# Calculus

## Applications of the Derivative # Optimization Problems in Physics

## Solved Problems

Click or tap a problem to see the solution.

### Example 9

A body is placed on an inclined plane and put in motion upwards with the initial velocity v0. The coefficient of friction between the body and the inclined plane is k. Determine the angle of inclination α at which the body travels upwards the smallest distance.

### Example 10

A ski jumper travels down a slope of the height H and leaves the ski track moving in the horizontal direction. The only force acting on the ski jumper is gravity. Find the height of the ski track h at which the ski jumper flies the longest distance.

### Example 11

Find the minimum initial speed of a champagne cork that travels a horizontal distance of $$L$$ meters.

### Example 12

A light source is located on the line segment joining the centers of the spheres with radii $${R_1}$$ and $${R_2}.$$ Determine the location of the source, at which the total area of the illuminated surface of the two spheres is the largest.

### Example 13

Two electrical charges labeled $$A$$ and $$B$$ are located a distance $$d$$ apart. Charge $$A$$ is positive of magnitude $${q_A}$$ and charge $$B$$ is negative of magnitude $${q_B}.$$ Where a positively charged particle $$P$$ should be placed so that the total Coulomb force due to the charges $$A$$ and $$B$$ is minimal?

### Example 14

A charged particle is placed on the axis of a charged ring of radius $$R$$. Find the distance $$x$$ from the centre of the ring at which the Coulomb force acting on the particle is the largest.

### Example 15

A vessel filled with liquid of height $$h$$ is on a horizontal surface. The vessel has a side opening, through which the liquid can flow out. At what position of the opening the liquid jet has the largest range?

### Example 16

Find the maximum possible temperature of an ideal gas undergoing the process $p = {p_0}{e^{ - \beta V}},$ where $$\beta$$ is a positive constant, $$V$$ is the molar volume.

### Example 17

A tourist wants to get across a round shaped lake from point $$A$$ to point $$B.$$ This can be done by sailing on a boat or moving on foot along the coastline. Determine the fastest route, if the boat speed is $$u$$ and walking speed is $$v.$$

### Example 18

Suppose you decide to walk from point $$A$$ to point $$B$$. The point $$A$$ is in the forest where you can walk only at $$v = 3\frac{{\text{km}}}{{\text{hr}}}.$$ However, along the road you can walk at $$u = 5\frac{{\text{km}}}{{\text{hr}}}.$$ The distance from point $$A$$ to the road is $$q = 4\,\text{km},$$ and the distance from $$S$$ to $$B$$ is $$p = 8\,\text{km}.$$ How to choose point $$M$$ so as to minimize the time $$T$$ it takes to get from $$A$$ to $$B?$$

### Example 19

Car $$B$$ is $$b = 100$$ miles directly east of car $$A$$ and begins moving north at $$v = 50$$ miles per hour. At the same moment, car $$A$$ begins moving north-east at the same speed $$v = 50$$ miles per hour. Find the time at which the distance between the two cars is minimum.

### Example 20

The first car $$A$$ moves in the northern direction at a speed $$u\left(\text{mph}\right)$$, and the second car $$B$$ moves to the eastern direction with a speed $$v\left(\text{mph}\right).$$ At the initial time, the car $$B$$ is located $$b\left(\text{miles}\right)$$ east from the car $$A$$. Determine the minimum distance between the cars.

### Example 21

The strength of a rectangular beam is directly proportional to the product of its width $$w$$ and the square of its depth $$d.$$ Find the dimensions of the strongest beam that can be cut from a cylindrical log of radius $$R.$$

### Example 9.

A body is placed on an inclined plane and put in motion upwards with the initial velocity $${v_0}.$$ The coefficient of friction between the body and the inclined plane is $$k.$$ Determine the angle of inclination $$\alpha$$ at which the body travels upwards the smallest distance.

Solution.

We use Newton's Second Law to write the motion equation for the body in vector form:

$m\vec{a} = {\vec{F}_f} + m\vec{g} + \vec{N},$

where $$\vec{a}$$ is the acceleration of the box, $$m$$ is its mass, $$\vec{F}_f$$ is the friction force, $$m\vec{g}$$ is the weight of the body, and $$\vec{N}$$ is the normal force.

The coordinate system is chosen so that the $$x$$-axis is parallel to the inclined plane and the $$y$$-axis is orthogonal to the $$x$$-axis.

