Calculus Rate of change problems and their 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*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
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).
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 hr = 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.
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
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.