# Calculus

## Applications of the Derivative # Optimization Problems in Physics

There are many different types of optimization problems we may encounter in physics and engineering. In such problems, it is often necessary to optimize some physical quantity such as distance, velocity, time, mass, acceleration, force, electric current, illuminance, etc.

The constraint equations can follow from physical laws and formulas.

The general approach for solving optimization problems remains the same. It is described on the page Optimization Problems in 2D Geometry.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

What should be the internal resistance in a battery to provide the maximum power transfer to the load (Figure $$1$$)?

### Example 2

The simplest electric circuit consists of a power supply with an electromotive force (emf) $$\varepsilon$$ and internal resistance $$r$$ and an external load of resistance $$R$$ (Figure $$1$$). Find under what conditions the power efficiency is the largest?

### Example 3

A body is thrown at an angle $$\alpha$$ with the horizontal. Neglecting air resistance, determine the angle $$\alpha,$$ at which the horizontal range is maximum.

### Example 4

A raindrop of initial mass $${m_0}$$ falls due to gravity. While falling, the drop evaporates so that its mass decreases with time by the linear law $m\left( t \right) = {m_0} - bt,$ where $$b$$ is the evaporation rate. Determine the point of time at which the kinetic energy of the drop is the greatest.

### Example 5

The potential energy of a particle in a field has the form $U = \frac{a}{{{r^2}}} - \frac{b}{r},$ where $$a$$ and $$b$$ are positive numbers, $$r$$ is the distance from the centre of the field. Find the maximum possible value of the attractive force.

### Example 6

A light bulb is to be hung above the center of a round table. What height gives the best illumination of the edge of the table?

### Example 7

Two light sources of intensities $${I_1}$$ and $${I_2}$$ are $$L$$ units apart. Find the point on the line segment joining the two sources that has the least illumination.

### Example 8

A body lies on a horizontal surface. The coefficient of friction between the body and the surface is $$k.$$ Determine the angle $$\alpha,$$ at which the force acting to the body and causing its movement is the least.

### Example 1.

What should be the internal resistance in a battery to provide the maximum power transfer to the load (Figure $$1$$)?

Solution.

The current in an electric circuit is determined by Ohm's law:

$I = \frac{\varepsilon }{{R + r}}.$

Then the power transferred to the external load is given by

$P = P\left( R \right) = {I^2}R = {\left( {\frac{\varepsilon }{{R + r}}} \right)^2}R = \frac{{{\varepsilon ^2}R}}{{{{\left( {R + r} \right)}^2}}}.$

This power is a function $$P\left( R \right).$$ We investigate it for extreme values. The derivative $$P'\left( R \right)$$ is written as

$P'\left( R \right) = \left( {\frac{{{\varepsilon ^2}R}}{{{{\left( {R + r} \right)}^2}}}} \right)^\prime = {\varepsilon ^2}\frac{{R'{{\left( {R + r} \right)}^2} - R{{\left( {{{\left( {R + r} \right)}^2}} \right)}^\prime }}}{{{{\left( {R + r} \right)}^4}}} = {\varepsilon ^2}\frac{{{{\left( {R + r} \right)}^2} - 2R\left( {R + r} \right)}}{{{{\left( {R + r} \right)}^4}}} = {\varepsilon ^2}\frac{{R + r - 2R}}{{{{\left( {R + r} \right)}^3}}} = {\varepsilon ^2}\frac{{r - R}}{{{{\left( {R + r} \right)}^3}}}.$

This shows that the derivative is zero if $$R = r,$$ and when passing through this value (with increasing $$R$$) the derivative changes sign from plus to minus. Hence, this value corresponds to the maximum of the function $$P\left( R \right).$$ Thus, we have proved the maximum power transfer theorem. The maximum value of the output power is given by the following expression:

${P_{\max }} = \frac{{{\varepsilon ^2}r}}{{{{\left( {r + r} \right)}^2}}} = \frac{{{\varepsilon ^2}\cancel{r}}}{{4{r^{\cancel{2}}}}} = \frac{{{\varepsilon ^2}}}{{4r}}.$

### Example 2.

The simplest electric circuit consists of a power supply with an electromotive force (emf) $$\varepsilon$$ and internal resistance $$r$$ and an external load of resistance $$R$$ (Figure $$1$$). Find under what conditions the power efficiency is the largest?

