 # Historical Activities for Calculus - Solutions to Exercises

Author(s):
Gabriela R. Sanchis (Elizabethtown College)

### Module 1: Curve Drawing Then and Now

##### Exercise 1.
The equation is  $\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=d.$ Solving for $y$ we obtain $y=\pm \frac{\sqrt{\left(-4 c^2+d^2\right) \left(d^2-4 x^2\right)}}{2 d}.$ The values of $c$ and $d$ for which the points $(0,\pm 5)$ and $(\pm 3,0)$ satisfy the equations(s) are $c=4$ and $d=10$. Substituting in these values and simplifying, the equations are
$y=\pm\frac{3}{10}\sqrt{100-4x^2}.$ ##### Exercise 2.
We must have $a+b=5$ and $a-b=3$, so $a=4$ and $b=1$. ##### Exercise 3.
We must have $\frac{a}{2}+b=5$ and $\frac{a}{2}-b=3,$ so $a=8$ and $b=1$. ##### Exercise 4.
We must have $a+b=5$ and $a=3,$ so $b=2$. ### Module 2: Tangent Lines Then and Now

##### Exercise 5.
(a)  Since the point $(4,5)$ is 4 units above the vertex, the $y$-intercept of the tangent line must be 4 units below the vertex, at $(0,-3)$. (b)  If the $x$-intercept $A$ of the tangent line has coordinates $(x,0)$, then Apollonius' equation $\frac{|AH|}{|AG|}=\frac{|BH|}{|BG|}$ becomes

$\frac{x-5}{x+5}=\frac{2}{8}$

Solving this for $x$ we get $x=\frac{25}{3}$ so the $x$-intercept is $(25/3,0)$. (c)  If the $x$-intercept $A$ of the tangent line has coordinates $(x,0)$, then Apollonius' equation $\frac{|AH|}{|AG|}=\frac{|BH|}{|BG|}$ becomes

$\frac{4-x}{x+4}=\frac{1}{9}$

Solving this for $x$ we get $x=3.2$ so the $x$ intercept is $(3.2,0)$. ##### Exercise 6.
(a) For the parabola, the tangent line is the angle bisector of $\measuredangle FPB$: (b)  For the ellipse, the tangent is the angle bisector of $\measuredangle {F_1^{\prime}}PF_2$ where ${F_1^{\prime}}$ is obtained by rotating $F_1$ about $P$ 180 degrees. (c)  For the hyperbola, the tangent line is the angle bisector of $\measuredangle F_1PF_2$. (d)  There are two ways to approach the tangent line to the cycloid. You can think of $P$ as being pulled by two forces, one parallel to $EG$ as the position of the circle moves forward, and another perpendicular to the radius $PC$ as $P$ rotates about $C$. So we construct a line through $P$ parallel to $EG$, and then another line through $P$ perpendicular to $PC$. The angle bisector of the angle between these two lines will then be the tangent.

Alternatively $P$ is being rotated about $A$, so we can just construct the perpendicular of $AP$. (e)  For the epicycloid, there are two forces pulling on $P$, one perpendicular to $CA$, and one perpendicular to $AB$. The tangent line is the angle bisector of the angle between these two lines. ##### Exercise 7.
(a)  The center of the circle at which $A$ and $B$ merge is $(3,0)$. The slope of the radius of $-1$, so the slope of the tangent line, which is perpendicular, is 1. Hence the equation of the tangent line is $y-2=1(x-1)$ or $y=x+1$. (b)  The center of the circle at which $A$ and $B$ merge is $(11,0)$. The slope of the radius of $-3$, so the slope of the tangent line, which is perpendicular, is ${\frac13}$. Hence the equation of the tangent line is $y-6={\frac13}(x-9)$ or $y=\frac13x+3$. ### Module 3: Optimization Problems Then and Now

#### Heron's “Shortest Distance" Problem

##### Exercise 8.
If $C=(x,0)$, then we want to minimize

$f(x)=|AC|+|CB|=\sqrt{(0-x)^2+(4-0)^2}+\sqrt{(10-x)^2+(12-0)^2}$ $=\sqrt{x^2 - 20x + 244} + \sqrt{x^2 + 16}.$

We use a CAS to compute the derivative:

$f^{\prime}(x)=\frac{x-10}{\sqrt{144+(x-10)^2}}+\frac{x}{\sqrt{16+x^2}}$

Then $f^{\prime}(x)=0\Rightarrow x=\frac{5}{2}$. So the point $C$ that minimizes $|AC|+|CB|$ is $C=\left(\frac52,0\right)$. #### Snell's Law and the Principle of Least Time