In components form, the above equation becomes

$\left\{ \begin{array}{l} ma = - {F_f} - mg\sin \alpha \;(x-\text{axis})\\ 0 = N - mg\cos \alpha \;\left( y-\text{axis} \right) \end{array} \right.$

As

${F_f} = kN = kmg\cos \alpha ,$

we get

$ma = - kmg\cos \alpha - mg\sin \alpha = - mg\left( {k\cos \alpha + \sin \alpha } \right).$

Hence, the magnitude (the absolute value) of the acceleration is given by

$a = g\left( {k\cos \alpha + \sin \alpha } \right).$

The distance and the velocity are defined by the equations

$s = {v_0}t - \frac{{a{t^2}}}{2},\;\;v = {v_0} - at,$

where $$a \gt 0.$$

The time of movement is

$t = \frac{{{v_0}}}{a}.$

Then the distance travelled for this time is equal to

$s = {v_0}t - \frac{{a{t^2}}}{2} = {v_0} \cdot \frac{{{v_0}}}{a} - \frac{{a{{\left( {\frac{{{v_0}}}{a}} \right)}^2}}}{2} = \frac{{v_0^2}}{a} - \frac{{v_0^2}}{{2a}} = \frac{{v_0^2}}{{2a}}.$

Substituting the acceleration $$a,$$ we obtain the function $$s\left( \alpha \right):$$

$s\left( \alpha \right) = \frac{{v_0^2}}{{2a}} = \frac{{v_0^2}}{{2g\left( {k\cos \alpha + \sin \alpha } \right)}}.$

The derivative of $$s\left( \alpha \right)$$ is given by

$s^\prime\left( \alpha \right) = \left( {\frac{{v_0^2}}{{2g\left( {k\cos \alpha + \sin \alpha } \right)}}} \right)^\prime = \frac{{v_0^2}}{{2g}} \cdot \frac{{k\sin \alpha - \cos \alpha }}{{{{\left( {k\cos \alpha + \sin \alpha } \right)}^2}}}.$

Determine the critical value of $$\alpha:$$

$s^\prime\left( \alpha \right) = 0,\;\; \Rightarrow k\sin \alpha - \cos \alpha = 0,\;\; \Rightarrow \tan \alpha = \frac{1}{k},\;\; \Rightarrow \alpha = \arctan \frac{1}{k}.$

Using the First Derivative Test, we can show that the angle $$\alpha = \arctan \frac{1}{k}$$ corresponds to the smallest distance travelled upwards.

### Example 10.

A ski jumper travels down a slope of the height $$H$$ and leaves the ski track moving in the horizontal direction. The only force acting on the ski jumper is gravity. Find the height of the ski track $$h$$ at which the ski jumper flies the longest distance.

Solution.

The horizontal take-off velocity $$v$$ is given by the formula:

$v = \sqrt {2g\left( {H - h} \right)} .$

We use the well-known equation $$h = \frac{{g{t^2}}}{2}$$ to find the airborne time. Then

$t = \sqrt {\frac{{2h}}{g}} .$

Hence, the horizontal distance (s) is expressed in the form

$s = vt = \sqrt {2g\left( {H - h} \right)} \sqrt {\frac{{2h}}{g}} = 2\sqrt {\left( {H - h} \right)h} = 2\sqrt {Hh - {h^2}} = s\left( h \right).$

Take the derivative:

$s^\prime\left( h \right) = \left( {2\sqrt {Hh - {h^2}} } \right)^\prime = \frac{{2\left( {H - 2h} \right)}}{{2\sqrt {Hh - {h^2}} }} = \frac{{H - 2h}}{{\sqrt {Hh - {h^2}} }}.$

Find the critical points:

$s^\prime\left( h \right) = 0,\;\; \Rightarrow \frac{{H - 2h}}{{\sqrt {Hh - {h^2}} }} = 0,\;\; \Rightarrow H - 2h = 0,\;\; \Rightarrow h = \frac{H}{2}.$

So, the critical points are

$h = 0,\,\frac{H}{2},\,H.$

Using the First Derivative Test, we can show that the range $$s$$ has a maximum at $$h = \frac{H}{2}.$$

### Example 11.

Find the minimum initial speed of a champagne cork that travels a horizontal distance of $$L$$ meters.

Solution.

First we determine the angle $$\alpha$$ that gives the maximum range $$L$$ for given initial speed.

