Mathematics, 04.08.2021 08:40 reagancunningham2004

Lisa is on the west shore of Mighty River, which is 1 mile wide and has two parallel shorelines running exactly north-to-south. She wishes to get to a point on the opposite shore that is 1 mile south of where she is now as quickly as possible. (So this point is sqrt (2) miles due southeast from her starting position.) Assuming that Lisa can walk twice as fast as she can swim, and that she wants to swim across the river first before walking any necessary distance along the opposite shoreline, at what course (in degrees) should she start swimming? (The angle is measured from north, so 90 degrees is due east, meaning she swims directly across the river, and 135 degrees is due southeast, meaning she swims directly to her destination point.)

Lisa is on the west shore of Mighty River, which is 1 mile wide and has two parallel shorelines run

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Answer from: MrKrinkle77

Lisa should start swimming at an angle of $120^{\circ}$ to reach her destination as quickly as possible.

Given:

The river is 1 mile wideThe end point is 1 mile south of the starting pointLisa's speed of walking is twice that of her speed of swimmingLisa wants to swim across the river first before walking the rest of the distance along the opposite shoreline

To find: The course (in degrees) at which she should start swimming so that she can reach the end point as quickly as possible

To solve this problem, we need to know the following:

Pythagoras theorem states that in a right angled triangle, the square of the length of hypotenuse is the sum of the squares of lengths of perpendicular and baseCosine of an angle is the ratio of base and hypotenuse in the right angled triangleA single variable function is minimized by a value for which, the first derivative of the function is zero and the second derivative of the function is positive

Let us assume that Lisa starts swimming at a course such that she needs to swim for 'x' miles to reach the opposite shore, as shown in the figure.

Labelling the points in the figure, we can say that,

AB = 1 mile (given)

BD = 1 mile (given)

AC = x miles (by our assumption)

It is clear that ABC forms a right angled triangle where,

Perpendicular: BC

Base: AB

Hypotenuse: AC

Using Pythagoras Theorem for triangle ABC, we have,

$(Hypotenuse)^{2} =(Perpendicular)^{2} +(Base)^{2}$ , which implies,

$(AC)^{2} =(BC)^{2} +(AB)^{2}$

Put AC = x, AB = 1 in the above equation to get,

$x^{2} =(BC)^{2} +1^{2}$

$x^{2} =(BC)^{2} +1$

$(BC)^{2}=x^{2} -1$

$BC=\sqrt{x^{2} -1}$

From the figure, we can say that,

CD=BD-BC

Put $BC=\sqrt{x^{2} -1}$ and BD=1 in the above equation to get,

$CD=1-\sqrt{x^{2} -1}$

According to our assumption, Lisa swims the distance of AC and walks the distance of CD, that is, she swims a distance of miles and walks a distance of $1-\sqrt{x^{2} -1}$ miles.

Now, let us assume that Lisa's speed of swimming is miles/hour. It is given that Lisa's speed of walking is twice that of her speed of swimming. Then, accordingly, Lisa's speed of walking must be miles/hour. We note that these speeds are constants for Lisa.

We know that,

$Speed=\frac{Distance}{Time}$

Then,

$Time=\frac{Distance}{Speed}$

This implies that,

Time spent by Lisa on swimming = $\frac{x}{k}$

Similarly, time spent by Lisa on walking = $\frac{1-\sqrt{x^{2} -1} }{2k}$

Then, total time taken by Lisa to travel the whole distance from the starting point to the end point is,

$\frac{1-\sqrt{x^{2} -1} }{2k} +\frac{x}{k}$

$\frac{1}{2k}( 1-\sqrt{x^{2} -1} +2x )$

Since Lisa wishes to get to the end point as quickly as possible, we must minimize the total time taken by her to travel the entire distance.

Total time taken: $T=\frac{1}{2k}( 1-\sqrt{x^{2} -1} +2x )$

Differentiating with respect to 'x', we have,

$T'=\frac{1}{2k}( -\frac{x}{\sqrt{x^{2} -1}} +2 )$

Differentiating with respect to 'x' again, we have,

$T''=\frac{1}{2k(\sqrt{x^{2} -1})^{3}}$

Equating the first derivative to 0, we have,

$\frac{1}{2k}( -\frac{x}{\sqrt{x^{2} -1}} +2 )=0$

$-\frac{x}{\sqrt{x^{2} -1}} +2 =0$

$\frac{x}{\sqrt{x^{2} -1}} =2$

$2\sqrt{x^{2} -1}= x$

Squaring both sides,

$4(x^{2} -1)= x^{2}$

$4x^{2} -4= x^{2}$

$3x^{2} =4$

$x^{2} =\frac{4}{3}$

$x=\frac{2}{\sqrt{3}}$

Note that we assumed the positive square root for 'x' because 'x' denotes a distance which cannot be negative.

Put $x=\frac{2}{\sqrt{3}}$ in the expression for second derivative to get,

$T''=\frac{1}{2k(\sqrt{(\frac{2}{\sqrt{3}} )^{2} -1})^{3}}$

$T''=\frac{1}{2k(\sqrt{\frac{4}{3} -1})^{3}}$

$T''=\frac{1}{2k(\sqrt{\frac{1}{3}})^{3}}0$

The last expression is positive because 'k' denotes a speed which is always positive.

This implies that the obtained value $x=\frac{2}{\sqrt{3}}$ minimizes the quantity of total time taken.

Now, from the figure, we can say that,

$cos(\angle BAC)=\frac{AB}{AC}$

Put AC = x, AB = 1 in the above equation to get,

$cos(\angle BAC)=\frac{1}{x}$

Put the obtained minimizing value, $x=\frac{2}{\sqrt{3}}$ in the above equation to get,

$cos(\angle BAC)=\frac{1}{\frac{2}{\sqrt{3}} }$

$cos(\angle BAC)=\frac{\sqrt{3}}{2 }$

$cos(\angle BAC)=cos(30^{\circ})$

Then,

$\angle BAC=30^{\circ}$

Since the angle is measured from the north, the required angle is, $90^{\circ}+30^{\circ}=120^{\circ}$

Thus, Lisa should start swimming at an angle of $120^{\circ}$ to reach her destination as quickly as possible.

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