Q. 70 - Calculus

Question

During World War II, a handful of German submarines were captured by the Allies to get access to German codes. On one of these submarines, the submarine’s depth had been plotted carefully as D(t), in meters, where t measures the number of hours after noon on that day. Unfortunately, during the capture, the information about D(t) was lost. However, the following graph of the rate of change D'(t) of the submarine’s depth was recovered:

(a) At 4:00 p.m., was the submarine rising or falling? Was it speeding up or slowing down? 

(b) Was the submarine closer to the surface at 2:00 p.m. or at 5:00 p.m.?

(c) At the end of the first two hours of the voyage, had the submarine risen or fallen from its position at the beginning? By how much?

(d) Approximately how much did the submarine rise or fall in the two minutes after 6:00 p.m.? (e) At what time was the submarine diving at the fastest rate? 

(f) When was the submarine at its highest point? 

(g) Suppose the submarine was on the surface when it was at its highest point. What was the greatest depth of the submarine during its voyage that day? 

Step-by-Step Solution

Answer

(a) The submarine falls at 4:00 P.M and it is slowing down. 

(b) The submarine is closer to the surface at 2:00 P.M. 

(c) The submarine rose at the end of the first two hours from its original position. 

(d) the submarine rises $0.02$meters per minute in 2 minutes after 6 PM

(e) At  $t=3$ the submarine drives at the fastest rate which is equal to 35 meters per hour.

(f)  ${\int }_{1.5}^{4.5}D\text{'}\left(t\right)d$was the submarine at its highest point

(g) The greatest depth of the submarine is ${\int }_{1.5}^{4.5}D\text{'}\left(t\right)dt=9$

Part (a) Step 1. Given information

During World War II, a handful of German submarines were captured by the Allies to get access to German codes 

On one of these submarines, the submarine’s depth had been plotted carefully as D(t), in meters, where t measures the number of hours of afternoon on that day. 

The given time is $4:00$PM.

Part (a) Step 2. State of the submarine at 4 : 00   PM.

The time t indicates the number of hours afternoon. Therefore, t=4 corresponds to the time 4:00 P.M.

From the graph, it is observed that the curve of the rate of change $D\text{'}\left(t\right)$of depth of a submarine falling after $t=4$which implies that at$t=4,D\text{'}\left(t\right)<0$.
Thus, the submarine falls at 4:00 P.M and it is slowing down.

Part (b) Step 1. Given information

The submarine is closer to the surface at 2:00 PM. or at 5:00 P.M.  

Part (b) Step 1. Time at which the submarine is closer to the surface.

The x-axis represents the surface of the submarine.
From the graph, it is observed that the position of the submarine at 2:00 P.M. is (2,10) which is closer to the surface. The position of the submarine at 5:00 P.M is $(5,-30)$which is farthest from the surface.
Thus, the submarine is closer to the surface at 2:00 P.M.  

Part (c) Step 1. submarine state at the end of the first two hours from its original position.

The position of the submarine at t=0 is $(0,-0.3)$and the position of the submarine at $t=2$is$(2,10)$.
Therefore, the submarine rose at the end of the first two hours from its original position.  

Part (d) Step 1. Given information

The given time is:

Two minutes after 6:00 p.m. 

Part (d) Step 2. State of the submarine in two minutes after 6:00 p.m.

The time t indicates the number of hours afternoon. Therefore, $t=6$corresponds to the time $6.00$P.M.

The position of the submarine at $t=6$is $(6,-50)$and the position of the submarine at $t=7$is$(7,-20)$.
The acceleration of a submarine between $t=6$is and $t=7$is $30$meters per hour.
Consider x as the position of the submarine two minutes after $6:00$P.M.
The acceleration is the same on the interval$\left[6,7\right]$. The acceleration is the ratio of speed and time.

$$\begin{gathered} \frac{-50-x}{2}=\frac{30}{(60.60)} \\ \frac{-50-x}{2}=\frac{1}{120} \\ -6000-120x=2 \end{gathered}$$

Move all the variables on the left side and constants on the right side of the equation.

$$\begin{gathered} -120x=2+6000 \\ -120x=6002 \\ x=-\frac{6002}{120} \\ \approx -50.02 \end{gathered}$$

therefore, the submarine rises $0.02$meters per minute in $2$minutes after $6:00$P.M.

Part (e) Step 1. Time at which the submarine dives at the fastest rate

The main objective is to determine the time when the submarine dives at the fastest rate.
From the graph, it is observed that the curve $D\text{'}\left(t\right)$is maximum at$t=3$.

At$t=3$, the value of the function is$35$.
Thus at $t=3$, the submarine drives at the fastest rate of$35$ meters per hour.  

Part (f) Step 1. highest point of the submarine

The given curve represents the speed of the submarine at the depth. The x-axis represents the surface.

On the x-axis, the speed of the submarine is zero which is the highest point of the submarine.
The speed of the submarine is zero at $t=1.5,t=4.5$and$t=8$.
Thus, the submarine is at its highest point at $t=1.5,t=4.5$and$t=8$.

Part (f) Step 1. The greatest depth of the submarine during its voyage that day.

The area under the curve $D\text{'}\left(t\right)$indicates the distance traveled. The largest area gives the greatest depth of the submarine.
From the graph, it is observed that the curve has the largest area on an interval$\left[1.5,4.5\right]$.
The area under the curve over an interval $\left[1.5,4.5\right]$is${\int }_{1.5}^{4.5}D\text{'}\left(t\right)dt$.
Thus, the greatest depth of the submarine is:

$$\begin{gathered} {\int }_{1.5}^{4.5}D\text{'}\left(t\right)dt={\left[\frac{D{\text{'}}^2}{2}\right]}_{1.5}^{4.5} \\ =\left[\frac{{(4.5)}^2}{2}-\frac{{\left(1.5\right)}^2}{2}\right] \\ =\frac{20.25}{2}-\frac{2.25}{2} \\ =\frac{18}{2} \\ =9 \end{gathered}$$