Question 1

Profit is the difference between revenue and cost. Revenue is the quantity sold (q) multiplied by the price(p) and the costs is given at $10 times the quantity sold, thus \[P(q,p) = R(q,p) - C(q) = qp - 10q\] The quantity sold is given as a function of the price \[q(p) = 80-2p\] So rewriting the equation for profit as a function of the price \[P(p) = R(p) - C(p)\] \[ (80-2p)p - 10(80-2p)\] \[ (80p-2p^2) - (800-20p)\] \[-2p^2 + 100p - 800 \]

Question 2

Given \(f(x)=8x^3+7x^2-5\):

Step 1

\[f(3) = 8(3)^3+7(3)^2-5 = (8\cdot27)+(7\cdot9)-5 = 216+63-5 = 274\]

Step 2

\[f(-1) = 8(-2)^3+7(-2)^2-5 = (8\cdot-8)+(7\cdot4)-5 = -64+28-5 = -41\]

Step 3

\[f(x+c) = 8(x+c)^3+7(x+c)^2-5\] Verifying my math with R:

f2 <- function(x){8*x**3+7*x**2-5}
f2(3)
## [1] 274
f2(-2)
## [1] -41

Question 3

Step 1

\[\lim_{x\to1^-} f(x) = 2\]

Step 2

\[\lim_{x\to1^+} f(x) = -5\]

Step 3

\[\lim_{x\to1} f(x) = \mbox{ Does not exist} \]

Question 4

Find the derivative of \(f(x)=-2x^3\)

\[f'(x)=-6x^2\] Verifying my math in R:

f4 <- function(x){-2*x**3}
Deriv(f4)
## function (x) 
## -(6 * x^2)

Question 5

\[f(x) = \frac{-8}{x^2}\] \[f'(x) = \frac{16}{x^3}\]

Verify with R:

f5 <-function(x){-8/(x**2)}
Deriv(f5)
## function (x) 
## 16/x^3

Question 6

\[g(x) 5\sqrt[3]{x}\]

\[g'(x) \frac{5}{3x^\frac{2}{3}}=\frac{5}{3}\cdot\frac{1}{x^\frac{2}{3}}\]

Verify with R:

f6 <- function(x){5*x**(1/3)}
Deriv(f6)
## function (x) 
## 1.66666666666667/x^0.666666666666667
5/3
## [1] 1.666667
2/3
## [1] 0.6666667

Question 7

\[y=-2x^\frac{9}{8}\]

\[y'=\frac{-9x^\frac{1}{8}}{4}=\frac{-9}{4}\cdot x^\frac{1}{8}\]

Verify with R:

f7 <- function(x){-2*x**(9/8)}
Deriv(f7)
## function (x) 
## -(2.25 * x^0.125)
-9/4
## [1] -2.25
1/8
## [1] 0.125

Question 8

\[\frac{35-40}{4-0}=\frac{-5}{4} = -1\frac{1}{4}\]

Question 9

\[C(x)=630+2.4x\] \[\mbox{Avg Cost} = \frac{\mbox{Cost}}{\mbox{Unit}}=\frac{C(x)}{x}=\frac{630+2.4x}{x}\]

Question 10

\[f(x)=(-2x^{-2} +1)(-5x + 9)\]

Applying the quotient rule where \(f(x)=-2x^{-2}+1\) and \(g(x)=(-5x+9)\)

\[f'(x) = \frac{4}{x^3}(-5x+9) + -5(2x^{-2} + 1) = \frac{36}{x^3} - \frac{20}{x^2} + \frac{10}{x^2} - 5=\frac{36}{x^3}-\frac{10}{x^2}-5\]

Verify with R:

f10 <- function(x){(-2*x**-2+1)*(-5*x+9)}
Deriv(f10)
## function (x) 
## 4 * ((9 - 5 * x)/x^3) - 5 * (1 - 2/x^2)
# Evaluating at 10
f10_prime <- function(x){(36/x**3)-(10/x**2)-5}
f10_prime(10)
## [1] -5.064
f10_prime <- Deriv(f10)
f10_prime(10)
## [1] -5.064

Question 11

\[f(x)=\frac{5x^\frac{1}{2}+7}{x^3 + 1}\]