The initial speed is written in vector form as

${\vec{v}_0} = {\vec{v}_{0x}} + {\vec{v}_{0y}},$

where

${v_{0x}} = {v_0}\cos \alpha,\;\;{v_{0y}} = {v_0}\sin \alpha .$

The $$y-$$coordinate of the cork is given by

$y = {v_{0y}}t - \frac{{g{t^2}}}{2} = t\left( {{v_{0y}} - \frac{{gt}}{2}} \right).$

The time of flight of the cork is determined from the condition $$y = 0.$$ Hence,

$t = \frac{{2{v_{0y}}}}{g} = \frac{{2{v_0}\sin \alpha }}{g}.$

Then the range $$L$$ is determined by the formula:

$L = {v_{0x}}t = {v_0}\cos \alpha \cdot \frac{{2{v_0}\sin \alpha }}{g} = \frac{{v_0^2\sin 2\alpha }}{g}.$

The range $$L$$ has a maximum value when

$\sin 2\alpha = 1,\;\; \Rightarrow 2\alpha = \frac{\pi }{2},\;\; \Rightarrow \alpha = \frac{\pi }{4} = 45^\circ.$

Thus, the maximum possible range is equal to

${L} = \frac{{v_0^2}}{g}.$

It follows from this equation that

${v_0} = \sqrt {Lg} .$

### Example 12.

A light source is located on the line segment joining the centers of the spheres with radii $${R_1}$$ and $${R_2}.$$ Determine the location of the source, at which the total area of the illuminated surface of the two spheres is the largest.

Solution.

Let the distance $${O_1}{O_2}$$ between the centers of the spheres be equal to $$L,$$ and the light source is at a distance $$x$$ from the center of the first sphere (Figure $$10$$).

The illuminated areas are the surface areas of spherical sectors. Their total area is given by the formula

$S = {S_1} + {S_2} = 2\pi {R_1}{h_1} + 2\pi {R_2}{h_2},$

where $${h_1},$$ $${h_2}$$ are the heights of the corresponding spherical caps.

Given the similarity of the right angled triangles $${AL{O_1}}$$ and $${{O_1}AI},$$ we can write:

${h_1} = {R_1} - {R_1}\sin {\alpha _1},\;\;\;\text{where}\;\;\sin {\alpha _1} = \frac{{{R_1}}}{x}.$

Hence,

${h_1} = {R_1} - {R_1} \cdot \frac{{{R_1}}}{x} = {R_1} - \frac{{R_1^2}}{x}.$

A similar expression for the second sphere is as follows:

${h_2} = {R_2} - \frac{{R_2^2}}{{L - x}}.$

So, the total illuminated area is

$S = S\left( x \right) = 2\pi {R_1}\left( {{R_1} - \frac{{R_1^2}}{x}} \right) + 2\pi {R_2}\left( {{R_2} - \frac{{R_2^2}}{{L - x}}} \right) = 2\pi R_1^2 - \frac{{2\pi R_1^3}}{x} + 2\pi R_2^2 - \frac{{2\pi R_2^3}}{{L - x}}.$

Find the derivative:

$S'\left( x \right) = \left( {2\pi R_1^2 - \frac{{2\pi R_1^3}}{x} + 2\pi R_2^2 - \frac{{2\pi R_2^3}}{{L - x}}} \right)^\prime = \frac{{2\pi R_1^3}}{{{x^2}}} - \frac{{2\pi R_2^3}}{{{{\left( {L - x} \right)}^2}}}.$

By equating the derivative to zero, we obtain the following solution

$S'\left( x \right) = 0,\;\; \Rightarrow \frac{{2\pi R_1^3}}{{{x^2}}} - \frac{{2\pi R_2^3}}{{{{\left( {L - x} \right)}^2}}} = 0,\;\; \Rightarrow \frac{{{x^2}}}{{{{\left( {L - x} \right)}^2}}} = \frac{{R_1^3}}{{R_2^3}}.$

When passing through zero the derivative changes sign from plus to minus, that is the above point is the maximum of the function $$S\left( x \right).$$

Thus, the illuminated area is the largest when the light source is located at a point, at which the ratio of the squares of the distances from the source to the centers of the spheres is as the ratio of their radii cubed.

### Example 13.

Two electrical charges labeled $$A$$ and $$B$$ are located a distance $$d$$ apart. Charge $$A$$ is positive of magnitude $${q_A}$$ and charge $$B$$ is negative of magnitude $${q_B}.$$ Where a positively charged particle $$P$$ should be placed so that the total Coulomb force due to the charges $$A$$ and $$B$$ is minimal?

Solution.

Suppose that the particle $$P$$ has a charge $$q.$$ Given that the Coulomb force is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between the charges, we can write:

${F_A} = k\frac{{{q_A}q}}{{{x^2}}},\;\;{F_B} = k\frac{{{q_B}q}}{{{{\left( {d - x} \right)}^2}}}.$

The total force is given by

$\vec F = \vec {F_A} + \vec {F_B},$
$F = {F_A} + {F_B} = k\frac{{{q_A}q}}{{{x^2}}} + k\frac{{{q_B}q}}{{{{\left( {d - x} \right)}^2}}} = F\left( x \right).$

Take the derivative:

