The scores on the test are 92,84,82,76,75,73,73,72,60,59,58,57,49,49,46,42,27,12,3 with μ=57.3 and σ=24.0. Individual scores are on Blackboard.
Here is a solution for Extra Credit 7.
Since \(f(x)=ax^2+bx+c\), we see \(f'(x)=2ax+b\). Using these, the three given conditions translate to \begin{align*} &f(0)=9 && a\cdot0^2+b\cdot0+c=9 && c=9\\ &f'(2)=0 &\implies\quad & 2a\cdot2+b=0 &\implies\quad& 2a+b=1\\ &f(2)=1 && a\cdot2^2+b\cdot2+c=1 && 4a+2b+c=1 \end{align*} Solving the rightmost system of equations gives \(a=2,b=-8,c=9\). Therefore, \[ f(x)=2x^2-8x+9. \]
The scores on Quiz 12 are 27,26,25,23,23,22,22,21,21,19,17,15,13,10,9,4,1 with μ=17.5 and σ=7.7. Individual scores are on Blackboard.
The Fundamental Theorem of Calculus was introduced.
Quiz 12 may include topics through §5.1.
Extra Credit 7. (Due April 20.) Find \(a\), \(b\) and \(c\) such that the graph of \(f(x)=a x^2+b x +c\) passes through \((0,9)\) and \(f\) has a minimum value of \(1\) at \(2\).
Quiz 11 had the following three problems from WebAssign.
The scores on Quiz 11 are 30,30,27,27,26,24,24,19,17,16,16,13,12,11,10,4,2,1,0,0 with μ=15.4 and σ=10.3. Individual scores are on Blackboard.
Here is a solution for Extra Credit 6.
Suppose the circle has radius \(r\) and is situated as in the figure. Then, we know \begin{align}\label{ec8:y} y=\sqrt{r^2-x^2} \tag{1} \end{align} and the perimeter is \begin{align}\label{ec8:p} p=2r+\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}.\tag{2} \end{align} Substitute (1) into (2) to eliminate \(y\): \begin{align*} p&=2r+\sqrt{(x+1)^2+r^2-x^2}+\sqrt{(x+1)^2+r^2-x^2}\\ &=2r+\sqrt{1+r^2+2x}+\sqrt{1+r^2-2x} \end{align*} We must find the maximum value of \(p\) for \(0\le x\le r\). To do so, find the derivative. \begin{align*} \frac{dp}{dx}&=\frac{1}{\sqrt{1+r^2+2x}}-\frac{1}{\sqrt{1+r^2-2x}}\\ &= \frac{\sqrt{1+r^2-2x}-\sqrt{1+r^2+2x}}{\sqrt{(1+r^2+2x)(1+r^2-2x)}} \end{align*} The derivative exists everywhere for \(0\le x\le r\), so we search for critical numbers by finding where the numerator is \(0\). \begin{align*} \sqrt{1+r^2-2x}-\sqrt{1+r^2+2x}=0\\ \sqrt{1+r^2-2x}=\sqrt{1+r^2+2x}\\ {1+r^2-2x}={1+r^2+2x}\\ x=0 \end{align*} This shows the isosceles triangle with vertex at \((0,r)\) in the figure has the largest perimeter.
Quiz 11 may include topics through §4.7.
Examples of optimization problems (§4.7) were presented.
The scores on Quiz 10 are 30,30,27,26,26,26,26,25,23,23,22,20,20,18,17,13,11,6,5,0 with μ=19.7 and σ=8.5. Individual scores are on Blackboard.
Check your scores on Blackboard to make sure you and I agree on the numbers.
l’Hospital’s rule was introduced (§4.4) and several examples showing different indeterminate forms were presented.
Extra Credit 6. (Due April 9.) A fixed circle lies in the plane. A triangle is drawn inside the circle with all three vertices on the circle and two of the vertices at the ends of a diameter. Where should the third vertex lie to maximize the perimeter of the triangle?