Solution.

The power efficiency of a circuit is determined as the ratio of power dissipated by the load $${W_R}$$ to the full power output of the source $$W:$$

$\eta = \frac{{{W_R}}}{W} = \frac{{{W_R}}}{{{W_R} + {W_r}}}.$

The expressions for the power components $${W_R},$$ $${W_r},$$ and $$W$$ follow from Ohm's law:

${W_R} = {I^2}R = {\left( {\frac{\varepsilon }{{R + r}}} \right)^2}R = \frac{{{\varepsilon ^2}R}}{{{{\left( {R + r} \right)}^2}}};$
${W_r} = {I^2}r = {\left( {\frac{\varepsilon }{{R + r}}} \right)^2}r = \frac{{{\varepsilon ^2}r}}{{{{\left( {R + r} \right)}^2}}};$
$W = {W_R} + {W_r} = \frac{{{\varepsilon ^2}R}}{{{{\left( {R + r} \right)}^2}}} + \frac{{{\varepsilon ^2}r}}{{{{\left( {R + r} \right)}^2}}} = \frac{{{\varepsilon ^2}\cancel{\left( {R + r} \right)}}}{{{{\left( {R + r} \right)}^{\cancel{2}}}}} = \frac{{{\varepsilon ^2}}}{{R + r}},$

where $$\varepsilon$$ is emf of the source, $$I$$ is the current in the circuit. Hence, we obtain the following formula for the efficiency $$\eta:$$

$\eta = \frac{{{W_R}}}{W} = \frac{{\frac{{{\varepsilon ^2}R}}{{{{\left( {R + r} \right)}^2}}}}}{{\frac{{{\varepsilon ^2}}}{{R + r}}}} = \frac{{R\cancel{\left( {R + r} \right)}}}{{{{\left( {R + r} \right)}^{\cancel{2}}}}} = \frac{R}{{R + r}}.$

In this expression, the efficiency $$\eta$$ depends on $$R,$$ i.e. it is a function $$\eta \left( R \right).$$ We investigate it using the derivative:

$\eta^\prime\left( R \right) = {\left( {\frac{R}{{R + r}}} \right)^\prime } = \frac{{R^\prime\left( {R + r} \right) - R{{\left( {R + r} \right)}^\prime }}}{{{{\left( {R + r} \right)}^2}}} = \frac{{\cancel{R} + r - \cancel{R}}}{{{{\left( {R + r} \right)}^2}}} = \frac{r}{{{{\left( {R + r} \right)}^2}}} \gt 0.$

Thus, the derivative $$\eta^\prime\left( R \right)$$ is always positive. In other words, the efficiency increases monotonically with increasing the external resistance $$R$$ (at a fixed internal resistance $$r$$). If we write the efficiency $$\eta$$ in the form

$\eta \left( R \right) = \frac{R}{{R + r}} = \frac{1}{{1 + \frac{r}{R}}},$

we can see that efficiency approaches the maximum possible value $$1$$ as $${\frac{r}{R}} \to 0.$$

### Example 3.

A body is thrown at an angle $$\alpha$$ with the horizontal. Neglecting air resistance, determine the angle $$\alpha,$$ at which the horizontal range is maximum.

Solution.

Suppose that the body is thrown with initial velocity $${v_0}$$ (Figure $$2$$).