Applying the quotient rule where \(f(x)=5\sqrt{x}+7\) and \(g(x)=x^3+1\) so \(f'(x)=\frac{5}{2\sqrt{x}}\) and \(g'(x)=3x^2\)

\[f'(x) = \frac{\Big(\frac{5}{2\sqrt{x}}\Big)(x^3+1)-(3x^2(5\sqrt{x}+7))}{(x^3+1)^2}\] \[f'(x) = \frac{\frac{5x^3+5}{2\sqrt{x}}-3x^2 (5\sqrt{x}+7)}{(x^3+1)^2}\]

Simplified to \[f'(x) = -\frac{42x^\frac{5}{2} + 25x^3 - 5}{2\sqrt{x}\cdot (x^3 + 1)^2}\]

Again checking with R:

f11 <- function(x){(5*x**(1/2)+7)/(x**3+1)}
Deriv(f11)
## function (x) 
## {
##     .e1 <- 1 + x^3
##     .e2 <- sqrt(x)
##     (2.5/.e2 - 3 * (x^2 * (5 * .e2 + 7)/.e1))/.e1
## }
# Evaluate at 5
fprime11 <- function(x){-1*(42*x**(5/2)+25*x**3-5)/((2*x**(1/2))*((x**3+1)**2))}
fprime11(5)
## [1] -0.07701267
fprime11 <- Deriv(f11)
fprime11(5)
## [1] -0.07701267

Question 12

\[f(x)=(3x^{-3}-8x+6)^\frac{4}{3}\]

\[f'(x) = \frac{4}{3}(-\frac{9}{x^4}-8) \Big(\frac{3}{x^3}-8x+6\Big)^\frac{1}{3}\]

Checking my math in R:

f12 <- function(x){(3*x**-3-(8*x)+6)**(4/3)}
Deriv(f12)
## function (x) 
## -(1.33333333333333 * ((3/x^3 + 6 - 8 * x)^0.333333333333333 * 
##     (8 + 9/x^4)))
# Evaluate at 1
f12_prime <- function(x){(4/3)*((-9/x**4)-8)*(((3/x**3)-8*x+6)**(1/3))}
f12_prime(1)
## [1] -22.66667
f12_prime <- Deriv(f12)
f12_prime(1)
## [1] -22.66667

Question 13

\[f(t) =\frac{550t^2}{\sqrt{t^2+15}}\]

After factoring out the 550 constant and applying the quotient rule where \(f(x)=t^2\) and \(g(x)=\sqrt{t^2+15}\)

\[550\cdot f'(x) = \frac{(2t)(\sqrt{t^2+15})-(t^2)\Big(\frac{t}{\sqrt{t^2+15}}\Big)}{t^2+15} = \frac{2t\sqrt{t^2+15} -\frac{t^3}{\sqrt{t^2+15}}}{t^2+15}\]

Multiplying by 550 and simplifying \[f'(x) = \frac{1100t\sqrt{t^2+15} -\frac{550t^3}{\sqrt{t^2+15}}}{t^2+15}=\frac{550t(t^2+30)}{(t^2+15)^\frac{3}{2}}\]

Verifying with R:

f13 <- function(t){(550*t**2)/((t**2+15)**(1/2))}
Deriv(f13)
## function (t) 
## {
##     .e1 <- t^2
##     .e2 <- 15 + .e1
##     550 * (t * (2 - .e1/.e2)/sqrt(.e2))
## }
f13_prime <- function(t){(550*t*(t**2+30))/((t**2+15)**(3/2))}
f13_prime(3)
## [1] 547.3079
f13_prime <- Deriv(f13)
f13_prime(3)
## [1] 547.3079

The answer is 547

Question 14

\[N(t) = 1000(6+0.1t)^\frac{1}{2}=1000\sqrt{0.1t+6}\]

Step 1

\[N(3) = 1000(6+0.1(3))^\frac{1}{2}=39690\]

Step 2

Let \(u=0.1t+6\) so \(\sqrt{u}\) becomes \(\frac{1}{2\sqrt{u}}\) because of the power rule

and due to the chain rule \(u'=0.1\). Substitute

\[1000\cdot N'(t) = \frac{0.1\cdot0.5}{\sqrt{0.1t+6}}=\frac{0.05}{\sqrt{0.1t+6}}\] \[ N'(t)=\frac{1000\cdot 0.05}{\sqrt{0.1t+6}}=\frac{50}{\sqrt{0.1t+6}}\] \[N'(5)=\frac{50}{\sqrt{0.1(5)+6}}=\frac{50}{\sqrt{6.5}}=19.6116135138\approx20 \] 20 is roughly how many additional people attend in the 5th week of the season