Quiz 10 may include topics through §4.3.
Here is a solution for Extra Credit 5.
Let \(h\) and \(r\) be as in the figure, \(V\) the volume of water in the tank and \(a\) be the rate at which water is being drained. Using similar triangles from the figure \begin{align} \frac{4}{16}=\frac{r}{h}\implies r=\frac{h}{4}.\tag{⚾} \end{align} The volume of water in the tank is \(V=\frac{1}{3}\pi r^2h\). Using (⚾) to eliminate \(r\), this becomes \begin{align*} V=\frac{\pi}{48}h^3\implies \frac{dV}{dt}=\frac{\pi}{16}h^2\frac{dh}{dt}. \end{align*} Since the rate of change of the volume of water in the tank is the rate at which it is entering minus the rate at which it is leaving, \(dV/dt=10-a\). Combining these two expressions for \(dV/dt\) gives \[ \frac{\pi}{16}h^2\frac{dh}{dt}=10-a\implies a=10-\frac{\pi}{16}h^2\frac{dh}{dt}. \] At the instant when \(h=12\) and \(dh/dt=1/3\), \[ a=10-\frac{\pi}{16}(12)^2\frac{1}{3}=10-3\pi\text{ ft\(^3\)/min}. \] (This is about \(0.575\)ft\(^3\)/min.)
The problems on Quiz 9 were copied from the following:
The scores on Quiz 9 are 30,30,28,26,25,20,19,15,15,14,13,13,12,12,10,8,8,4,1,0 with μ=15.2 and σ=9.1. Individual scores are on Blackboard.
Three topics were discussed:
The problems on Quiz 8 were all from homework assignments:
The scores on Quiz 8 are 30,28,28,26,24,23,20,19,15,15,14,13,13,10,5,3,2,1,0,0 with μ=14.5 and σ=10.1. Individual scores are on Blackboard.
Quiz 9 may contain topics through §4.1.
Extra Credit 5. (Due Mar 29.) A water tank has the shape of a cone \(16\) ft high with a radius of \(4\) ft at the top. Water is being pumped in at \(10\) ft\(^3\) per minute and is being removed from the bottom. At the time the water is \(12\) ft deep, the depth is increasing at \(4\) inches per minute. How fast is the water being removed?
Quiz 8 has been moved to Friday.
Several examples of finding the extreme values of continuous functions on closed intervals were presented (§4.1). After this, Rolle’s Theorem and the Mean Value Theorem were introduced (§4.2).
Here is a solution for Extra Credit 4.
Solution. Let \(a\) be the distance of the car from the intersection, \(b\) be the distance of the truck from the intersection and \(s\) be the distance between them. By assumption \begin{align}\label{ec3eq1} \frac{da}{dt}=-50\text{ and }\frac{db}{dt}=-40. \tag{1} \end{align} From the Law of Cosines, \[ s^2=a^2+b^2-2ab\cos\left(\frac{\pi}{3}\right). \] This can be simplified. \begin{align}\label{ec3eq2} s^2=a^2+b^2-ab \tag{2} \end{align} We want to find \(ds/dt\) when \(a=b=s=2\). To do this, differentiate \eqref{ec3eq2} and substitute \eqref{ec3eq1}. \begin{align*} &2s\frac{ds}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}-\frac{da}{dt}b-a\frac{db}{dt}\\ &\frac{ds}{dt}=-\frac{30a+15b}{s} \end{align*} Wnen \(a=b=s=2\), it follows that \(\displaystyle \frac{ds}{dt}=-45\) mph.
Quiz 8 may include topics through §3.10.
We finished the section on linear approximations by considering differential error (§3.10) and moved on to considering the extreme values of a function (§4.1). Notice we’re skipping §3.11.
The scores on Quiz 7 are 30,30,30,25,24,23,22,22,21,17,15,14,8,8,6,4,2,1,1 with μ=15.9 and σ=10.2. Individual scores are on Blackboard.
