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).

tangent lines to the graph of y = x<sup>3</sup> - 3x


Solution to Problem 1:
  • The volume V of water in the tank is given by.
    V = w*L*H
  • We know the rate of change of the volume dV/dt = 20 liter /sec. We need to find the rate of change of the height H of water dH/dt. V and H are functions of time. We can differentiate both side of the above formula to obtain
    dV/dt = W*L*dH/dt
  • note 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 dH/dt as follows.
    dH/dt = dV/dt / W*L
  • We need to convert liters into cubic cm and meters into cm as follows
    1 litter = 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.
    dH/dt = dV/dt / W*L
    = ( 20*1000 cm 3 / sec ) / (100 cm * 200 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).

tangent lines to the graph of y = x<sup>3</sup> - 3x


Solution to Problem 2:
  • The airplane is flying horizontally at the rate of dx/dt = 500 km/hr. We need a relationship between angle a and distance x. From trigonometry, we can write
    tan a = h/x
  • angle a and distance x are both functions of time t. Differentiate both sides of the above formula with respect to t.
    d(tan a)/dt = d(h/x)/dt
  • We now use the chain rule to further expand the terms in the above formula
    d(tan a)/dt = (sec 2 a) da/dt
    d(h/x)/dt = h*(-1 / x 2) dx/dt.
    (note: height h is constant)
  • Substitute the above into the original formula to obtain
    (sec 2 a) da/dt = h*(-1 / x 2) dx/dt
  • The above can be written as
    da/dt = [ h*(-1 / x 2) dx/dt ] / (sec 2 a)
  • We now use the first formula to find x in terms of a and h follows
    x = h / tan a
  • Substitute the above into the formula for da/dt and simplify
    da/dt = [ h*(- tan 2a / h 2) dx/dt ] / (sec 2 a)
    = [ (- tan 2a / h) dx/dt ] / (sec 2 a)
    = (- sin 2a / h) dx/dt
  • Use the values for a, h and dx/dt to approximate da/dt with the right conversion of units: 1km = 1000 m and 1 hour = 3600 sec.
    da/dt = [- sin 2(25 deg)/5000 m]*[500 000 m/3600 sec]
    = -0.005 radians/sec
    = -0.005 * [ 180 degrees / Pi radians] /sec
    = -0.3 degrees/sec

Problem 3: If two resistors with resistances R1 and R2 are connected in parallel as shown in the figure below, their electrical behavior is equivalent to a resistor of resistance R such that
1 / R = 1 / R1 + 1 / R2

If R1 changes with time at a rate r = dR1/dt and R2 is constant, express the rate of change dR / dt of the resistance of R in terms of dR1/dt, R1 and R2.

tangent lines to the graph of y = x<sup>3</sup> - 3x


Solution to Problem 3:
  • We start by differentiating, with respect to time, both sides of the given formula for resistance R, noting that R2 is constant and d(1/R2)/dt = 0
    (-1/R 2)dR/dt = (-1/R1 2)dR1/dt
  • Arrange the above to obtain
    dR/dt = (R/R1) 2dR1/dt
  • From the formula 1 / R = 1 / R1 + 1 / R2, we can write
    R = R1*R2 / (R1 + R2)
  • Substitute R in the formula for dR/dt and simplify
    dR/dt = (R1*R2 / R1*(R1 + R2)) 2dR1/dt
    = (R2 / (R1 + R2)) 2dR1/dt

Exercises
1 - Find a formula for the rate of change dV/dt of the volume of a balloon being inflated such that it radius R increases at a rate equal to dR/dt.
2 - Find a formula for the rate of change 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 s1 and s2. Find a formula for the rate of change of the distance D between the two cars.

solutions to the above exercises
1 -    dV/dt = 4*Pi*R
2dR/dt
2 -    dA/dt = 4x cm
2/sec
3 -    dD/dt = sqrt( s1
2 + s2 2 )
More references on
calculus problems