$F^\prime\left( x \right) = \left( {k\frac{{{q_A}q}}{{{x^2}}} + k\frac{{{q_B}q}}{{{{\left( {d - x} \right)}^2}}}} \right)^\prime = - \frac{{2k{q_A}q}}{{{x^3}}} - \frac{{2k{q_B}q}}{{{{\left( {d - x} \right)}^3}}} \cdot \left( { - 1} \right) = 2kq\left( {\frac{{{q_B}}}{{{{\left( {d - x} \right)}^3}}} - \frac{{{q_A}}}{{{x^3}}}} \right).$

Determine the point where the derivative is zero:

$F^\prime\left( x \right) = 0,\;\; \Rightarrow 2kq\left( {\frac{{{q_B}}}{{{{\left( {d - x} \right)}^3}}} - \frac{{{q_A}}}{{{x^3}}}} \right) = 0,\;\; \Rightarrow \frac{{{q_B}}}{{{{\left( {d - x} \right)}^3}}} = \frac{{{q_A}}}{{{x^3}}},\;\; \Rightarrow \frac{{{q_B}}}{{{q_A}}} = {\left( {\frac{{d - x}}{x}} \right)^3},\;\; \Rightarrow \frac{{{q_B}}}{{{q_A}}} = {\left( {\frac{d}{x} - 1} \right)^3},\;\; \Rightarrow \frac{d}{x} - 1 = \sqrt{{\frac{{{q_B}}}{{{q_A}}}}},\;\; \Rightarrow \frac{d}{x} = 1 + \sqrt{{\frac{{{q_B}}}{{{q_A}}}}},\;\; \Rightarrow x = \frac{d}{{1 + \sqrt{{\frac{{{q_B}}}{{{q_A}}}}}}}.$

Using the First Derivative Test, one can show that the point $$x = \frac{d}{{1 + \sqrt{{\frac{{{q_B}}}{{{q_A}}}}}}}$$ is a point of minimum.

### Example 14.

A charged particle is placed on the axis of a charged ring of radius $$R$$. Find the distance $$x$$ from the centre of the ring at which the Coulomb force acting on the particle is the largest.

Solution.

Consider an infinitesimal piece of the ring with charge $$dQ.$$ The magnitude of the force at a point $$x$$ due to the charge is given by Coulomb's law:

$dF = k\frac{{dQ \cdot q}}{{{x^2} + {R^2}}}.$

The projection of the elementary force along the $$x$$-axis is written in the form

$d{F_x} = dF\cos \alpha = \frac{{kqdQ}}{{{x^2} + {R^2}}} \cdot \frac{x}{{\sqrt {{x^2} + {R^2}} }} = \frac{{kqdQx}}{{{{\left( {{x^2} + {R^2}} \right)}^{\frac{3}{2}}}}}.$

Integrating over the ring, one can show that the magnitude of the total force acting on the charge $$q$$ at the point $$x$$ is equal to

$F\left( x \right) = \frac{{kqQx}}{{{{\left( {{x^2} + {R^2}} \right)}^{\frac{3}{2}}}}}.$

The derivative of the function $$F\left( x \right)$$ is given by

$F^\prime\left( x \right) = \left( {\frac{{kqQx}}{{{{\left( {{x^2} + {R^2}} \right)}^{\frac{3}{2}}}}}} \right)^\prime = kqQ\frac{{{R^2} - 2{x^2}}}{{{{\left( {{x^2} + {R^2}} \right)}^{\frac{5}{2}}}}}.$

The derivative is equal to zero at the following point:

$F^\prime\left( x \right) = 0,\;\; \Rightarrow kqQ\frac{{{R^2} - 2{x^2}}}{{{{\left( {{x^2} + {R^2}} \right)}^{\frac{5}{2}}}}} = 0,\;\; \Rightarrow {R^2} - 2{x^2} = 0,\;\; \Rightarrow {x^2} = \frac{{{R^2}}}{2},\;\; \Rightarrow x = \frac{R}{{\sqrt 2 }}.$

Using the First Derivative Test, one can show that the point $$x = \frac{R}{{\sqrt 2 }}$$ corresponds to the maximum of the electrostatic force.

### Example 15.

A vessel filled with liquid of height $$h$$ is on a horizontal surface. The vessel has a side opening, through which the liquid can flow out. At what position of the opening the liquid jet has the largest range?

Solution.

Suppose that the liquid flow rate through the opening is described by Torricelli's Law:

$v = \sqrt {2g\left( {h - x} \right)} ,$

where $$x$$ is the height of the opening over the horizontal surface, $$h - x$$ is the height of the liquid column above the opening, $$g$$ is gravitational acceleration (Figure $$13$$).

Calculate the range of the jet $$y$$ assuming that the fluid motion is described by the same equations as the motion of a material point.

