Solve Rate of Change Problems in Calculus
Rate of change calculus problems and their detailed solutions are presented.
Problem 1
A rectangular water tank (see figure below) is being filled at the constant rate of \(20\) liters / second. The base of the tank has dimensions \(w = 1\) meter and \(L = 2\) meters. What is the rate of change of the height of water in the tank? (express the answer in cm / sec).
Solution to Problem 1:
The volume \(V\) of water in the tank is given by.
\(V = w \times L \times H\)
We know the rate of change of the volume \(\dfrac{{dV}}{{dt}} = 20\) liter /sec.
We need to find the rate of change of the height \(H\) of water \(\dfrac{{dH}}{{dt}}\). \(V\) and \(H\) are functions of time.
Differentiate both sides of the above volume formula to obtain
\(\dfrac{{dV}}{{dt}} = w \times L \times \dfrac{{dH}}{{dt}}\)
Note that \(w\) and \(L\) do not change with time and are therefore considered as constants in the above operation of differentiation.
We now find a formula for \(\dfrac{{dH}}{{dt}}\) as follows.
\(\dfrac{{dH}}{{dt}} = \dfrac{{\dfrac{{dV}}{{dt}}}}{{w \times L}}\)
We need to convert liters into cubic cm and meters into cm as follows
\(1\) liter = \(1\) cubic decimeter
\(= 1000\) cubic centimeters
\(= 1000\) cm^3
and \(1\) meter = \(100\) centimeter.
We now evaluate the rate of change of the height \(H\) of water.
\(\dfrac{{dH}}{{dt}} = \dfrac{{\dfrac{{dV}}{{dt}}}}{{w \times L}}\)
\(= \dfrac{{(20 \times 1000 \, \text{cm}^3/\text{sec})}}{{(100 \, \text{cm} \times 200 \, \text{cm})}}\)
\(= 1\) cm / sec.
Problem 2
An airplane is flying in a straight direction and at a constant height of \(5000\) meters (see figure below). The angle of elevation of the airplane from a fixed point of observation is \(a\). The speed of the airplane is \(500\) km / hr. What is the rate of change of angle \(a\) when it is \(25\) degrees? (Express the answer in degrees / second and round to one decimal place).
Solution to Problem 2:
The airplane is flying horizontally at the rate of \(\dfrac{{dx}}{{dt}} = 500\) km/hr. We need a relationship between angle \(a\) and distance \(x\). From trigonometry, we can write
\( \tan \, a = \dfrac{{h}}{{x}}\)
angle \(a\) and distance \(x\) are both functions of time \(t\). Differentiate both sides of the above formula with respect to \(t\).
\(\dfrac{{d( \tan \, a)}}{{dt}} = \dfrac{{d(\dfrac{{h}}{{x}})}}{{dt}}\)
We now use the chain rule to further expand the terms in the above formula
\(\dfrac{{d( \tan \, a)}}{{dt}} = (sec^2 \, a) \dfrac{{da}}{{dt}}\)
\(\dfrac{{d(\dfrac{{h}}{{x}})}}{{dt}} = h \left( \dfrac{{-1}}{{x^2}} \right) \dfrac{{dx}}{{dt}}\).
(note: height \(h\) is constant)
Substitute the above into the original formula to obtain
\(( \sec^2 \, a) \dfrac{{da}}{{dt}} = h \left( \dfrac{{-1}}{{x^2}} \right) \dfrac{{dx}}{{dt}}\)
The above can be written as
\(\dfrac{{da}}{{dt}} = \left[ h \left( \dfrac{{-1}}{{x^2}} \right) \dfrac{{dx}}{{dt}} \right] / ( \sec^2 \, a)\)
We now use the first formula to find \(x\) in terms of \(a\) and \(h\) follows
\(x = \dfrac{{h}}{{ \tan \, a}}\)
Substitute the above into the formula for \(\dfrac{{da}}{{dt}}\) and simplify
\(\dfrac{{da}}{{dt}} = \left[ h \left( \dfrac{{- \tan^2 a}}{{h^2}} \right) \dfrac{{dx}}{{dt}} \right] / ( \sec^2 \, a)\)
\(= \left[ \left( \dfrac{{- \tan^2 a}}{{h}} \right) \dfrac{{dx}}{{dt}} \right] / (sec^2 \, a)\) Use the values for \(a\), \(h\) and \(\dfrac{{dx}}{{dt}}\) to approximate \(\dfrac{{da}}{{dt}}\) with the right conversion of units: \(1 \, \text{km} = 1000 \, \text{m}\) and \(1 \, \text{hour} = 3600 \, \text{sec}\).