Checking my math with R:

f14 <- function(t){1000*(6+0.1*t)**(1/2)}
Deriv(f14)
## function (t) 
## 50/sqrt(0.1 * t + 6)
f14_prime <- function(t){50/(6.5**(1/2))}
f14_prime(5)
## [1] 19.61161
f14_prime <- Deriv(f14)
f14_prime(5)
## [1] 19.61161

Question 15

\[3x^3+4y^3=77\]

Step 1

\[\frac{d}{dx}(3x^3+4y^3)=\frac{d}{dx}(77)\]

\[\frac{d}{dx}(3x^3)+\frac{d}{dx}(4y^3)=0\] \[9x^2+12y^2\frac{dy}{dx}=0\]

\[12y^2\frac{dy}{dx}=-9x^2\]

\[\frac{dy}{dx}=\frac{-9x^2}{12y^2} = -\frac{3x^2}{4y^2}\]

Step 2

The slope of the tangent line at (3,-1) is

\[-\frac{3(3)^2}{4(-1)^2}=-\frac{3(9)}{4(1)}=-\frac{7}{4}\]

Question 16

\[f(x)=\frac{x+3}{x-8}\]

Differentiate using the quotient rule where \(f(x)=x+3\) and \(g(x)=x-8\)

\[f'(x)=\frac{(1)(x-8)-(1)(x+3)}{(x-8)^2}=\frac{(x-8)-(x+3)}{(x-8)^2}=\frac{x-8-x-3}{(x-8)^2}=-\frac{11}{(x-8)^2}\]

Checking my work with R:

f16 <- function(x){(x+3)/(x-8)}
Deriv(f16)
## function (x) 
## {
##     .e1 <- x - 8
##     (1 - (3 + x)/.e1)/.e1
## }
f16_prime <- function(x){-11/((x-8)^2)}
f16_prime(20)
## [1] -0.07638889
f16_prime <- Deriv(f16)
f16_prime(20)
## [1] -0.07638889

\[f''(x)= \frac{22}{(x-8)^3}\]

The critical point is when x=8. f(x) is decreasing when x < 8 and increasing when x > 8.

Question 17

\[F(t) = 14 + \frac{367t^2}{t^2+100}\]

Step 1

\[367 \cdot F'(t) = \frac{t^2}{(t^2+100)^2}=\frac{(2t(t^2+100))(-2t^2\cdot t)}{(t^2+100)^2}=\frac{2t(t^2+100)(-2t^3)}{(t^2+100)^2}\]

Multiply it by 367

\[f'(x)=367 \cdot \frac{2t(t^2+100)(-2t^3)}{(t^2+100)^2}=\frac{734t(t^2+100)-734t^3}{(t^2+100)^2}=\frac{73400t}{(t^2+100)^2}\]

In the interest of turning my assignment on time some of the steps have been excluded

\[f''(x)=-\frac{73400(3t^2-100)}{(t^2+100)^3}\]

\(F'(t) > 0\) from \((0,\infty)\) and \(F'(t) < 0\) from \((-\infty, 0)\) which doesn’t make sense in this case.

Step 2

\[\lim_{t\to\infty} F(t) = 381^\circ F\]

Question 18

\[f(x)=\frac{x^3}{3340000}-\frac{7x^2}{9475}+\frac{42417727x}{1265860000}+\frac{1}{33}\]

Again in the interest of time I will skip the steps of the calculation:

\[f'(x) = \frac{3x^2}{3340000}-\frac{14x}{9475}+\frac{42417727}{1265860000}\]

\[0 = \frac{3x^2}{3340000}-\frac{14x}{9475}+\frac{42417727}{1265860000}\]

f18 <- function(x){}
f18_prime <- function(x){(3*x**2/3340000)-(14*x/9475)+(42417727/1265860000)}
x <- seq(0, 30, by=.01)
y <- f18_prime(x)
i <- stats::uniroot(f18_prime,c(-30,30))
plot(y~x,type="l", col="blue")
abline(h=0, col="black")
abline(v=0, col="black")
abline(v=i$root, col="red")
points(i$root, 0, col="red", pch=21)

i$root
## [1] 23

Question 19

\[f(x) = 7x^2+28x-35\]

\[f'(x) = 14x+28\]

\[0 = 14x+28 \]

\[-28 = 14x\]

\[x=-\frac{28}{14}=-2\]

\[f(-2) = (7(-2)^2)+(28(-2))-35 = 28-56-35=-63\]

\[\mbox minimum = (-2,-63)\]