In this case, the time for which the liquid flowing out of the opening reaches the ground surface is given by

$t = \sqrt {\frac{{2x}}{g}} ,$

and the jet range is

$y = vt = v\sqrt {\frac{{2x}}{g}} .$

Substitute the expression for the velocity $$v$$ from Torricelli's Law:

$y = y\left( x \right) = \sqrt {2g\left( {h - x} \right)} \sqrt {\frac{{2x}}{g}} = 2\sqrt {\frac{{\cancel{g}x\left( {h - x} \right)}}{\cancel{g}}} = 2\sqrt {hx - {x^2}} .$

We have obtained an expression for the jet range $$y$$ depending on the height of the opening $$x.$$ We investigate the function for extreme points:

$y'\left( x \right) = \left( {2\sqrt {hx - {x^2}} } \right)^\prime = 2 \cdot \frac{1}{{2\sqrt {hx - {x^2}} } \cdot \left( {hx - {x^2}} \right)^\prime } = \frac{{h - 2x}}{{\sqrt {hx - {x^2}} }} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {h - 2x = 0}\\ {h - x \ne 0} \end{array}} \right.,\;\; \Rightarrow x = \frac{h}{2}.$

It is clear that this point is a maximum, because when passing through this value the derivative changes sign from plus to minus.

Thus, the opening for the flowing liquid must be positioned exactly in the middle of the liquid column. In this case, the range of the jet will be the greatest and equal to the height of the liquid column:

${y_{\max }} = 2\sqrt {h \cdot \frac{h}{2} - {{\left( {\frac{h}{2}} \right)}^2}} = 2\sqrt {\frac{{{h^2}}}{2} - \frac{{{h^2}}}{4}} = 2\sqrt {\frac{{{h^2}}}{4}} = \frac{{\cancel{2}h}}{\cancel{2}} = h.$

### Example 16.

Find the maximum possible temperature of an ideal gas undergoing the process $p = {p_0}{e^{ - \beta V}},$ where $$\beta$$ is a positive constant, $$V$$ is the molar volume.

Solution.

We write the ideal gas equation for $$1$$ mole is written in the form

$\frac{{pV}}{T} = R,$

where $$R$$ is the universal gas constant.

Substituting the expression for the pressure $$p,$$ we have

$\frac{{{p_0}{e^{ - \beta V}}V}}{T} = R,\;\; \Rightarrow T\left( V \right) = \frac{{{p_0}}}{R}{e^{ - \beta V}}V.$

Calculate the derivative of $$T\left( V \right)$$ and equate it to zero:

$T^\prime\left( V \right) = \left( {\frac{{{p_0}}}{R}{e^{ - \beta V}}V} \right)^\prime = \frac{{{p_0}}}{R}\left( { - \beta {e^{ - \beta V}}V + {e^{ - \beta V}}} \right) = \frac{{{p_0}{e^{ - \beta V}}}}{R}\left( {1 - \beta V} \right);$
$T^\prime\left( V \right) = 0,\;\; \Rightarrow \frac{{{p_0}{e^{ - \beta V}}}}{R}\left( {1 - \beta V} \right) = 0,\;\; \Rightarrow V = \frac{1}{\beta }.$

Using the First Derivative Test, we can see from the sign of $$T^\prime\left( V \right)$$ on each interval that the point $$V = \frac{1}{\beta }$$ is a point of maximum.

Now we can compute the maximum temperature of the gas:

${T_{\max }} = T\left( {V = \frac{1}{\beta }} \right) = \frac{{{p_0}}}{R}{e^{ - \beta \cdot \frac{1}{\beta }}} \cdot \frac{1}{\beta } = \frac{{{p_0}}}{{R\beta e}}.$

### Example 17.

A tourist wants to get across a round shaped lake from point $$A$$ to point $$B.$$ This can be done by sailing on a boat or moving on foot along the coastline. Determine the fastest route, if the boat speed is $$u$$ and walking speed is $$v.$$

Solution.

Let the radius of the lake be $$R.$$ Consider an arbitrary combined route when a part of the path runs across the lake and the other part along the coast. The route of movement is determined by the inscribed angle $$\alpha = \angle CAB$$ (Figure $$14$$).