The equations of motion for the freely falling body in the coordinate system $$xOy$$ have the form

$\left\{ \begin{array}{l} x = {v_{x0}}t\\ y = {v_{y0}}t - \frac{{g{t^2}}}{2} \end{array} \right.,$

where $$g$$ is acceleration of gravity, $$t$$ is time, and the components of the initial velocity $${v_0}$$ are as follows:

$v_{x0} = {v_0}\cos \alpha ,\;\;\;v_{y0} = {v_0}\sin\alpha.$

Then we can write:

$\left\{ \begin{array}{l} x = {v_0}\cos \alpha \,t\\ y = {v_0}\sin \alpha \,t - \frac{{g{t^2}}}{2} \end{array} \right..$

At the moment the body hits the ground, the $$y$$-coordinate is zero. Therefore, the total time of flight is

$y = 0,\;\; \Rightarrow {v_0}\sin\alpha\,t - \frac{{g{t^2}}}{2} = 0,\;\; \Rightarrow t\left( {{v_0}\sin\alpha - \frac{{gt}}{2}} \right) = 0,\;\; \Rightarrow \frac{{gt}}{2} = {v_0}\sin\alpha ,\;\; \Rightarrow t = \frac{{2{v_0}\sin\alpha }}{g}.$

Putting this value of $$t$$ in the first equation of the system, we find the horizontal range $$L:$$

$L = {v_0}\cos\alpha \,t = \frac{{2v_0^2\sin\alpha \cos \alpha }}{g} = \frac{{v_0^2\sin2\alpha }}{g}.$

The quantity $$L$$ is a function of the throw angle $$\alpha:$$ $$L = L\left( \alpha \right).$$ One can immediately note that the throw range $$L$$ reaches its maximum value when

$\sin2\alpha = 1,\;\; \Rightarrow 2\alpha = \frac{\pi }{2},\;\; \Rightarrow \alpha = \frac{\pi }{4} = 45^\circ.$

The same result can be obtained more rigorously by examining the function $$L\left( \alpha \right)$$ with the help of the derivative. Differentiating $$L\left( \alpha \right),$$ we get:

$L'\left( \alpha \right) = \left( {\frac{{v_0^2\sin2\alpha }}{g}} \right)^\prime = \frac{{v_0^2}}{g}{\left( {\sin2\alpha } \right)^\prime } = \frac{{2v_0^2}}{g}\cos2\alpha .$

By equating the derivative to zero, we find the extremum of the function $$L\left( \alpha \right):$$

$L'\left( \alpha \right) = 0,\;\; \Rightarrow \frac{{2v_0^2}}{g}\cos2\alpha = 0,\;\; \Rightarrow \cos2\alpha = 0,\;\; \Rightarrow 2\alpha = \frac{\pi }{2},\;\; \Rightarrow \alpha = \frac{\pi }{4}.$

The function $$\cos2\alpha$$ is positive to the left of the point $$\alpha = \frac{\pi }{4}$$ and negative to the right, i.e. when passing through this point the derivative changes sign from plus to minus. Consequently, $$\alpha = {\frac{\pi }{4}}$$ is a maximum point.

Thus, the maximum range is reached when the body is thrown at an angle of $$\alpha = {\frac{\pi }{4}} = 45^\circ.$$ This value does not depend on the initial velocity $${v_0}.$$ The maximum range itself $${L_{\max }}$$ depends on $${v_0}$$ and is determined by the formula

${L_{\max }} = L\left( {\frac{\pi }{4}} \right) = \frac{{v_0^2}}{g}\sin\frac{\pi }{2} = \frac{{v_0^2}}{g}.$

### Example 4.

A raindrop of initial mass $${m_0}$$ falls due to gravity. While falling, the drop evaporates so that its mass decreases with time by the linear law $m\left( t \right) = {m_0} - bt,$ where $$b$$ is the evaporation rate. Determine the point of time at which the kinetic energy of the drop is the greatest.

Solution.

