Q. 60 - Calculus

Question

Jimmy is doing some arithmetic problems. As the evening wears on, he gets less and less effective, so that $t\ge 1$minutes after the start of his study session he can solve $r\left(t\right)=1+\frac{1}{t}$ problems per minute. Assume that it takes $1$ minute for Jimmy to set up his work area; thus he is not doing arithmetic problems in the first minute.

Part (a): How many arithmetic problems per minute can Jimmy do when he first begins his studies at time $t = 1$? What about at $$\begin{gathered} t=4, \\ t=20 \end{gathered}$$?

Part (b): Make a rough estimate of the number of arithmetic problems Jimmy will have completed after $4$ minutes.

Part (c): Use an integral to express the number of arithmetic problems completed after t minutes, and interpret this definite integral as a logarithm. Calculate the number of problems Jimmy can complete in $10$ minutes and in $20$ minutes.

Part (d): Approximately how long will it take for Jimmy to finish his arithmetic homework if he must complete $40$ problems?

Step-by-Step Solution

Answer

Part (a): The person solves $2$ problems in the first minute when the person begins.

The person solves the number of problems at,

$t=4$ is $1.25$

$t=20$ is $1.05$

Part (b): An approximated value of number of arithmetic problems that the person solve after $4$ minutes is $4$.

Part (c): The no. of problems that can be solved by Jimmy after $10,20$ minutes after $11,22$.

Part (d): The approximated time to solve $40$ problems for Jimmy is $37$ minutes.

Part (a) Step 1. Given information.

 Consider the given question,

The function is $r\left(t\right)=1+\frac{1}{t}$.

Where $t\ge 1$.

Part (a) Step 2. Determine the no. of arithmetic problems per minute that Jimmy can complete.

The function $r\left(t\right)$ is not defined for $t=0$.

Substitute $t=1$ in the given function,

$$\begin{gathered} r\left(1\right)=1+\frac{1}{1} \\ =2 \end{gathered}$$

The person solves $2$ problems in the first minute when the person begins.

Substitute $t=4$ in the given function,

$$\begin{gathered} r\left(4\right)=1+\frac{1}{4} \\ =1+0.25 \\ =1.25 \end{gathered}$$

Substitute $t=20$ in the given function,

$$\begin{gathered} r\left(20\right)=1+\frac{1}{20} \\ =1+0.05 \\ =1.05 \end{gathered}$$

The person solves the number of problems at,

$t=4$ is $1.25$

$t=20$ is $1.05$

Part (b) Step 1. Determine the no. of problems Jimmy can complete after 4 minutes.

Integrate the given function $r\left(t\right)$ from $1 to 4$ with respect to t.

${\int }_1^4\left(1+\frac{1}{t}\right)dt={\int }_1^{4}1 dt+{\int }_1^4 \frac{1}{t} dt$

For a function $f\left(x\right)$ which is continuous on an interval $\left[a,b\right]$, the fundamental theorem of calculus states that ${\int }_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$, where F  is the anti-derivative of the function f.

Use the fundamental theorem of calculus in the integral ${\int }_1^{4}1 dt+{\int }_1^4 \frac{1}{t} dt$,

$$\begin{gathered} {\int }_1^{4}1 dt+{\int }_1^4 \frac{1}{t} dt={{\left[t\right]}^4}_1+{{\left[\ln t\right]}^4}_1 \\ =\left[4-1\right]+\left[\ln 4-\ln 1\right] \\ =3+\ln 4-0 \\ \approx 4 \end{gathered}$$

Part (c) Step 1. Calculate the no. of problems Jimmy can complete in 10 , 20 minutes.

Integrate the given function $r\left(t\right)$ from $1$ to t with respect to t,

${\int }_1^t\left(1+\frac{1}{t}\right)dt={\int }_1^t 1 dt+{\int }_1^t\frac{1}{t} dt$

For a function $f\left(x\right)$ which is continuous on an interval $\left[a,b\right]$, the fundamental theorem of calculus states that ${\int }_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$, where F  is the anti-derivative of the function f.

Use the fundamental theorem of calculus in the integral ${\int }_1^{t}1 dt+{\int }_1^t \frac{1}{t} dt$,

$$\begin{gathered} {\int }_1^{t}1 dt+{\int }_1^t \frac{1}{t} dt={{\left[t\right]}^t}_1+{{\left[\ln t\right]}^t}_1 \\ =\left[t-1\right]+\left[\ln t-\ln 1\right] \\ =t+\ln t-1 ...... \left(i\right) \end{gathered}$$

Part (c) Step 2. Determine the no. of problems Jimmy can complete after 4 minutes.

Substitute $t=10$ in equation (i),

$$\begin{gathered} =10+\ln 10-1 \\ =9+\ln 10 \\ \approx 11 \end{gathered}$$

Substitute $t=20$ in equation (i),

$$\begin{gathered} =20+\ln 20-1 \\ =19+\ln 20 \\ \approx 22 \end{gathered}$$

Therefore, the no. of problems that can be solved by Jimmy after $10,20$ minutes are  $11,22$.

Part (d) Step 1. Determine how long it will take Jimmy to complete 40 problems.

The no. of solved problems after t minutes is ${\int }_1^{t}r\left(t\right) dt=t+\ln t-1$.

Equate the function with $40$ to calculate the value of t,

$$\begin{gathered} 40=t+\ln t-1 \\ 41=t+\ln t \\ t+\ln t-41=0 \end{gathered}$$

Consider a function $f\left(t\right)=t+\ln t-41 ......\hspace{0.17em}\left(ii\right)$

Substitute $t=4$ in equation (ii),

$$\begin{gathered} f\left(4\right)=4+\ln 4-41 \\ =-35.61 \end{gathered}$$

Make the table of values of the function for different values of t,

Part (d) Step 2. Plot the graph.

On plotting the graph,

From the graph, it is observed that the graph of the function passes through the point $\left(37.379,0\right)$.

Therefore, an approximated value of time to solve $40$ problems is $37$ minutes.