Checking with R:

f19 <- function(x){7*x**2+28*x-35}
f19_prime <- Deriv(f19)
i <- stats::uniroot(f19_prime,c(-30,30))
i$root
## [1] -2
f19(i$root)
## [1] -63
x <- seq(-30, 30, by=.01)
y <- f19(x)
i <- stats::uniroot(f19_prime,c(-30,30))
plot(y~x,type="l", col="blue")
abline(h=0, col="black")
abline(v=0, col="black")
points(i$root, 0, col="red", pch=21)

Question 20

\[f(x)=-6x^3+27x^2+180x\]

\[f'(x)=-18x^2+54x+180\]

\[0=-18x^2+54x+180\]

\[0=-18(x^2-3x-10)\]

\[x^2-3x-10=0\]

\[(x+2)(x-5)=0\]

The function is minimized/maximized when \(x = -2\) or \(x = 5\). Plugging in and solving for y:

\[f(-2)=-6(-2)^3+27(-2)^2+180(-2)=48+108-360=-204\] \[f(-2)=-6(5)^3+27(5)^2+180(5)=-750+650+900=800\]

\[f''(x)=-36x+54\] \[f''(-2)=-36(-2)+54=72+54=126>0\] \[f''(5)=-36(5)+54=180+54=234>0\]

So the critical points are \((-2, -204)\) and \((5, 800)\). \((-2, -204)\) is a local min and \((5, 800)\)

f20 <- function(x){(-6*x**3)+(27*x**2)+(180*x)}
f20_prime <- Deriv(f20)
x <- seq(-30, 30, by=.01)
y <- f20(x)
#i <- stats::uniroot(f20_prime,c(-30,30))
plot(y~x,type="l", col="blue")
abline(h=0, col="black")
abline(v=0, col="black")

#points(i$root, 0, col="red", pch=21)

Question 21

Inventory cost is defined as the cost of holding stock and the cost of ordering. The inventory cost as a function of the iron lot size is:

\[C(i)=\frac{120}{i}(6+4.5(i))+(1.60\frac{i}{2})\] This function assumes the irons are selling at a constant rate (no seasonality) throughout the year. Simplifing the function:

\[C(i)=\frac{620}{i}+ \frac{540i}{i}+(0.8i) = \frac{620}{i}+0.8i+540\]

Since we want to minimize this cost we take the first derivative and set it equal to zero.

\[C'(i)=-\frac{720}{i^2} +0.8\]

\[0=-\frac{720}{i^2} +0.8\] \[\frac{720}{i^2}=0.8\] \[720=0.8i^2\] \[i^2=\frac{720}{0.8}=900\] \[i=\sqrt{900}=30\] So the lot size is 30 irons. Since we are expecting to sell 120 irons per year this would require 4 orders.

Question 22

\[Volume = 18432ft^2 = x\cdot x \cdot y \Rightarrow y=\frac{18432}{x^2}\] The cost is the sum of the sides, top and bottom. The sides and top material cost $3.00 per square foot and the bottom is $5.00.

\[C(x,y) = 3(4xy) + 3x^2 + 5x^2 = 12xy+8x^2\] Substituting in y: \[ C(x)=\frac{12x\cdot 18432}{x^2}+8x^2=\frac{221184}{x}+8x^2\] Minimize the cost function: \[ C'(x)=-\frac{221184}{x^2}+16x\] \[ 0=-\frac{221184}{x^2}+16x\Rightarrow 0=-221184+16x^3 \Rightarrow 0=-13824 + x^3 \] \[ x^3 = 13824 \Rightarrow x=\sqrt[3]{13824} = 24\] Plugging 24 into the volume equation and solving for y: \[18432 = x^2y\Rightarrow 18432=(24)^2 y\Rightarrow 18432=576 y \Rightarrow y=\frac{18432}{576} \Rightarrow y = 32\] Now checking everything with R:

f22 <- function(x){
  (3*(4*x*(18432/(x**2)))) + (3*x**2) + (5*x**2)
}
f22_prime <- Deriv(f22)

x22 <- uniroot.all(f22_prime, c(0,100))
x22
## [1] 24
y22 <- function(x){18432/(x**2)}
y22(x22)
## [1] 32

Question 23

\[Area=xy=1056 \Rightarrow y=\frac{1056}{x}\]