The kinetic energy of a body falling from rest is given by

$K = \frac{{m{v^2}}}{2} = \frac{{m{{\left( {gt} \right)}^2}}}{2} = \frac{{m{g^2}{t^2}}}{2}.$

Putting the law of mass change in this formula, we get the function $$K\left( t \right):$$

$K\left( t \right) = \frac{{m\left( t \right){g^2}{t^2}}}{2} = \left( {{m_0} - bt} \right)\frac{{{g^2}{t^2}}}{2} = \frac{{{m_0}{g^2}{t^2}}}{2} - \frac{{b{g^2}{t^3}}}{2}.$

We calculate the derivative and find the critical points of the function $$K\left( t \right):$$

$K'\left( t \right) = {\left( {\frac{{{m_0}{g^2}{t^2}}}{2} - \frac{{b{g^2}{t^3}}}{2}} \right)^\prime } = {m_0}{g^2}t - \frac{{3b{g^2}{t^2}}}{2} = {g^2}t\left( {{m_0} - \frac{3}{2}bt} \right);$
$K'\left( t \right) = 0,\;\; \Rightarrow {g^2}t\left( {{m_0} - \frac{3}{2}bt} \right) = 0,\;\; \Rightarrow {t_1} = 0,\;{t_2} = \frac{{2{m_0}}}{{3b}}.$

The derivative $$K'\left( t \right)$$ is a quadratic function. The first root $${t_1} = 0$$ is a minimum point, and the second root $${t_2} = {\frac{{2{m_0}}}{{3b}}}$$ is a maximum point of the function $$K\left( t \right).$$ Hence, the drop will have the greatest kinetic energy at the time $$t = {\frac{{2{m_0}}}{{3b}}}.$$ At this point, the kinetic energy is equal to

${K_{\max }} = \frac{{{m_0}{g^2}{{\left( {\frac{{2{m_0}}}{{3b}}} \right)}^2}}}{2} - \frac{{b{g^2}{{\left( {\frac{{2{m_0}}}{{3b}}} \right)}^3}}}{2} = \frac{1}{2}\left( {\frac{{4m_0^3{g^2}}}{{9{b^2}}} - \frac{{8m_0^3{g^2}}}{{27{b^2}}}} \right) = \frac{1}{2} \cdot \frac{{4m_0^3{g^2}}}{{27{b^2}}} = \frac{{2m_0^3{g^2}}}{{27{b^2}}}.$

### Example 5.

The potential energy of a particle in a field has the form $U = \frac{a}{{{r^2}}} - \frac{b}{r},$ where $$a$$ and $$b$$ are positive numbers, $$r$$ is the distance from the centre of the field. Find the maximum possible value of the attractive force.

Solution.

Calculate the force acting on the particle in the field:

$F = - \frac{{dU}}{{dr}} = - \frac{d}{{dr}}\left( {\frac{a}{{{r^2}}} - \frac{b}{r}} \right) = - \left( { - \frac{{2a}}{{{r^3}}} + \frac{b}{{{r^2}}}} \right) = \frac{{2a}}{{{r^3}}} - \frac{b}{{{r^2}}} .$

Take the derivative of $$F\left( r \right)$$ and equate it to zero to determine the critical points:

$F^\prime\left( r \right) = \left( {\frac{{2a}}{{{r^3}}} - \frac{b}{{{r^2}}}} \right)^\prime = - \frac{{6a}}{{{r^4}}} + \frac{{2b}}{{{r^3}}} = \frac{{2br - 6a}}{{{r^4}}};$
$F^\prime\left( r \right) = 0,\;\; \Rightarrow \frac{{2br - 6a}}{{{r^4}}} = 0,\;\; \Rightarrow 2br - 6a = 0,\;\; \Rightarrow r = \frac{{3a}}{b}.$

Using the first derivative test we can show that $$r = \frac{{3a}}{b}$$ is a point of minimum of $$F\left( r \right).$$

Compute the value of the force at this point:

