# 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)  $$r\approx 2.86$$ 