The cost function is

\[C(x,y)=(14.40\cdot 2y)+(14.40 \cdot 2x) + (12.00 \cdot 2x)\] Substituting y in: \[C(x)=(14.40\cdot 2(\frac{1056}{x}))+(14.40 \cdot 2x) + (12.00 \cdot 2x)\] Simplified \[C(x)=\frac{30412.8}{x}+52.8x\]

Minimize it

\[C'(x)=-\frac{30412.8}{x^2}+52.8\]

\[0=-\frac{30412.8}{x^2}+52.8 \] \[\frac{30412.8}{x^2}=52.8 \] \[ \frac{30412.8}{52.8}=x^2 \] \[x^2=576 \] \[ x=\sqrt{576} \Rightarrow x=\pm 24\] Negative 24 is meaningless in this context. Now substituting x in the area equation: \[y=\frac{1056}{24} = 44\]

f23 <- function(x){
  (28.8*(1056/x))+(52.8*x)
}
f23_prime <- Deriv(f23)

x23 <- uniroot.all(f23_prime, c(0,100))
x23
## [1] 24
y23 <- function(x){1056/x}
y23(x23)
## [1] 44

Question 24

\[Value(time)=base \cdot e^{rate \cdot time}\] Enter in the information from the question:

\[37000=67000 \cdot e^{rate \cdot 7}\]

Solving for rate:

\[\frac{37000}{67000} = e^{rate \cdot 7}\]

\[ln\Big(\frac{37000}{67000}\Big) = rate \cdot 7 \Rightarrow rate = \frac{ln\Big(\frac{37000}{67000}\Big)}{7} \approx -0.08482496\]

Now substitue and solve for the value at year 9

\[Value(9) = 67000 \cdot e^{-0.08482496 \cdot (9)} = 31226.53\] Checking my math with R:

f24 <- function(t){
  e <- exp(1)
  67000*e**(t*((log(37000/67000)/7)))
}
f24(9)
## [1] 31226.53

Question 25

Given \(p = D(x) = 23.2-0.4x\) price and Revenue is \(R = p \cdot D(x)\) the maximizing level of production is:

\[ R = p \cdot D(x) \Rightarrow R(x) = D(x) \cdot D(x) \Rightarrow R(x) = D(x)^2\]

Using the chain rule: \[R'(x) = 0.32x -18.56\]

\[0 = 0.32x -18.56 \Rightarrow 0.32x = 18.56 \Rightarrow x = \frac{18.56}{0.32} = 58\]

Checking the math in R:

f25 <- function(x){
  (23.2-0.4*x)*(23.2-0.4*x)
}

f25_prime <- Deriv(f25)
x25 <- uniroot.all(f25_prime, c(0,100))
x25
## [1] 58

Question 26

Start with the exponential growth formula: \[Cost(time)=base \cdot e^{rate \cdot time}\]

And enter what we know and solve for rate: \[426.8=400 \cdot e^{rate \cdot 11} \Rightarrow \frac{426.80}{400}=e^{rate \cdot 11} \Rightarrow rate\cdot 11 = ln\Big(\frac{426.8}{400}\Big) \Rightarrow rate = \frac{ln\Big(\frac{426.8}{400}\Big)}{11} \approx 0.005895543\]

Substitute rate back into the equation: \[Cost(year) = 400 \cdot e^{0.005895543 \cdot (year - 1995)}\] \[Cost(2017) = 400 \cdot e^{0.005895543 \cdot (2017 - 1995)} \approx 455.40\] Verifying with R:

f26 <- function(year){
  e <- exp(1)
  400*e**((log(426.8/400)/11)*(year-1995))
}
f26(2017)
## [1] 455.3956

Question 27

Given: \[P'(x) = 380-4x \] \[P(x) = 380x - 2x^2+C\]

Step 1

\[1700 = 380(38)-2(38)^2 + C\] Solving for C

\[ 1700 = 14440 - 2888 + C \Rightarrow C = 17000 - 14440 + 2888 \Rightarrow C = -9852\] The profit function is: \[ P(x)= 380x - 2x^2-9852\]

Step 4

\[ P(56)= 380(56) - 2x^2-9852 = 5156\]

f27 <- function(x){
  (380*x) - (2*x**2) - 9852
}
f27(56)
## [1] 5156

Question 28

Solve: \[ \int\frac{-5(ln(y))^3}{y}dy = -5 \int \frac{1}{y} ln(y)^3\] Substitute \(u = ln(y)\) and \(du = \frac{1}{y}\)

\[-5 \int du \cdot u^3 \Rightarrow -5 \cdot \frac{u^4}{4}+c \Rightarrow \frac{-5}{4}ln(y)^4 + c\]