$F\left( {r = \frac{{3a}}{b}} \right) = \frac{{2a}}{{{{\left( {\frac{{3a}}{b}} \right)}^3}}} - \frac{b}{{{{\left( {\frac{{3a}}{b}} \right)}^2}}} = \frac{{2a}}{{\frac{{27{a^3}}}{{{b^3}}}}} - \frac{b}{{\frac{{9{a^2}}}{{{b^2}}}}} = \frac{{2a{b^3}}}{{27{a^3}}} - \frac{{{b^3}}}{{9{a^2}}} = \frac{{2{b^3} - 3{b^3}}}{{27{a^2}}} = - \frac{{{b^3}}}{{27{a^2}}}.$

The negative value corresponds to the attractive force (directed to the centre of the field). Hence, the maximum value of the attractive force is equal to

${F_{\max }} = \frac{{{b^3}}}{{27{a^2}}}.$

### Example 6.

A light bulb is to be hung above the center of a round table. What height gives the best illumination of the edge of the table?

Solution.

We introduce the following notations (see Figure $$4$$): $$R$$ is the radius of the table, $$h$$ is the height at which the bulb is suspended, $$L$$ is the distance from the light bulb to the edge of the table, $$I$$ is the intensity of illumination, $$\alpha$$ is the angle between the straight line $$L$$ and normal to the surface of the table.

The illuminance at the edge of the table is determined by the formula

$E = \frac{I}{{{L^2}}}\cos \alpha .$

Since

${L^2} = {h^2} + {R^2}\;\;\text{and}\;\;\;\cos \alpha = \frac{h}{L} = \frac{h}{{\sqrt {{h^2} + {R^2}} }},$

the illuminance $$E$$ can be written as a function $$E\left( h \right)$$ in the form

$E = E\left( h \right) = \frac{I}{{{h^2} + {R^2}}} \cdot \frac{h}{{\sqrt {{h^2} + {R^2}} }} = \frac{{Ih}}{{{{\left( {{h^2} + {R^2}} \right)}^{\frac{3}{2}}}}}.$

Calculate the derivative:

$E'\left( h \right) = {\left( {\frac{{Ih}}{{{{\left( {{h^2} + {R^2}} \right)}^{\frac{3}{2}}}}}} \right)^\prime } = I \cdot \frac{{{h^2} + {R^2} - 3{h^2}}}{{{{\left( {{h^2} + {R^2}} \right)}^{\frac{5}{2}}}}} = \frac{{I\left( {{R^2} - 2{h^2}} \right)}}{{{{\left( {{h^2} + {R^2}} \right)}^{\frac{5}{2}}}}}.$

Find the roots of the derivative:

$E'\left( h \right) = 0,\;\; \Rightarrow \frac{{I\left( {{R^2} - 2{h^2}} \right)}}{{{{\left( {{h^2} + {R^2}} \right)}^{\frac{5}{2}}}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{R^2} - 2{h^2} = 0}\\ {{h^2} + {R^2} \ne 0} \end{array}} \right.,\;\; \Rightarrow {h^2} = \frac{{{R^2}}}{2},\;\; \Rightarrow h = \frac{R}{{\sqrt 2 }}.$

When passing through the point $$h = \frac{R}{{\sqrt 2 }}$$ the derivative changes sign from plus to minus. Therefore, it is a maximum point. The maximum possible illuminance is given by

${E_{\max }} = \frac{{I\frac{R}{{\sqrt 2 }}}}{{{{\left( {\frac{{{R^2}}}{2} + {R^2}} \right)}^{\frac{3}{2}}}}} = \frac{{IR}}{{\sqrt 2 {{\left( {\frac{3}{2}{R^2}} \right)}^{\frac{3}{2}}}}} = \frac{{{2^{\frac{3}{2}}}IR}}{{{2^{\frac{1}{2}}} \cdot {3^{\frac{3}{2}}}{R^3}}} = \frac{{2I}}{{{3^{\frac{3}{2}}}{R^2}}} = \frac{{2I}}{{\sqrt {27} {R^2}}}.$

### Example 7.

Two light sources of intensities $${I_1}$$ and $${I_2}$$ are $$L$$ units apart. Find the point on the line segment joining the two sources that has the least illumination.

Solution.