The central angle $$\angle COB$$ is equal to $$2\alpha.$$ The length of the portion of the path across the lake is

$\left| {AC} \right| = 2\left| {AD} \right| = 2R\cos \alpha ,$

and the length of the portion of the path along the coast is

$\left| {CB} \right| = 2\alpha R,$

where the angle $$\alpha$$ is measured in radians and varies in the range $$0 \le \alpha \le {\frac{\pi }{2}}.$$ Then the total time of motion is written as follows:

$t = t\left( \alpha \right) = \frac{{\left| {AC} \right|}}{u} + \frac{{\left| {CB} \right|}}{v} = \frac{{2R\cos \alpha }}{u} + \frac{{2\alpha R}}{v} = 2R\left( {\frac{{\cos \alpha }}{u} + \frac{\alpha }{v}} \right).$

Differentiating, we find:

$t'\left( \alpha \right) = \left[ {2R\left( {\frac{{\cos \alpha }}{u} + \frac{\alpha }{v}} \right)} \right]^\prime = 2R\left( { - \frac{{\sin \alpha }}{u} + \frac{1}{v}} \right) = 2R\left( {\frac{1}{v} - \frac{{\sin \alpha }}{u}} \right).$

The derivative is equal to zero when

$\frac{1}{v} - \frac{{\sin \alpha }}{u} = 0,\;\; \Rightarrow \sin \alpha = \frac{u}{v},\;\; \Rightarrow \alpha = \arcsin \frac{u}{v}.$

We investigate the type of extremum using the second derivative:

$t^{\prime\prime}\left( \alpha \right) = {\left[ {2R\left( {\frac{1}{v} - \frac{{\sin \alpha }}{u}} \right)} \right]^\prime } = - \frac{{2R\cos \alpha }}{u} \lt 0.$

Since the second derivative is always negative for $$0 \le \alpha \le {\frac{\pi }{2}},$$ the above point is a maximum of the function $$t\left( \alpha \right),$$ which does not fit for us. It is clear that the function $$t\left( \alpha \right)$$ is convex upward in the interval $$\left( {0,{\frac{\pi }{2}}} \right)$$ and, hence, reaches its lower value in one of the endpoints of the interval. Calculate the values of $$t\left( \alpha \right)$$ at $$\alpha = 0$$ and $${\alpha = {\frac{\pi }{2}}}:$$

$t\left( {\alpha = 0} \right) = 2R\left( {\frac{{\cos 0}}{u} + \frac{0}{v}} \right) = \frac{{2R}}{u},$
$t\left( {\alpha = \frac{\pi }{2}} \right) = 2R\left( {\frac{{\cos \frac{\pi }{2}}}{u} + \frac{\pi }{{2v}}} \right) = \frac{{\pi R}}{v}.$

The case $$\alpha = 0$$ corresponds to the motion on the water along the straight line $$AOB,$$ and the case $${\alpha = \frac{\pi }{2}}$$ describes the movement on foot along the semicircle $$ACB.$$ Which of these options has the least time? This depends on the relationship between the speeds $$u$$ and $$v.$$ If

$\frac{u}{v} \gt \frac{2}{\pi },$

then boating (with the speed $$u$$) will take less time than walking on the ground (with the speed $$v\text{).}$$ And conversely, if

$\frac{u}{v} \lt \frac{2}{\pi },$

the movement along the coastline will be faster.

In case of equality of these ratios, both routes in terms of elapsed time will be equivalent.

### Example 18.

Suppose you decide to walk from point $$A$$ to point $$B$$. The point $$A$$ is in the forest where you can walk only at $$v = 3\frac{{\text{km}}}{{\text{hr}}}.$$ However, along the road you can walk at $$u = 5\frac{{\text{km}}}{{\text{hr}}}.$$ The distance from point $$A$$ to the road is $$q = 4\,\text{km},$$ and the distance from $$S$$ to $$B$$ is $$p = 8\,\text{km}.$$ How to choose point $$M$$ so as to minimize the time $$T$$ it takes to get from $$A$$ to $$B?$$

Solution.

The total time $$T$$ (in hours) is given by the equation

$T = \frac{{\left| {AM} \right|}}{v} + \frac{{\left| {MB} \right|}}{u} = \frac{{\left| {AM} \right|}}{3} + \frac{{\left| {MB} \right|}}{5}.$

Let the distance $${\left| {SM} \right|}$$ be equal to $$x.$$ Using the Pythagorean theorem, we have

$\left| {AM} \right| = \sqrt {{{\left| {AS} \right|}^2} + {x^2}} = \sqrt {{4^2} + {x^2}} = \sqrt {16 + {x^2}} ;$
$\left| {MB} \right| = \left| {SB} \right| - x = 8 - x.$

Then

$T = \frac{{\sqrt {16 + {x^2}} }}{3} + \frac{{8 - x}}{5} = T\left( x \right).$

Take the derivative:

$T^\prime\left( x \right) = \left( {\frac{{\sqrt {16 + {x^2}} }}{3} + \frac{{8 - x}}{5}} \right)^\prime = \frac{{2x}}{{6\sqrt {16 + {x^2}} }} - \frac{1}{5} = \frac{{5x - 3\sqrt {16 + {x^2}} }}{{15\sqrt {16 + {x^2}} }}.$