Extra Credit 4. (Due Mar 19.) Two roads meet at an angle of \(60^\circ\). A truck is approaching the intersection at 40 mph and a car is approaching the intersection on the other road, but on the same side of the intersection as the truck, at 50 mph. Suppose, at a particular time, both are \(2\) miles from the intersection. How fast is the distance between them changing at that time?
Five examples of related rate problems were solved (§3.9).
Quiz 7 may include topics through §3.8.
Here is a solution for Extra Credit 3.
In the figure given above, the red and green lines are the tangents to \(y=x^3\) and \(xy=1\), respectively. Let \(\theta_1\) be the angle between the red tangent and the \(x\)-axis and \(\theta_2\) be the angle between the green tangent and the \(x\)-axis. Using derivatives, it’s easy to see \(\tan\theta_1=3\) and \(\tan\theta_2=-1\). If \(\theta\) is the angle between the lines, then \[ \tan\theta=\tan(\theta_2-\theta_1)=\frac{\tan\theta_2-\tan\theta_1}{1+\tan\theta_2\tan\theta_1} =2 \] Therefore, \(\theta=\tan^{-1} 2\approx1.1075\) radians, or about \(63.4349^\circ\).
The scores on Quiz 6 are 30,30,29,27,27,26,25,24,22,21,16,14,13,12,12,11,4,2,0 with μ=18.2 and σ=9.6. Individual scores are on Blackboard.
Quiz 6 may include topics through §3.6.
Extra Credit 3. (Due Mar 2.) The curves \(xy=1\) and \(y=x^3\) cross in the first quadrant. What is the angle between their tangent lines where they cross?
Derivatives of the inverse trigonometric functions and \(\ln x\) were derived (§3.5, §3.6).
The scores on Quiz 5 are 30,29,29,27,26,26,25,25,24,23,20,19,17,17,16,15,14,14,12,8,7,6,2,0 with μ=18.0 and σ=8.8. Individual scores are on Blackboard.
More was done with implicit differentiation and the inverse trigonometric functions were reviewed.
Here is a solution for Extra Credit 2.
From the chain rule and the product rule \begin{align*} \frac{d}{dx}(F^3+&G^3+H^3-3FGH)\\ &= 3F^2F'+3G^2G'+3H^2H'-3(F'GH+FG'H+FGH')\\ &= 3F^2G+3G^2H+3H^2F-3(G^2H+FH^2+F^2H)\\ &= 3F^2G+3G^2H+3H^2F-3G^2H-3FH^2-3F^2H\\ &= 0. \end{align*}
Quiz 5 may include topics through §3.4
The main topic of the day was the chain rule. Implicit differentiation was introduced near the end of the class period.
After Quiz 4, the Chain Rule was introduced.
The scores on Quiz 4 are 30,30,30,30,30,29,29,28,28,27,27,25,25,24,24,22,20,19,18,10,10,9,8,5 with μ=22.4 and σ=8.1. Individual scores are on Blackboard.
Extra Credit 2. (Due Feb 19.)
Given functions \(F\), \(G\) and \(H\) satisfying
\[
F'=G,\quad G'=H,\quad H'=F
\]
determine
\[
\frac{d}{dx}(F^3+G^3+H^3-3FGH).
\]
Here is a solution for Extra Credit 1.
This equation can be written as one quadratic in form as follows. \begin{align*} 5^x+2=3(5^{-x})\\ 5^{2x}+2(5^x)=3\\ 5^{2x}+2(5^x)-3=0 \tag{1}\label{p1:line3}\\ \left(5^x+3\right)\left(5^x-1\right)=0\\ 5^x=-3,1 \end{align*} Since \(5^x>0\) for all \(x\), the first is clearly an extraneous solution, so \(5^x=1\) is the only solution to the quadratic. This only happens when \(x=0\). Therefore, \(x=0\) is the only solution to the original equation.