The illumination at the point $$x$$ is given by the formula

$E = \frac{{{I_1}}}{{{x^2}}} + \frac{{{I_2}}}{{{{\left( {L - x} \right)}^2}}}.$

Differentiate the function $$E\left( x \right):$$

$E^\prime\left( x \right) = \left( {\frac{{{I_1}}}{{{x^2}}} + \frac{{{I_2}}}{{{{\left( {L - x} \right)}^2}}}} \right)^\prime = \left( {\frac{{{I_1}}}{{{x^2}}}} \right)^\prime + \left( {\frac{{{I_2}}}{{{{\left( {L - x} \right)}^2}}}} \right)^\prime = - \frac{{2{I_1}}}{{{x^3}}} - \frac{{2{I_2}}}{{{{\left( {L - x} \right)}^3}}} \cdot \left( { - 1} \right) = - \frac{{2{I_1}}}{{{x^3}}} + \frac{{2{I_2}}}{{{{\left( {L - x} \right)}^3}}} = \frac{{2\left[ {{I_2}{x^3} - {I_1}{{\left( {L - x} \right)}^3}} \right]}}{{{x^3}{{\left( {L - x} \right)}^3}}}.$

Assuming that $$x \ne 0$$ and $$x \ne L,$$ we have the following critical point:

$E^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{2\left[ {{I_2}{x^3} - {I_1}{{\left( {L - x} \right)}^3}} \right]}}{{{x^3}{{\left( {L - x} \right)}^3}}} = 0,\;\; \Rightarrow {I_2}{x^3} - {I_1}{\left( {L - x} \right)^3} = 0,\;\; \Rightarrow \frac{{{{\left( {L - x} \right)}^3}}}{{{x^3}}} = \frac{{{I_2}}}{{{I_1}}},\;\; \Rightarrow \frac{{L - x}}{x} = \sqrt{{\frac{{{I_2}}}{{{I_1}}}}},\;\; \Rightarrow \frac{L}{x} - 1 = \sqrt{{\frac{{{I_2}}}{{{I_1}}}}},\;\; \Rightarrow \frac{L}{x} = 1 + \sqrt{{\frac{{{I_2}}}{{{I_1}}}}},\;\; \Rightarrow x = \frac{L}{{1 + \sqrt{{\frac{{{I_2}}}{{{I_1}}}}}}}.$

Using the First Derivative Test, one can show that $$x = \frac{L}{{1 + \sqrt{{\frac{{{I_2}}}{{{I_1}}}}}}}$$ is a point of minimum.

In the particular trivial case, when $${I_1} = {I_2},$$ the least illumination is attained at the center of the segment line:

$x = \frac{L}{{1 + \sqrt{{\frac{{{I_2}}}{{{I_1}}}}}}} = \frac{L}{{1 + \sqrt{1}}} = \frac{L}{2}.$

When $${I_2} = 2{I_1},$$ the point of minimum illumination has the coordinate

$x = \frac{L}{{1 + \sqrt{{\frac{{{I_2}}}{{{I_1}}}}}}} = \frac{L}{{1 + \sqrt{2}}} \approx \frac{L}{{2.26}} \approx 0.44L$

### Example 8.

A body lies on a horizontal surface. The coefficient of friction between the body and the surface is $$k.$$ Determine the angle $$\alpha,$$ at which the force acting to the body and causing its movement is the least.

Solution.

There are four forces acting to the body (Figure $$6$$) the force of gravity $$\mathbf{mg},$$ the normal force $$\mathbf{N},$$ the frictional force $$\mathbf{F_{\text{fr}}},$$ and an external force $$\mathbf{F}.$$

The angle $$\alpha$$ may vary in the interval $$0 \le \alpha \le {\frac{\pi }{2}}.$$

The equation of motion of the body in vector form is written as follows:

$\mathbf{ma} = \mathbf{F} + \mathbf{mg} + \mathbf{F_\text{fr}} + \mathbf{N},$

In the limit case (when the body is still at rest), all the forces are balanced.

