Q. 68 - Calculus

Question

Water is evaporating from an open vase. The rate of evaporation changes according to the amount of exposed surface area at the top of the container in such a way that the rate of change of the volume of water in the vase is given by 

$v\left(t\right)=0.00015(t-5)(t-10)(t-15)(t-20)$

cubic centimeters per hour. The vase initially contains 200 cubic centimeters of water. Use the Net Change Theorem to determine the amount of time it will take for all of the water in the vase to evaporate 

Step-by-Step Solution

Answer

The net distance travelled in 5 seconds $\approx 5.27$

The net distance travelled in 10 seconds $\approx 3.125$

The net distance travelled in 20 seconds $\approx 0$

Step 1. Given information

The given rate of change of the volume of water in the vase is given by   

$v\left(t\right)=0.00015(t-5)(t-10)(t-15)(t-20)$

Step 2. Finding the net distance travelled in 5 and 10 seconds

The velocity of the machine is,

$v\left(t\right)=0.00015(t-5)(t-10)(t-15)(t-20)$

The mechanics of the music box are programmed to move a figurine back and forth with the velocity $\left(v\left(t\right)\right)$ inches per second over 20 seconds.
The objective is to describe the direction of motion of the figurine using the graph shown in the book during various parts of the music program.
Considering that the "positive" direction is right and "negative" direction is left, and then the figurine moves right along the track for the first 5 seconds then to the left for the next five seconds and continues.
The objective is to determine the net distance travelled by the figurine in the first 5 seconds, 10 seconds and the full 20 seconds music program.
The net distance travelled in 5 seconds is,

$$\begin{gathered} {\int }_0^{5}v\left(t\right)dt={\int }_0^{5}0.00015(t-5)(t-10)(t-15)(t-20) dt \\ \approx 5.27 \end{gathered}$$

The net distance travelled in 10 seconds is, 

$$\begin{gathered} {\int }_0^{10}v\left(t\right)dt={\int }_0^{10}0.00015(t-5)(t-10)(t-15)(t-20) dt \\ \approx 3.125 \end{gathered}$$

Step 3. Finding the net distance travelled in 20 seconds

The net distance travelled in 20 seconds is,

$$\begin{gathered} {\int }_0^{20}v\left(t\right)dt={\int }_0^{20}0.00015(t-5)(t-10)(t-15)(t-20) dt \\ \approx 0 \end{gathered}$$

Therefore,the net distance travelled are $\approx 5.27;\approx 3.125;\approx 0$