Differential Calculus: Time Rates Problems with Solutions
Introduction to Time Rates Problems
Differential calculus provides powerful tools for solving problems involving rates of change. Time rates problems are particularly important in physics, engineering, economics, and other sciences, as they help us understand how quickly quantities change with respect to time. This document explores common time rates problems and their solutions using differential calculus techniques.
Basic Concepts and Formulas
When solving time rates problems, we often use these key formulas:
- Related Rates: If two variables x and y both depend on time t, and are related by an equation, then their rates of change are related by differentiating the equation with respect to t.
- Chain Rule: dy/dt = (dy/dx)(dx/dt)
- Pythagorean Theorem: a + b = c (for right triangles)
- Surface Areas:
- Sphere: SA = 4r
- Cube: SA = 6s
- Cylinder (lateral): SA = 2rh
- Volumes:
- Sphere: V = (4/3)r
- Cube: V = s
- Cone: V = (1/3)rh
- Cylinder: V = rh
Example Problems with Solutions
Problem 1: Expanding Circle
The radius of a circle is increasing at a rate of 3 cm/min. How fast is the area of the circle increasing when the radius is 10 cm?
Solution:
Given: dr/dt = 3 cm/min
Unknown: dA/dt when r = 10 cm
A = r
dA/dt = d/dt(r)
dA/dt = 2r(dr/dt)
dA/dt = 2(10)(3)
dA/dt = 60 cm/min 188.5 cm/min
Answer: The area is increasing at a rate of 60 cm/min (approximately 188.5 cm/min) when the radius is 10 cm.
Problem 2: Filling a Water Tank
A water tank has the shape of an inverted circular cone with a height of 10 m and a radius of 5 m at the top. If water is being pumped into the tank at a rate of 2 m/min, how fast is the water level rising when the water is 6 m deep?
Solution:
Given: dV/dt = 2 m/min, height of cone H = 10 m, radius of cone R = 5 m
Unknown: dh/dt when h = 6 m
Volume of cone: V = (1/3)rh
r/h = R/H = 5/10 = 1/2
r = h/2
V = (1/3)(h/2)h = (1/12)h
dV/dt = d/dt[(1/12)h]
dV/dt = (1/4)h(dh/dt)
2 = (1/4)(6)(dh/dt)
2 = 9(dh/dt)
dh/dt = 2/(9) m/min 0.071 m/min
Answer: The water level is rising at a rate of 2/(9) m/min (approximately 0.071 m/min) when the water is 6 m deep.
Problem 3: Moving Particle
A particle moves along the curve y = x such that its x-coordinate is increasing at a rate of 2 units per second. How fast is the y-coordinate changing when x = 9?
Solution:
Given: dx/dt = 2 units/s
Unknown: dy/dt when x = 9
y = x = x^(1/2)
dy/dt = d/dt[x^(1/2)]
dy/dt = (1/2)x^(-1/2)(dx/dt)
dy/dt = (1/(2x))(dx/dt)
dy/dt = (1/(29))(2)
dy/dt = (1/6)(2)
dy/dt = 1/3 units/s
Answer: The y-coordinate is increasing at a rate of 1/3 units per second when x = 9.
Problem 4: Separating Cars
Car A is traveling due north at 60 km/h, and Car B is traveling due east at 80 km/h. Both cars pass through the same intersection at the same time. At what rate is the distance between the cars changing 2 hours after they pass through the intersection?
Solution:
Given: dx/dt = 80 km/h (eastward velocity of Car B), dy/dt = 60 km/h (northward velocity of Car A)
Unknown: dD/dt when t = 2 hours
D = x + y
Where D is the distance between the cars, x is the eastward distance, and y is the northward distance.
x = 80 2 = 160 km
y = 60 2 = 120 km
D = (x + y) = (160 + 120) = (25,600 + 14,400) = 40,000 = 200 km
2D(dD/dt) = 2x(dx/dt) + 2y(dy/dt)
dD/dt = (x(dx/dt) + y(dy/dt))/D
dD/dt = (160(80) + 120(60))/200
dD/dt = (12,800 + 7,200)/200
dD/dt = 20,000/200
dD/dt = 100 km/h
Answer: The distance between the cars is increasing at a rate of 100 km/h 2 hours after they pass through the intersection.