Consequently,

$\mathbf{F} + \mathbf{mg} + \mathbf{F_\text{fr}} + \mathbf{N} = \mathbf{0}.$

In projections on the coordinate axes, we get the following system of equations:

$\left\{ \begin{array}{l}F\cos \alpha - {F_\text{fr}} = 0\\F\sin \alpha - mg + N = 0\end{array} \right..$

Next, we take into account that the frictional force in absolute value is equal to $${F_\text{fr}} = kN.$$ Then

$\left\{ \begin{array}{l} F\cos \alpha - kN = 0\\ F\sin \alpha - mg + N = 0 \end{array} \right..$

From this system we express the force $$F$$ in terms of the angle $$\alpha$$ and other quantities. It follows from the second equation that

$N = mg - F\sin \alpha .$

Substituting this into the first equation, we find the function $$F\left( \alpha \right):$$

$F\cos \alpha - k\left( {mg - F\sin \alpha } \right) = 0,\;\; \Rightarrow F\cos \alpha + kF\sin \alpha - kmg = 0,\;\; \Rightarrow F\left( {\cos \alpha + k\sin \alpha } \right) - kmg = 0,\;\; \Rightarrow F = F\left( \alpha \right) = \frac{{kmg}}{{\cos \alpha + k\sin \alpha }}.$

Differentiating this expression with respect to $$\alpha$$ yields:

$F'\left( \alpha \right) = \left( {\frac{{kmg}}{{\cos \alpha + k\sin \alpha }}} \right)^\prime = - \frac{{kmg}}{{{{\left( {\cos \alpha + k\sin \alpha } \right)}^2}}} \cdot {\left( {\cos \alpha + k\sin \alpha } \right)^\prime } = - \frac{{kmg}}{{{{\left( {\cos \alpha + k\sin \alpha } \right)}^2}}} \cdot \left( { - \sin\alpha + k\cos \alpha } \right) = \frac{{kmg\left( {\sin\alpha - k\cos \alpha } \right)}}{{{{\left( {\cos \alpha + k\sin \alpha } \right)}^2}}}.$

It is evident that the derivative is zero under the following condition:

$\sin\alpha - k\cos \alpha = 0,\;\; \Rightarrow \tan\alpha - k = 0,\;\; \Rightarrow \tan\alpha = k,\;\; \Rightarrow \alpha = \arctan k.$

Thus, the critical angle is $$\alpha = \arctan k.$$ The function $$\tan \alpha$$ is increasing with increasing $$\alpha.$$ Therefore, when passing through this critical value the derivative changes sign from minus to plus, i.e. we have a minimum of the function $$F\left( \alpha \right)$$ at this point. Hence, at the specified angle $$\alpha$$ the force $$F$$ will be the least.

Calculate the value of the smallest force. Putting the above angle $$\alpha,$$ we obtain:

${F_{\min }} = {\frac{{kmg}}{{\cos \left( {\arctan k} \right)} + {k\sin \left( {\arctan k} \right)}}}.$

Convert the trigonometric functions in the denominator:

$\cos \left( {\arctan k} \right) = \frac{1}{{\sqrt {1 + {{\tan }^2}\left( {\arctan k} \right)} }} = \frac{1}{{\sqrt {1 + {k^2}} }},$
$\sin\left( {\arctan k} \right) = \frac{{\tan \left( {\arctan k} \right)}}{{\sqrt {1 + {{\tan }^2}\left( {\arctan k} \right)} }} = \frac{k}{{\sqrt {1 + {k^2}} }}.$

The result is as follows:

${F_{\min }} = \frac{{kmg}}{{\frac{1}{{\sqrt {1 + {k^2}} }} + k \cdot \frac{k}{{\sqrt {1 + {k^2}} }}}} = \frac{{kmg\sqrt {1 + {k^2}} }}{{1 + {k^2}}} = \frac{{kmg}}{{\sqrt {1 + {k^2}} }}.$

See more problems on Page 2.