Q. 83 - Calculus

Question

In another science fiction novel, gravity on planet Xillian again acts very strangely. The height $s\left(t\right)$ of a falling object on Xillian is always a cubic polynomial function. Suppose a kiwi fruit is dropped (with initial velocity of zero) from the top of a Xillian radio tower. After $5$ seconds, the kiwi fruit is $100$ feet from the ground and falling at a rate of $-200$ feet per second. The acceleration of the kiwi fruit at that moment is $-46$ feet per second per second.

Part (a): Use the values of $s\text{'}\left(0\right)$ (note the kiwi initially has velocity zero), $s\left(5\right), s\text{'}\left(5\right), s\text{'}\text{'}\left(5\right)$ given in the preceding description to find a formula for the height $s\left(t\right)$ of the kiwi fruit t seconds after being dropped from the Xillian radio tower. Specifically, use these four data points to solve for the coefficients of a cubic polynomial $\mathrm{s}\left(\mathrm{t}\right) = {\mathrm{at}}^3 + {\mathrm{bt}}^2 + \mathrm{ct} + \mathrm{d}$.

Part (b): Verify that the function$s\left(t\right)$ you just found produces the correct values for $s\text{'}\left(0\right),s\left(5\right),s\text{'}\left(5\right),s\text{'}\text{'}\left(5\right)$ in this exercise.

Part (c): How high is the Xillian radio tower from which the kiwi fruit was dropped?

Part (d): On Earth, acceleration due to gravity is given by a constant $-32$ feet per second per second, that is, on Earth we always have $a\left(t\right)=-32$ for falling objects. What is the function $a\left(t\right)$ for acceleration due to Xillian “gravity”? Is this acceleration constant? What are the physical implications of gravity on Xillian?

Step-by-Step Solution

Answer

Part (a): The cubic polynomial for the distance function is $s\left(t\right)=-\frac{2}{5}{t}^3-17t^2+575$.

Part (b): It has been verified that the correct values for $s\left(5\right),s\text{'}\left(0\right),s\text{'}\left(5\right),\mathrm{s}\text{'}\text{'}\left(5\right)$ is $100,0,-200,-46$.
Part (c): The height from which the fruit was dropped is $575 ft$.
Part (d): The function $a\left(t\right)$ for acceleration due to Xillian “gravity” is $s\text{'}\text{'}\left(t\right)=-\frac{12t}{5}-34$.

The function is not a constant value.

The gravity on the planet acts downward. Then due to gravity, the planet moves downward.

Part (a) Step 1. Given information.

Consider the given question,

A cubic polynomial is of the form $s\left(t\right)=a{t}^3+b{t}^2+ct+d$, where a, b, c, d are real numbers.

The height of the fruit from the ground after $5$secs is $100 ft$.

Part (a) Step 2. Find the derivative of the distance function.

Substitute $t=5$ in function $s\left(t\right)$,

$$\begin{gathered} s\left(5\right)=a{\left(5\right)}^3+b{\left(5\right)}^2+c\left(5\right)+d \\ 100=125a+25b+5c+d ....... \left(i\right) \end{gathered}$$

The derivative of the distance function is the velocity function.

Differentiate both sides of the distance function with respect to t,

$$\begin{gathered} \frac{d}{dt}\left[s\left(t\right)\right]=\frac{d}{dt}\left(a{t}^3\right)+\frac{d}{dt}\left(b{t}^2\right)+\frac{d}{dt}\left(ct\right)+\frac{d}{dt}\left(d\right) \\ s\text{'}\left(t\right)=a\frac{d}{dt}\left(t^3\right)+b\frac{d}{dt}\left(t^2\right)+c\frac{d}{dt}\left(t\right)+0 \\ s\text{'}\left(t\right)=a\left(3t^{3-1}\right)+b\left(2t^{2-1}\right)+c\left(1\right) \\ s\text{'}\left(t\right)=3a{t}^2+2bt+c \end{gathered}$$

The velocity of the falling fruit at $t=0$ is $0$ and the velocity of fruit after $5 secs$ is equal to $-200 ft per second$.

Part (a) Step 3. Substitute the values of t in the function s t .

Substitute $$\begin{gathered} t=0, \\ s\text{'}\left(0\right)=0 \end{gathered}$$ in the function $s\left(t\right)$,

$$\begin{gathered} s\text{'}\left(0\right)=3a\left(0\right)+2b\left(0\right)+c \\ 0=0+0+c \\ c=0 \end{gathered}$$

Substitute $$\begin{gathered} t=5, \\ s\text{'}\left(5\right)=-200 \end{gathered}$$ in the function $s\left(t\right)$,

$$\begin{gathered} s\text{'}\left(5\right)=3a{\left(5\right)}^2+2b\left(5\right)+c \\ -200=75a+10b+0 \\ 15a+2b-40 ...... \left(ii\right) \end{gathered}$$

The derivation of the velocity function is the acceleration function.

Part (a) Step 4. Find the derivative of the acceleration function.