Determine the critical points:

$T^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{5x - 3\sqrt {16 + {x^2}} }}{{15\sqrt {16 + {x^2}} }} = 0,\;\; \Rightarrow 5x - 3\sqrt {16 + {x^2}} = 0,\;\; \Rightarrow 5x = 3\sqrt {16 + {x^2}} ,\;\; \Rightarrow \frac{{25{x^2}}}{9} = 16 + {x^2},\;\; \Rightarrow \frac{{16{x^2}}}{9} = 16,\;\; \Rightarrow {x^2} = 0,\;\; \Rightarrow x = 3.$

Using the First Derivative Test, we can show that $$x = 3\text{ km}$$ is a point of minimum.

### Example 19.

Car $$B$$ is $$b = 100$$ miles directly east of car $$A$$ and begins moving north at $$v = 50$$ miles per hour. At the same moment, car $$A$$ begins moving north-east at the same speed $$v = 50$$ miles per hour. Find the time at which the distance between the two cars is minimum.

Solution.

The coordinates of car $$A$$ are given by

$A = \left( {{x_A},{y_A}} \right) = \left( {\frac{{\sqrt 2 }}{2}vt,\frac{{\sqrt 2 }}{2}vt} \right).$

Similarly, for car $$B:$$

$B = \left( {{x_B},{y_B}} \right) = \left( {b,vt} \right).$

The distance $$d$$ between the cars is defined by the formula

$d = \sqrt {{{\left( {{x_B} - {x_A}} \right)}^2} + {{\left( {{y_B} - {y_A}} \right)}^2}} .$

Substituting the coordinates, we have

$d = \sqrt {{{\left( {b - \frac{{\sqrt 2 }}{2}vt} \right)}^2} + {{\left( {vt - \frac{{\sqrt 2 }}{2}vt} \right)}^2}} = \sqrt {{b^2} - \sqrt 2 bvt + \left( {2 - \sqrt 2 } \right){v^2}{t^2}} = d\left( t \right).$

The derivative of the function $$d\left( t \right)$$ has the form:

$d^\prime\left( t \right) = \left( {\sqrt {{b^2} - \sqrt 2 bvt + \left( {2 - \sqrt 2 } \right){v^2}{t^2}} } \right)^\prime = \frac{{\left( {4 - 2\sqrt 2 } \right){v^2}t - \sqrt 2 bv}}{{2\sqrt {{b^2} - \sqrt 2 bvt + \left( {2 - \sqrt 2 } \right){v^2}{t^2}} }}.$

The derivative is equal to zero at the following point:

$d^\prime\left( t \right) = 0,\;\; \Rightarrow \left( {4 - 2\sqrt 2 } \right){v^2}t - \sqrt 2 bv = 0,\;\; \Rightarrow t = \frac{{\sqrt 2 b}}{{\left( {4 - 2\sqrt 2 } \right)v}} = \frac{2}{{4\sqrt 2 - 4}}\frac{b}{v} = \frac{1}{{2\left( {\sqrt 2 - 1} \right)}}\frac{b}{v}.$

Using the First Derivative Test, one can show that this point corresponds to the minimum of the distance between the cars.

Substituting $$b = 100$$ miles and $$v = 50$$ miles per hour, we get the particular value of $$t$$ (hours):

$t = \frac{1}{{2\left( {\sqrt 2 - 1} \right)}}\frac{b}{v} = \frac{1}{{2\left( {\sqrt 2 - 1} \right)}} \cdot \frac{{100}}{{50}} = \frac{1}{{\sqrt 2 - 1}} = \frac{{\sqrt 2 + 1}}{{{{\left( {\sqrt 2 } \right)}^2} - {1^2}}} = \sqrt 2 + 1.$

### Example 20.

The first car $$A$$ moves in the northern direction at a speed $$u\left(\text{mph}\right)$$, and the second car $$B$$ moves to the eastern direction with a speed $$v\left(\text{mph}\right).$$ At the initial time, the car $$B$$ is located $$b\left(\text{miles}\right)$$ east from the car $$A$$. Determine the minimum distance between the cars.

Solution.

Consider the $$xy$$-plane (Figure $$17$$).