Another way to see this is the only solution is to look at the left-hand side of \eqref{p1:line3} and let \(f(x)=5^{2x}+2(5^x)-3\). The solutions of the equation are where the graph of \(f\) crosses the \(x\)-axis. Examining the graph, given below, shows the only solution is \(x=0\).
Quiz 4 may include topics through §3.3.
The limits \[ \lim_{\theta\to0}\frac{\sin\theta}{\theta}=1 \text{ and } \lim_{\theta\to0}\frac{\cos\theta-1}{\theta}=0 \] were found and used to find the derivatives of the trigonometric functions (§3.3).
Things accomplished included more examples using the product rule, the quotient rule with examples and a few examples showing all the rules mixed together (§3.2). On Friday the trigonometric functions will be differentiated (§3.3).
The scores on Quiz 3 are 30,30,30,30,30,30,30,28,26,24,23,21,20,20,19,13,13,10,6,3,3,2,2,0 with μ=18.5 and σ=11.0. Individual scores are on Blackboard.
The number \(e\) was introduced. It was shown that if \(f(x)=e^x\), then \(f'(x)=f(x)\). After that, the product rule was derived.
Extra Credit 1. (Due Feb 12.) Find all solutions of the equation \(5^x+2=3(5^{-x})\). Show your work.
Quiz 3 will be given on Wednesday, as scheduled. It may include topics through §3.1.
The problems on Quiz 2 are exact copies or very similar clones of the following.
The scores on Quiz 2 are 30,30,30,30,30,30,28,27,26,26,26,20,20,20,16,16,15,10,10,8,8,3,3,0 with μ=19.3 and σ=10.1. Individual scores are on Blackboard.
Please no texting during class!
After Quiz 2, the constant multiple and sum rules for derivatives were derived (§3.1). On Friday, derivatives of the exponential function will be considered along with the product rule (§3.2).
Quiz 2 may include topics through §2.7.
A few more examples of limits at infinity were presented (§2.6). Then the definition of the derivative was presented (§2.7).
The scores on Quiz 1 are 30,30,30,30,30,30,30,30,28,28,27,26,26,25,21,20,20,19,19,18,10,8,0,0 with μ=22.3 and σ=9.3. Individual scores are on Blackboard.
The Spring schedule for the Math Resource Center has been announced:
Monday-Thursday 9:00 AM–7:00 PM
Friday 9:00 AM–7:00 PM
The MRC also has online tutoring through GoBoard. Sessions can be scheduled Monday-Saturday. To set up a tutoring session, complete the online request form.
Some examples of the \(\e\)-\(\d\) definition of limit were presented (§2.4). Continuity was defined (§2.5)
Quiz 1 on Wednesday may include topics through §2.3.
More examples of calculating limits were done and the \(\e\)-\(\d\) definition of limit was presented (§2.3–2.4).
The basic properties of limits were presented (§2.2–2.3).
Please don’t text during class!
Because of all the weather complications, the first quiz will be delayed a week to Wednesday, January 22. In the meantime, please read §2.2–2.3.
The tangent line and velocity problems from §2.1 were presented.
Most of the WebAssign problems correspond to problems in the text. For example, consider the following screen shot from the §2.2 problems in WebAssign.
The SCalcET8 is the code for the textbook, Stewart Calculus, Early Transcendentals. The numbers 2.2.004 mean it is Exercise 4 from §2.2 in the text. When you ask about a problem in class, make note of its coordinates in the text.
Read §2.1–2.2
.During the weather-shortened class period, all that was accomplished was a look at the syllabus.
1. Read the syllabus. It is on Blackboard and can be found elsewhere on this Web site.
2. Get squared away with WebAssign. You will need a WebAssign account and you must purchase access to the online textbook. The access code for this course is louisville 3434 0175. If you are new to WebAssign, then you should watch some of the introductory videos they provide.
3. 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!