Differentiate both sides of the velocity function $s\text{'}\left(t\right)=3a{t}^2+2bt+c$,

$$\begin{gathered} \frac{d}{dt}\left[s\text{'}\left(t\right)\right]=\frac{d}{dt}\left(3a{t}^2\right)+\frac{d}{dt}\left(2at\right)+\frac{d}{dt}\left(c\right) \\ s\text{'}\text{'}\left(t\right)=3a\frac{d}{dt}\left(t^2\right)+2b\frac{d}{dt}\left(t\right)+0 \\ s\text{'}\text{'}\left(t\right)=6at+2b \end{gathered}$$

Substitute $$\begin{gathered} t=5, \\ s\text{'}\text{'}\left(5\right)=-46 \end{gathered}$$ in the velocity function,

$$\begin{gathered} s\text{'}\text{'}\left(5\right)=6a\left(5\right)+2b \\ -46=30a+2b \\ 15a+b=-23 ...... \left(iii\right) \end{gathered}$$

Subtracting equations (ii) and (iii),

$$\begin{gathered} \left(15a+b\right)-\left(15a+b\right)=-40-\left(-23\right) \\ b=-17 \end{gathered}$$

Part (a) Step 5. Subtract equation (ii) and (iii).

Substitute $b=-17$ in equation (ii),

$$\begin{gathered} 15a+\left(-17\right)=-23 \\ a=-\frac{2}{5} \end{gathered}$$

Substitute $$\begin{gathered} a=-\frac{2}{5}, \\ b=-17, \\ c=0 \end{gathered}$$ in equation (i),

$$\begin{gathered} 125\left(-\frac{2}{5}\right)+25\left(-17\right)+5\left(0\right)+d=100 \\ d=575 \end{gathered}$$

Substitute the values in function $s\left(t\right)$,

$$\begin{gathered} s\left(t\right)=-\frac{2}{5}{t}^3-17t^2+0\left(t\right)+575 \\ s\left(t\right)=-\frac{2}{5}{t}^3-17t^2+575 \end{gathered}$$

Therefore, the cubic polynomial for the distance function is $s\left(t\right)=-\frac{2}{5}{t}^3-17t^2+575$.

Part (b) Step 1. Substitute the values in the polynomial function.

Substitute $t=5$in the polynomial function,

$$\begin{gathered} s\left(5\right)=-\frac{2}{5}{\left(5\right)}^3-17{\left(5\right)}^2+575 \\ =-50-425+575 \\ =100 \end{gathered}$$

Differentiate both sides of the polynomial function,

$$\begin{gathered} \frac{d}{dt}\left[s\left(t\right)\right]=-\frac{2}{5}\frac{d}{dt}\left(t^3\right)-17\frac{d}{dt}\left(t^2\right)+\frac{d}{dt}\left(575\right) \\ s\text{'}\left(t\right)=-\frac{6}{5}{t}^2-34t \end{gathered}$$

Substitute $t=0$ in $s\text{'}\left(t\right)$,

$$\begin{gathered} s\text{'}\left(0\right)=-\frac{6}{5}{\left(0\right)}^2-34\left(0\right) \\ =0 \end{gathered}$$

Part (b) Step 2. Substitute t = 5 in s ' t .

Substitute $t=5$ in $s\text{'}\left(t\right)$,

$$\begin{gathered} s\text{'}\left(5\right)=-\frac{6}{5}{\left(5\right)}^2-34\left(5\right) \\ =-\frac{6}{5}\left(25\right)-170 \\ =-200 \end{gathered}$$

Differentiate both sides of the polynomial function,

$$\begin{gathered} \frac{d}{dt}\left[s\text{'}\left(t\right)\right]=\left(-\frac{6}{5}\right)\frac{d}{dt}\left(t^2\right)-34\frac{d}{dt}\left(t\right) \\ s\text{'}\text{'}\left(t\right)=-\frac{12}{5}t-34 \end{gathered}$$

Part (b) Step 3. Substitute t = 5 in s ' ' t .

Substitute $t=5$ in $s\text{'}\text{'}\left(t\right)$,

$$\begin{gathered} s\text{'}\text{'}\left(5\right)=-\frac{12}{5}\left(5\right)-34 \\ =-12-34 \\ =-46 \end{gathered}$$

From the calculation, it is observed that the values of $s\left(5\right),s\text{'}\left(0\right),s\text{'}\left(5\right),\mathrm{s}\text{'}\text{'}\left(5\right)$ are $100,0,-200,-46$.

Thus, it is verified that the cubic polynomial produces the same values of $s\left(5\right),s\text{'}\left(0\right),s\text{'}\left(5\right),\mathrm{s}\text{'}\text{'}\left(5\right)$.

Part (c) Step 1. Substitute t = 0 in the function s t .

The fruit was dropped from the tower at $t=0$.

Substitute $t=0$ in the function $s\left(t\right)$,

$$\begin{gathered} s\left(0\right)=-\frac{2}{5}{\left(0\right)}^3-17{\left(0\right)}^2+575 \\ =575 \end{gathered}$$

Therefore, the height from which the fruit was dropped is $575 ft$.

Part (d) Step 1. Determine the function a t .

Consider the previous part,

The second derivative of the function $s\left(t\right)=-\frac{12}{5}t-34$.

The second derivative of the height function $s\text{'}\text{'}\left(t\right)$ is the acceleration function. Thus, the acceleration function $s\text{'}\text{'}\left(t\right)=-\frac{12t}{5}-34$.

The acceleration function depends on the variable t. Therefore, the function $s\text{'}\text{'}\left(t\right)$ is not a constant value.

The acceleration is the rate of change of the velocity with time. The gravity on the planet acts downward. Due to gravity, the planet moves downward.