssign. If you are new to WebAssign, then you should watch some of the introductory videos they provide.
4. Watch the eclipse! Louisville will only get 96% obscurity of the sun, but it will still be spectacular. (Be sure to use appropriate eyewear or a pinhole camera.) There will be events surrounding the eclipse on campus at the planetarium.