\(\dfrac{{da}}{{dt}} = \left[ - sin^2(25 \, \text{deg}) \dfrac{{500 \, \text{000} \, \text{m}}}{{3600 \, \text{sec}}} \right]\)
\(= -0.005 \, \text{radians/sec}\)
\(= -0.005 \times \left[ \dfrac{{180 \, \text{degrees}}}{{\pi \, \text{radians}}} \right] / \text{sec}\)
\(= -0.3 \, \text{degrees/sec}\)
Problem 3
If two resistors with resistances \(R_1\) and \(R_2\) are connected in parallel as shown in the figure below, their electrical behavior is equivalent to a resistor of resistance \(R\) such thatIf \(R_1\) changes with time at a rate \(r = \dfrac{{dR_1}}{{dt}}\) and \(R_2\) is constant, express the rate of change \(\dfrac{{dR}}{{dt}}\) of the resistance of \(R\) in terms of \(\dfrac{{dR_1}}{{dt}}\), \(R_1\) and \(R_2\).
Solution to Problem 3:
We start by differentiating, with respect to time, both sides of the given formula for resistance \(R\), noting that \(R_2\) is constant and \(\dfrac{{d(1/R_2)}}{{dt}} = 0\)
\(\left( \dfrac{{-1}}{{R^2}} \right) \dfrac{{dR}}{{dt}} = \left( \dfrac{{-1}}{{R_1^2}} \right) \dfrac{{dR_1}}{{dt}}\)
Arrange the above to obtain
\(\dfrac{{dR}}{{dt}} = \left( \dfrac{{R}}{{R_1}} \right)^2 \dfrac{{dR_1}}{{dt}}\)
From the formula \(\dfrac{{1}}{{R}} = \dfrac{{1}}{{R_1}} + \dfrac{{1}}{{R_2}}\), we can write
\(R = \dfrac{{R_1 \times R_2}}{{R_1 + R_2}}\)
Substitute \(R\) in the formula for \(\dfrac{{dR}}{{dt}}\) and simplify
\(\dfrac{{dR}}{{dt}} = \left( \dfrac{{R_1 \times R_2}}{{R_1 \times (R_1 + R_2)}} \right)^2 \dfrac{{dR_1}}{{dt}}\)
\(= \left( \dfrac{{R_2}}{{R_1 + R_ 2}} \right)^2 \dfrac{{dR_1}}{{dt}}\)
Exercises
1 - Find a formula for the rate of change \(\dfrac{{dV}}{{dt}}\) of the volume of a balloon being inflated such that it radius \(R\) increases at a rate equal to \(\dfrac{{dR}}{{dt}}\).2 - Find a formula for the rate of change \(\dfrac{{dA}}{{dt}}\) of the area \(A\) of a square whose side \(x\) centimeters changes at a rate equal to \(2\) cm/sec.
3 - Two cars start moving from the same point in two directions that makes \(90\) degrees at the constant speeds of \(s_1\) and \(s_2\). Find a formula for the rate of change of the distance \(D\) between the two cars.
Solutions to the Above Exercises
1 - \(\dfrac{{dV}}{{dt}} = 4 \times \pi \times R^2 \times \dfrac{{dR}}{{dt}}\)2 - \(\dfrac{{dA}}{{dt}} = 4x \, \text{cm}^2/\text{sec}\)
3 - \(\dfrac{{dD}}{{dt}} = \sqrt{{s_1^2 + s_2^2}}\)