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Related Rates Problems: Calculus Applications in Physics and Engineering - Prof. Aaron D. , Study notes of Calculus

Solutions to various related rates problems involving calculus concepts such as rates of change of volumes, distances, areas, and thicknesses. The problems cover topics like spherical snowball melting, train moving towards a camera, water tank, oil slick expansion, and hemispherical bowl. These problems help students understand the concept of related rates and its applications in physics and engineering.

Typology: Study notes

Pre 2010

Uploaded on 08/18/2009

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Mth 201
Related Rates Practice Problems
1. A spherical snowball is melting. Its radius is decreasing at 0.2 cen-
timeters per hour when the radius is 15cm. How fast is the volume
decreasing at this time?
Solution:
Variables Volume: V, radius: rand time: t
We want dV/dt - rate of change of volume with respect to time.
We know dr/dt =0.2 - the rate of change of radius with respect
to time (it is negative since the radius is decreasing).
Equations relating variables: V= 4πr3/3 (volume of a sphere in
terms of radius).
Solving the problem: We want dV/dt, so we need to differentiate
both sides with respect to t. Differentiating, we get
dV
dt = 4πr2dr
dt .
Plugging in r= 15 and dr/dt =0.2, we get dv/dt = 4π(15)2(0.2)
565.5cm3/hr.
2. A train is traveling at 0.8km/min along a straight track, moving in the
direction as shown below. A movie camera, 0.5km away from the track
is focused on the train.
Train
Camera
0.8kn/hr
Theta z
x
0.5km
(a) Express z, the distance between the camera and the train as a
function of x.
Solution: Using Pythagoras Theorem, we have x2+ (0.5)2=z2,
so
z=rx2+1
4.
1
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Mth 201 Related Rates Practice Problems

  1. A spherical snowball is melting. Its radius is decreasing at 0.2 cen- timeters per hour when the radius is 15cm. How fast is the volume decreasing at this time? Solution: - Variables Volume: V , radius: r and time: t - We want dV /dt - rate of change of volume with respect to time. - We know dr/dt = − 0 .2 - the rate of change of radius with respect to time (it is negative since the radius is decreasing). - Equations relating variables: V = 4πr^3 /3 (volume of a sphere in terms of radius). - Solving the problem: We want dV /dt, so we need to differentiate both sides with respect to t. Differentiating, we get

dV dt

= 4πr^2

dr dt

Plugging in r = 15 and dr/dt = − 0 .2, we get dv/dt = 4π(15)^2 (− 0 .2) ∼ − 565. 5 cm^3 /hr.

  1. A train is traveling at 0.8km/min along a straight track, moving in the direction as shown below. A movie camera, 0.5km away from the track is focused on the train.

Train

Camera

0.8kn/hr

Theta z

x

0.5km

(a) Express z, the distance between the camera and the train as a function of x. Solution: Using Pythagoras Theorem, we have x^2 + (0.5)^2 = z^2 , so z =

x^2 +

(b) How fast is the distance from the camera to the train changing when the train is 1km from the camera? Give units. Solution:

  • Variables: The distance from the camera to the train: z, the distance x (which is the distance from where the train directly passes the camera), time: t
  • We want dz/dt - the rate of change of distance from the cam- era to the train.
  • We know dx/dt = 0.8 since the train is moving at 0. 8 km/hr (so the distance x is increasing at a rate of 0. 8 km/hr.
  • Equations relating variables:

x^2 +

= z^2.

  • Solving the problem: Since we want dz/dt, we differentiate the expression we have found with respect to t:

2 x

dx dt

= 2z

dz dt giving dz dt

x z

dx dt

When z = 1, we have x =

3 /2 so dz/dt = 0. 8

(c) How fast is the camera rotating (in radians/min) at the moment when the train is 1km from the camera? Solution:

  • Variables: In this case, instead of the variable z, we introduce the variable θ - all other variables are the same as those for the first question.
  • We want dθ/dt
  • We know dx/dx = 0. 8
  • Equations relating variables: Using simple trigonometry, we have tan (θ) = 2x.
  • Solving the problem: Differentiating with respect to time t, we have 1 cos^2 (θ)

dθ dt

dx dt or dθ dt

= 2 cos^2 (θ)

dx dt

When z = 1 and x = 0 .5, then θ = π/3 (using simple trigonometry), so dθ/dt = 2 · (0.5)^2 · 0 .8 = 0. 4 rads/min.

  • Variables: We shall need radius r (since we are told dr/dt), thickness T and time t.
  • We want dT /dt
  • We know dr/dt
  • Equations relating variables: Since the volume is always fixed, we have V = π · (150)^2 · 0 .2 = 14137. In general, the volume will be equal to π · r^2 · T , so we have π · r^2 · T = 14137.
  • Solving the problem: Differentiating the equation for volume (using the product rule), we have

2 π · r · T

dr dt

  • πr^2

dT dt

or dT dt

2 T

r

dr dt

Evaluating when T = 0.2, r = 150 and dr/dt = 0.1, we have dT /dt = − 2 ∗ 0. 2 / 150 ∗ 0 .1 = − 0. 000267 cm/hr.

  1. (a) A hemispherical bowl of radius 10cm contains water to a depth of hcm. Find the radius of the surface of the water as a function of h. The following sketch shows how we can relate r and h - specifically, we shall use the triangle illustrated. We have

r^2 + (10 − h)^2 = r^2 + h^2 − 20 h + 100 = 10^2 = 100

or r =

(20h − h^2 ).

10−h

10

10

h

10

r

(b) The water level drops at a rate of 0.1cm per hour. At what rate is the radius of water decreasing when the depth is 5cm? Solution:

  • Variables: r, h, t
  • We want: dr/dt
  • We know dh/dt = − 0. 1
  • Equations relating variables r^2 = 20h − h^2
  • Solving the problem: Differentiating the last equation with respect to t, we get

2 r

dr dt

dh dt

− 2 h

dh dt or dr dt

r

dh dt

h r

dh dt

When the depth is 5, we have h = 5 and r =

√^100 −^ 25 =

  1. This gives

dr dt

  1. A gas station stands at the interesection of a north-south road and an east-west road. A police car is traveling toward the gas station from the east, chasing a stolen truck which is traveling north away from the gas station. The speed of the police car is 100mph at the moment it is 3 miles from the gas station. At the same time , the truck is 4 miles from the gas station going at 80mph. At this moment:

(a) Is the distance between the between the car and truck increasing or decreasing? How fast? (Distance is measured along a straight line joining the car and truck). Solution: This question is almost identical to the one we did in class. (b) How does your answer change of the truck is going 70mph instead of 80mph? Solution: This question is almost identical to the one we did in class.