##### Exercise 9.
(a)  If $C=(x,0)$, then we want to minimize

$f(x)=\frac{|AC|}{v_1}+\frac{|BC|}{v_2}=\frac{\sqrt{(x-1)^2+9}}{4}+\frac{\sqrt{(x-4)^2+4}}{5}$

We use a CAS to compute the derivative:

$f^{\prime}(x)=\frac{x-1}{4\sqrt{9+(x-1)^2}}+\frac{x-4}{5\sqrt{4+(x-4)^2}}$

Then $f^{\prime}(x)=0\Rightarrow x=2.57363$. (b)  We need to find the value of $x$ that minimizes

$f(x)=\frac{\sqrt{(x+3)^2+16}}{1}+\frac{\sqrt{(x-2)^2+25}}{2}$

Taking a derivative, setting it equal to 0, and solving, we find the solution $x=-1.743575120$. (c)  Let $C=(x,-x^2)$. We need to find the value of $x$ that minimizes $f(x)=|AC|+|BC|=\sqrt{(0-x)^2+(4+x^2)^2}+\sqrt{(5-x)^2+(-3+x^2)^2}.$ Using a CAS to compute the derivative and solve $f^{\prime}(x)=0$ numerically, we obtain the solution $x=0.726582$. #### L'Hôpital's Pulley Problem and the Principle of Least Potential Energy

##### Exercise 10. (a)  $|GA|=d-x$.

(b)  $|GC|=\sqrt{r^2-(d-x)^2}$.

(c)  $|BC|=\sqrt{x^2+r^2-(d-x)^2}$.

(d)  $|CD|=l-\sqrt{x^2+r^2-(d-x)^2}$.

(e)  $|GD|=\sqrt{r^2-(d-x)^2}+l-\sqrt{x^2+r^2-(d-x)^2}$.

(f)  Using a CAS to differentiate the expression with respect to $x$ and setting it equal to 0, we obtain the solutions
$x=\frac{4d^2 - r^2}{4d} \pm \frac{|r|\sqrt{8d^2 + r^2}}{4d}.$

(g)  Plugging in $d=12$, $r=7$, and $l=14$ above we obtain $x=5.925247326$.

(h)  Yes, the maximum value of $|GD|$ is $10.60747,$ and it occurs when $x=|GB|=5.925247,$ which agrees (to four decimal places) with the answer obtained above. #### Regiomontanus' Hanging Picture Problem

##### Exercise 11. We need to find the value of $x$ that maximizes $\tan\theta=\frac{\frac{b}{x}-\frac{a}{x}}{1+\frac{b}{x}\cdot\frac{a}{x}}.$ Using a CAS to take a derivative and set it equal to zero, we obtain the solution $x=\sqrt{ab}.$

##### Exercise 12.
(a)  The six-foot tall adult's eyes are two feet below the bottom of the painting. So $a=2$ and $b-a=12$. Hence $b=14$ and so he should stand $\sqrt{2\cdot 14}\approx5.291502622$ feet away from the painting.

The three-foot tall child's eyes are five feet below the bottom of the painting. So $a=5$ and $b-a=12$. Hence $b=17$ and so he should stand $\sqrt{5\cdot 17}\approx9.219544457$ feet away from the painting.

(b)  Plugging $a=2$, $b=14$, and $x=\sqrt{28}$ into the expression for $\tan\theta$ we get $\tan\theta=1.133893419,$ so $\theta=0.8480620789$ radians. To convert to degrees we multiply by $\frac{180\mbox{ deg}}{\pi\mbox{ radians}}$. Thus $\theta=\frac{0.8480620789\cdot 180}{\pi}\approx 48.59037788\mbox{ degrees}.$

Plugging $a=5$, $b=17$, and $x=\sqrt{85}$ into the expression for $\tan\theta$ we get $\tan\theta=0.979067$, so $\theta=0.774821$ radians. To convert to degrees we multiply by $\frac{180\mbox{ deg}}{\pi\mbox{ radians}}$. Thus $\theta=\frac{0.774821\cdot 180}{\pi}\approx 44.394\mbox{ degrees}.$

c)  For the 6-foot tall adult: For the 3-foot tall child: #### Galileo and the Brachistochrone Problem

##### Exercise 13.
(a)  $T(0,0,5,-5,0)=1.428571428$ seconds.

(b)  $T(0,0,1,-2,0)+T(1,-2,5,-5,V(0,0,1,-2,0))=1.333079013$ seconds.