The car $$A$$ moves along he positive $$y$$-axis. Its coordinates at any time $$t$$ are $$\left( {0,ut} \right).$$ The car $$B$$ moves along the negative $$x$$-axis and its coordinates at time $$t$$ are, respectively, $$\left( {b- vt,0} \right).$$

The distance between the cars is calculated by the Pythagorean theorem:

$d = \sqrt {{{\left| {OA} \right|}^2} + {{\left| {OB} \right|}^2}} = \sqrt {{{\left( {0 - \left( {b - vt} \right)} \right)}^2} + {{\left( {ut - 0} \right)}^2}} = \sqrt {{b^2} - 2bvt + {v^2}{t^2} + {u^2}{t^2}}\;\text{(miles)}.$

Consider the extreme values of the function $$d\left( t \right).$$ Compute the derivative:

$d'\left( t \right) = {\left( {\sqrt {{b^2} - 2bvt + {v^2}{t^2} + {u^2}{t^2}} } \right)^\prime } = \frac{{ - 2bv + 2\left( {{v^2} + {u^2}} \right)t}}{{2\sqrt {{b^2} - 2bvt + {v^2}{t^2} + {u^2}{t^2}} }} = \frac{{\left( {{v^2} + {u^2}} \right)t - bv}}{{\sqrt {{b^2} - 2bvt + {v^2}{t^2} + {u^2}{t^2}} }}.$

The derivative is equal to zero when

$d'\left( t \right) = 0,\;\; \Rightarrow \frac{{\left( {{v^2} + {u^2}} \right)t - bv}}{{\sqrt {{b^2} - 2bvt + {v^2}{t^2} + {u^2}{t^2}} }} = 0,\;\; \Rightarrow \left( {{v^2} + {u^2}} \right)t - bv = 0,\;\; \Rightarrow t = \frac{{bv}}{{{v^2} + {u^2}}}\;\text{(hours)}.$

The denominator of the derivative is twice the distance $$d\left( t \right)$$ and is always positive. The numerator and hence the entire derivative changes sign from minus to plus when passing through this value. Consequently, the obtained value of $$t$$ is a point of minimum. At this point, the distance between the cars will be the smallest. It is determined by the following formula:

${d_{\min }} = \sqrt {{b^2} - 2bv\left( {\frac{{bv}}{{{v^2} + {u^2}}}} \right) + \left( {{v^2} + {u^2}} \right){{\left( {\frac{{bv}}{{{v^2} + {u^2}}}} \right)}^2}} = \sqrt {{b^2} - \frac{{2{b^2}{v^2}}}{{{v^2} + {u^2}}} + \frac{{{b^2}{v^2}}}{{{v^2} + {u^2}}}} = \sqrt {{b^2} - \frac{{{b^2}{v^2}}}{{{v^2} + {u^2}}}} = b\sqrt {\frac{{\cancel{v^2} + {u^2} - \cancel{v^2}}}{{{v^2} + {u^2}}}} = \frac{{bu}}{{\sqrt {{v^2} + {u^2}} }}\;\text{(miles).}$

### Example 21.

The strength of a rectangular beam is directly proportional to the product of its width $$w$$ and the square of its depth $$d.$$ Find the dimensions of the strongest beam that can be cut from a cylindrical log of radius $$R.$$

Solution.

The strength of the beam $$S$$ is given by the formula

$S = kw{d^2},$

where $$k$$ is a constant of proportionality.

The constraint equation follows from the Pythagorean theorem:

${d^2} + {w^2} = {\left( {2R} \right)^2}$

or

${d^2} + {w^2} = 4{R^2}.$

We can substitute $${d^2} = 4{R^2} - {w^2}$$ into the objective function $$S:$$

$S = kw{d^2} = kw\left( {4{R^2} - {w^2}} \right) = 4k{R^2}w - k{w^3} = S\left( w \right).$

Now the strength of the beam $$S$$ (our objective function) is written as a function of the width $$w.$$

Differentiate it to find the critical points:

$S^\prime\left( w \right) = \left( {4k{R^2}w - k{w^3}} \right)^\prime = 4k{R^2} - 3k{w^2};$
$S^\prime\left( w \right) = 0,\;\; \Rightarrow 4k{R^2} - 3k{w^2} = 0,\;\; \Rightarrow {w^2} = \frac{{4{R^2}}}{3},\;\; \Rightarrow w = \frac{{2R}}{{\sqrt 3 }} = 2R\sqrt {\frac{1}{3}} .$

We can use the Second Derivative Test:

$S^{\prime\prime}\left( w \right) = \left( {4k{R^2} - 3k{w^2}} \right)^\prime = - 6kw \lt 0.$

As $$S^{\prime\prime}\left( w \right)$$ is always negative, we deal with a point of maximum.

Determine the depth $$d$$ of the strongest beam:

$d = \sqrt {4{R^2} - {w^2}} = \sqrt {4{R^2} - \frac{{4{R^2}}}{3}} = \sqrt {\frac{{8{R^2}}}{3}} = 2R\sqrt {\frac{2}{3}} .$

So, the strongest beam has the following dimensions:

$w = 2R\sqrt {\frac{1}{3}},\,d = 2R\sqrt {\frac{2}{3}} .$