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

A battery with the optimal internal resistance provides the maximum power transfer to the load

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.

A body thrown at the optimal angle has the maximum horizontal range

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