Problem 5: Shrinking Spherical Balloon
A spherical balloon is being deflated at a rate of 10 cm/s. How fast is the radius of the balloon decreasing when the radius is 8 cm?
Solution:
Given: dV/dt = -10 cm/s (negative because the volume is decreasing)
Unknown: dr/dt when r = 8 cm
V = (4/3)r
dV/dt = d/dt[(4/3)r]
dV/dt = 4r(dr/dt)
-10 = 4(8)(dr/dt)
-10 = 256(dr/dt)
dr/dt = -10/(256) cm/s
dr/dt = -5/(128) cm/s -0.0124 cm/s
Answer: The radius of the balloon is decreasing at a rate of 5/(128) cm/s (approximately 0.0124 cm/s) when the radius is 8 cm.
Problem 6: Growing Sand Pile
Sand is pouring from a chute at a rate of 30 cubic meters per minute, forming a conical pile whose altitude is always twice the radius. How fast is the radius of the pile increasing when the radius is 5 meters?
Solution:
Given: dV/dt = 30 m/min, h = 2r
Unknown: dr/dt when r = 5 m
V = (1/3)rh
V = (1/3)r(2r) = (2/3)r
dV/dt = d/dt[(2/3)r]
dV/dt = 2r(dr/dt)
30 = 2(5)(dr/dt)
30 = 50(dr/dt)
dr/dt = 30/(50) m/min
dr/dt = 3/(5) m/min 0.191 m/min
Answer: The radius of the pile is increasing at a rate of 3/(5) m/min (approximately 0.191 m/min) when the radius is 5 meters.
Problem 7: Rising Temperature
The temperature of a coffee cup is given by T(t) = 80 + 75e^(-0.1t), where T is in degrees Fahrenheit and t is in minutes. How fast is the temperature changing when t = 5 minutes?
Solution:
Given: T(t) = 80 + 75e^(-0.1t)
Unknown: dT/dt when t = 5
dT/dt = d/dt[80 + 75e^(-0.1t)]
dT/dt = 75 d/dt[e^(-0.1t)]
dT/dt = 75 (-0.1)e^(-0.1t)
dT/dt = -7.5e^(-0.1t) F/min
dT/dt|_(t=5) = -7.5e^(-0.15)
dT/dt|_(t=5) = -7.5e^(-0.5)
dT/dt|_(t=5) = -7.5 0.607
dT/dt|_(t=5) -4.55 F/min
Answer: The temperature is decreasing at a rate of approximately 4.55F per minute when t = 5 minutes.
Practice Problems
- A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius of the outer ripple is increasing at a constant rate of 3 feet per second. When the radius is 8 feet, at what rate is the total area of the disturbed water changing?
- A ladder 10 meters long rests against a wall. The bottom of the ladder slides away from the wall at a rate of 2 meters per second. How fast is the top of the ladder sliding down the wall when the bottom is 6 meters from the wall?
- A spherical snowball is melting at a rate of 4 cm/hour. At what rate is the radius decreasing when the radius is 6 cm?
- Water is leaking out of an inverted conical tank at a rate of 10,000 cm/min while water is being pumped into the tank at a constant rate. The tank has a height of 6 m and a diameter at the top of 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.
- A kite 100 feet above the ground moves horizontally at a speed of 8 feet per second. At what rate is the string being released when 200 feet of string have been released?
Conclusion
Differential calculus provides a systematic approach to solving time rates problems. Through the examples provided, we've seen how to apply differentiation to real-world scenarios involving changing quantities. The key steps typically involve:
- Identifying the given and unknown quantities
- Establishing relationships between the quantities using appropriate formulas or equations
- Differentiating with respect to time
- Substituting known values and solving for the unknown rate
With practice, these problems become more intuitive, and you'll develop an appreciation for the power of differential calculus in understanding and modeling the changing world around us.
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