Question 29

Given: \[P'(t)=75-9t^\frac{1}{2}\]

\[P(t) = 75t-6t^\frac{3}{2}\]

\[\int_0^9 P'(t) = P(9) - P(0) = (75(9)-6(9)^\frac{3}{2})-(0-0) = (675-162) -0 = 513\]

Question 30

\[ \int_{-7}^2 x\sqrt{x+7}dx\]

For this problem \(u=x\) and \(dv=\sqrt{x+7}dx\). Calculate \(du\) and \(v\):

\(du=1dx\) and \(v = \int \sqrt{x+7}dx = \frac{2}{3}(x+7)^\frac{3}{2}\)

Plugging it into the integration by parts formula:

\[ \int_a^b udv = uv]_a^b -\int_a^b vdu\]

\[ \int_{-7}^2x\sqrt{x+7}dx]_{-7}^2 -\int_{-7}^2 \frac{2}{3}(x+7)^\frac{3}{2}\cdot 1dx\] Integrate the right hand side

\[ \frac{2x}{3}(x+7)^\frac{3}{2}]_{-7}^2 -\frac{4}{15}(x+7)^\frac{5}{2}]_{-7}^2 \]

Plugging in the -7 and 2 yeilds -28.8 as shown below:

f30 <- function(a, b){
  left <- function(x){
    ((2*x)/3)*(x+7)**(3/2)
  }
  right <- function(y){
    (4/15)*(y+7)**(5/2)
  }
  (left(b)-left(a))-(right(b)-right(a))
}

f30(-7,2)
## [1] -28.8

Question 31

Step 1

In the interest of time I will solve this question by building a simple model

bounce <- function(total_times, drop_height, debug=F){
  rebound = .22 # Amount of the rebound as a share of drop height
  times_ball_hit_the_ground = 1
  total_distance = drop_height
  if(debug){
    message(paste(total_distance, 'meters after ball hits the ground',times_ball_hit_the_ground,'times'))
  }
  while(times_ball_hit_the_ground < total_times){
    drop_height = rebound * drop_height 
    total_distance = total_distance + (2 * drop_height) #going Up and down
    times_ball_hit_the_ground = times_ball_hit_the_ground + 1
    if(debug){message(paste(total_distance, 'meters after ball hits the ground',times_ball_hit_the_ground,'times'))}
  }
  if(!debug){
  message(paste(total_distance, 'meters after ball hits the ground',times_ball_hit_the_ground,'times'))
  }
}
  
bounce(20, 46, T)
## 46 meters after ball hits the ground 1 times
## 66.24 meters after ball hits the ground 2 times
## 70.6928 meters after ball hits the ground 3 times
## 71.672416 meters after ball hits the ground 4 times
## 71.88793152 meters after ball hits the ground 5 times
## 71.9353449344 meters after ball hits the ground 6 times
## 71.945775885568 meters after ball hits the ground 7 times
## 71.9480706948249 meters after ball hits the ground 8 times
## 71.9485755528615 meters after ball hits the ground 9 times
## 71.9486866216295 meters after ball hits the ground 10 times
## 71.9487110567585 meters after ball hits the ground 11 times
## 71.9487164324869 meters after ball hits the ground 12 times
## 71.9487176151471 meters after ball hits the ground 13 times
## 71.9487178753324 meters after ball hits the ground 14 times
## 71.9487179325731 meters after ball hits the ground 15 times
## 71.9487179451661 meters after ball hits the ground 16 times
## 71.9487179479365 meters after ball hits the ground 17 times
## 71.948717948546 meters after ball hits the ground 18 times
## 71.9487179486801 meters after ball hits the ground 19 times
## 71.9487179487096 meters after ball hits the ground 20 times

So the ball traveled 71.94872 meters when it hit the ground on the 20th time. I would note that the additional distance is very small after the 7th

Step 2

To evaluate the total distance as the number of bounces heads to infinity I will cheat and plug in a really large number and see what it tends toward.

bounce(1000000, 46)
## 71.9487179487179 meters after ball hits the ground 1e+06 times

So it is tending toward 72 meters.

Question 32

Checking my math with R:

f32 <- function(x){x*((x+7)**(1/2))}
integrate(Vectorize(f32),-7, 2)
## -28.8 with absolute error < 6.1e-08

Question 34

I currently don’t know how to do find the Taylor polnomial of degree 5 near x=4 for \(y=3e^{5x-3}\)