Q. 62 - Calculus

Question

While you are on a camping trip, your tent accidentally catches fire. At the time, you and the tent are both 50 feet from a stream and you are 200 feet away from the tent, as shown in the diagram. You have a bucket with you, and need to run to the stream, fill the bucket, and run to the tent as fast as possible. You can run only half as fast while carrying the full bucket as you can empty handed, and thus any distance travelled with the full bucket is effectively twice as long. Complete parts (a)–(f) to determine how you can put out the fire as quickly as possible.

(a) Let x represent the distance from the point on the stream directly “below” you to the point on the stream that you run to. Sketch the path that you would follow to run from your starting position, to the point x along the stream, to the tent.

(b) Let D(x) represent the effective distance (counting twice any distance travelled while carrying a full bucket) you have to run in order to collect water and get to the tent. Write a formula for D(x).

(c) Determine the interval I of x-values on which D(x) should be minimized. Explain in practical terms what happens at the endpoints of this interval, and calculate the value of D(x) at these endpoints.

(d) Find D'(x), and simplify as much as possible. Are there any points (in the intervaI) at which D'(x) is undefined?

(e) It is difficult to find the zeroes of D'(x) by hand. Use a graphing utility to approximate any zeroes of D'(x) in the interval I, and test these zeroes by evaluating D' at each one.

(f) Use the preceding information to determine the minimum value of D(x), and then use this value to answer the original word problem.

Step-by-Step Solution

Answer

(a) The path that would follow to run from the starting position is:

(b) The formula for D(x) is $D\left(x\right)=\sqrt{x^2+2500}+\sqrt{{(200-x)}^2+{\left(50\right)}^2}$.

(c) The interval I of x-values on which D(x) should be minimized is $[0,200]$ and the value at these intervals is $256.16$.

(d) The value is $D\text{'}\left(x\right)=\frac{x}{\sqrt{(x^2+2500)}}-\frac{x}{\sqrt{({(200-x)}^2+{\left(50\right)}^2)}}$ and there are no points in I such that the expression is undefined.

(e) The zeroes of D'(x) is $x=100$.

(f) The minimum value of D(x) is $253.60 feet$

Part (a) Step 1. Given Information.

The distance between me, the tent is 50 feet from the stream.

The distance between me and the tent is 200 feet

Part (a) Step 2. Draw the diagram.

Let A be the position of the person, D is the position of the tent and F is the point on the stream, where the water is filled in the bucket.

Then $BE=x; EC=200-x$.

Part (b) Step 1. Find the distance.

Consider $∆ABE$,

$$\begin{gathered} AE=\sqrt{{\left(BE\right)}^2+{\left(AB\right)}^2} \\ =\sqrt{x^2+{50}^2} \\ =\sqrt{x^2+2500} \\ In ∆ECD, \\ ED=\sqrt{{(200-x)}^2+{\left(50\right)}^2} \\ AE+ED=\sqrt{x^2+2500}+\sqrt{{(200-x)}^2+{\left(50\right)}^2} \\ D\left(x\right)=\sqrt{x^2+2500}+\sqrt{{(200-x)}^2+{\left(50\right)}^2} \end{gathered}$$

Part (c) Step 1.

At the left end point, the person will run due south from his initial position towards the stream, fill the water in the bucket and run diagonally from the stream towards the tent.

At the right end point , the person will run diagonally from his initial position along the stream and fill the water in a bucket from a point which is due south from the tent.and then run north from that point to the tent. 

Substitute $x=0$ in the function,

$$\begin{gathered} D\left(x\right)=\sqrt{x^2+2500}+\sqrt{{(200-x)}^2+{\left(50\right)}^2} \\ D\left(0\right)=\sqrt{0^2+2500}+\sqrt{{(200-0)}^2+{\left(50\right)}^2} \\ =50+206.16 \\ =256.16 \end{gathered}$$

Part (c) Step 2. Substitute x = 200 in the function.

Substitute $x=200$ in the function,

$$\begin{gathered} D\left(200\right)=\sqrt{{200}^2+2500}+\sqrt{{(200-200)}^2+{\left(50\right)}^2} \\ =256.16 \end{gathered}$$

The value of D(x) at both the interval is $[0,200]$.is $256.16$.

So the interval is $[0,200]$

Part (d) Step 1. Differentiate the function.

Differentiate the function with respect to x,

$$\begin{gathered} D\left(x\right)=\sqrt{x^2+2500}+\sqrt{{(200-x)}^2+{\left(50\right)}^2} \\ =\frac{1}{2}{(x^2+2500)}^{-\frac{1}{2}}.2x+\frac{1}{2}({(200-x)}^2+{\left(50\right)}^2{)}^{-\frac{1}{2}}.-2x \\ =\frac{x}{{(x^2+2500)}^{{\displaystyle \frac{1}{2}}}}-\frac{x}{({(200-x)}^2+{\left(50\right)}^2{)}^{\frac{1}{2}}} \\ =\frac{x}{\sqrt{(x^2+2500)}}-\frac{x}{\sqrt{({(200-x)}^2+{\left(50\right)}^2)}} \end{gathered}$$

Part (d) Step 2. Points possible?

The term becomes undefined only when denominator is zero.

Since the variables in the expression are squared, so the expressions will never become zero.

So there are no points in I such that the value is undefined.

Part (e) Step 1. Use the calculator to find zeroes.

By using the calculator, the zeroes of the function $D\text{'}\left(x\right)=\frac{x}{\sqrt{(x^2+2500)}}-\frac{x}{\sqrt{({(200-x)}^2+{\left(50\right)}^2)}}$ is $x=100$.

For verification, substitute x=100,

$$\begin{gathered} D\text{'}\left(100\right)=\frac{x}{\sqrt{{\left(100\right)}^2+2500)}}-\frac{x}{\sqrt{({(200-100)}^2+{\left(50\right)}^2)}} \\ =\frac{x}{\sqrt{({100}^2+2500)}}-\frac{x}{\sqrt{({100}^2+2500)}} \\ =0 \end{gathered}$$

Part (f) Step 1. Minimum value of x .

Substitute $x=100$ in the original function,

$$\begin{gathered} D\left(x\right)=\sqrt{x^2+2500}+\sqrt{{(200-x)}^2+{\left(50\right)}^2} \\ D\left(100\right)=\sqrt{{100}^2+2500}+\sqrt{{(200-100)}^2+{\left(50\right)}^2} \\ =2\sqrt{12500} \\ =223.60 feet \end{gathered}$$