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# Calculus

Applications of the Derivative

# Optimization Problems in Physics

Page 1
Problems 1-4
Page 2
Problems 5-10

What is the optimal system configuration or the optimal mode of operation? What are the extreme values of physical quantities characterizing a system? These questions often arise in the study of physical phenomena and systems. Such analysis can be performed by using derivatives. A variety of physical problems on this topic are considered below.

## Solved Problems

Click on problem description to see 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

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 6

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 7

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 8

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 9

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 10

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

Figure 1.

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

### 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:
${\eta^\prime\left( R \right) = {\left( {\frac{R}{{R + r}}} \right)^\prime } } = {\frac{{R^\prime\left( {R + r} \right) – R{{\left( {R + r} \right)}^\prime }}}{{{{\left( {R + r} \right)}^2}}} } = {\frac{{\cancel{R} + r – \cancel{R}}}{{{{\left( {R + r} \right)}^2}}} } = {\frac{r}{{{{\left( {R + r} \right)}^2}}} > 0.}$ 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 $${\large\frac{r}{R}\normalsize} \to 0.$$

### 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.

Figure 2.

#### 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:

${v_{x0}} = {v_0}\cos \alpha ,\;\;\;\kern-0.3pt{v_{y0}} = {v_0}\sin\alpha .$ 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:
${L’\left( \alpha \right) = {\left( {\frac{{v_0^2\sin2\alpha }}{g}} \right)^\prime } } = {\frac{{v_0^2}}{g}{\left( {\sin2\alpha } \right)^\prime } } = {\frac{{2v_0^2}}{g}\cos2\alpha .}$ 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 $$\alpha = \large\frac{\pi }{4}\normalsize$$ and negative to the right, i.e. when passing through this point the derivative changes sign from plus to minus. Consequently, $$\alpha = {\large\frac{\pi }{4}\normalsize}$$ is a maximum point.

Thus, the maximum range is reached when the body is thrown at an angle of $$\alpha = {\large\frac{\pi }{4}\normalsize} = 45^\circ.$$ 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 $${t_2} = {\large\frac{{2{m_0}}}{{3b}}\normalsize}$$ is a maximum point of the function $$K\left( t \right).$$ Hence, the drop will have the greatest kinetic energy at the time $$t = {\large\frac{{2{m_0}}}{{3b}}\normalsize}.$$ At this point, the kinetic energy is equal to
${{K_{\max }} } = {\frac{{{m_0}{g^2}{{\left( {\frac{{2{m_0}}}{{3b}}} \right)}^2}}}{2} – \frac{{b{g^2}{{\left( {\frac{{2{m_0}}}{{3b}}} \right)}^3}}}{2} } = {\frac{1}{2}\left( {\frac{{4m_0^3{g^2}}}{{9{b^2}}} – \frac{{8m_0^3{g^2}}}{{27{b^2}}}} \right) } = {\frac{1}{2} \cdot \frac{{4m_0^3{g^2}}}{{27{b^2}}} } = {\frac{{2m_0^3{g^2}}}{{27{b^2}}}.}$

Page 1
Problems 1-4
Page 2
Problems 5-10