(c)  Taking a derivative, setting it equal to $0$, and solving we find that $x = 0.4491013975$. The $y$-value is $-\sqrt{25-(x-5)^2}=-2.071067818$. So $C=(0.4491013975,-2.071067818)$.
The time of descent along this path is
$T(0,0,0.4491013975,-2.071067818,0)$ $+\,T(0.4491013975,-2.071067818,5,-5,V(0,0,0.4491013975,-2.071067818,0))$
$=1.330475856{\mbox{ seconds}}.$

This is less than $1.428571428$, the time along the path $(0,0)\to (5,-5)$.

(d)  Let $v_1=V(0,0,0.4491013975,-2.071067818,0)=6.371258058{\mbox{ m/sec}},$ the terminal velocity along $(0,0)\to(0.4491013975,-2.071067818)$. We need to find $x$ that minimizes

$T(0.4491013975,-2.0710678181,x,-\sqrt{25-(x-5)^2},v_1)$
$+\,T(x,-\sqrt{25-(x-5)^2},5,-5,V(0.4491013975,-2.0710678181,x,-\sqrt{25-(x-5)^2},v_1)).$

Taking a derivative and setting it equal to 0, we find $x=2.043116864$. The corresponding $y$-coordinate is $-\sqrt{25-(x-5)^2}=-4.031977445$, so $D=(2.043116864,-4.031977445)$.

The terminal velocity as the object reaches point $D$ is $v_2=V(0.4491013975,-2.071067818,2.043116864,-4.031977445,6.371258047)=8.889699541{\mbox{ m/sec}}.$

The total time along the path $(0,0)\to C\to D\to (5,-5)$ is $T(0,0,0.4491013975,-2.0710678181,0)$
$+\,T(0.4491013975,-2.0710678181,2.043116864,-4.031977445,6.371258058)$
$+\,T(2.043116864,-4.031977445,5,-5,8.889699541)$ $=1.327598538{\mbox{ seconds}}.$
##### Exercise 14.
(a)  We need to find the value of $x$ that minimizes $T(0,0,x,-1,0)+T(x,-1,5,-5,V(0,0,x,-1,0)).$ Taking the derivative and setting it equal to zero, the solution is $x=0.2434120319$. So $C=(0.2434120319,-1)$.
The terminal velocity as the object reaches point $C$ is $V(0,0,0.2434120319,-1,0)=4.427188724.$

(b)  We need to find the value of $x$ that minimizes $T(0.2434120319,-1,x,-2,4.427188724)+T(x,-2,5,-5,V(0.2434120319,-1,x,-2,4.427188724).$ Taking the derivative and setting it equal to zero, the solution is $x=0.876358533$. So $D=(0.876358533,-2)$.
The terminal velocity as the object reaches point $D$ is $V(0.2434120319,-1,0.876358533,-2,4.427188724)=6.260990337.$

(c)  We need to find the value of $x$ that minimizes  $T(0.876358533,-2,x,-3,6.260990337)+T(x,-3,5,-5,V(0.876358533,-2,x,-3,6.260990337)).$ Taking the derivative and setting it equal to zero, the solution is $x=1.786631272$. So $E=(1.786631272,-3)$.
The terminal velocity as the object reaches point $E$ is $V(0.876358533,-2,-3,6.260990337)=7.668115805.$

(d)  We need to find the value of $x$ that minimizes  $T(1.786631272,-3,x,-4,7.668115805)+T(x,-4,5,-5,V(1.786631272,-3,x,-4,7.668115805)).$ Taking the derivative and setting it equal to zero, the solution is $x=3.049531298$. So $F=(3.049531298,-4)$.
The terminal velocity as the object reaches point $F$ is $V(1.786631272,-3,3.049531298,-4,7.668115805)=8.854377448.$

(e)  The time of descent is

$T(0, 0, 0.2434120319,-1, 0) + T(0.2434120319,-1, 0.876358533,-2, 4.427188724)$
$+\,T(0.876358533,-2, 1.786631272,-3, 7.668115805) + T(1.786631272,-3, 3.049531298,-4, 8.854377448)$
$+\,T(3.049531298,-4, 5, -5, 8.854377448)$ $=1.309306399\mbox{ seconds}.$

This is less than $1.327598538,$ the time along the polygonal path from the previous exercise.

(f)  $r\approx 2.86$ Gabriela R. Sanchis (Elizabethtown College), "Historical Activities for Calculus - Solutions to Exercises," Convergence (July 2014)

## Dummy View - NOT TO BE DELETED

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