Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Calculus I - Fall 2014: Solved Problems on Changing Areas, Volumes, and Distances, Study notes of Calculus

Solutions to various calculus problems from a fall 2014 calculus i class. The problems involve finding the rates of change of areas, volumes, and distances in different scenarios, such as isosceles right triangles, swimming pools, inverted conical water tanks, and hot-air balloons. The solutions use calculus concepts like derivatives and integrals.

Typology: Study notes

2021/2022

Uploaded on 09/12/2022

jugnu900
jugnu900 🇺🇸

4.4

(7)

236 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Math 180: Calculus I Fall 2014
October 23
TA: Brian Powers
1. The hypotenuse of an isosceles right triangle is decreasing in length at a rate of 4 m/s.
(a) At what rate is the area of the triangle changing when the legs are 5m long?
SOLUTION: First we should define the variables. Let xbe the length of the legs, and hthe
length of the hypotenuse.
Since the legs of the isosceles right triangle are in ratio of 1 : 2 to the hypotenuse, the rate of
change of the legs will be proportional, i.e. x=h/2. Since dh
dt =4,
dx
dt =dh
dt
1
2=4
2=22
Because the area A=x2/2,
dA
dt =dA
dx
dx
dt =x(22 = 22xm/s
When x= 5, dA/dt =225 = 102 m2/s.
(b) At what rate are the length of the legs of the triangle changing?
SOLUTION: The legs are decreasing at a constant rate of 22 m/s.
(c) At what rate is the area of the triangle changing when the area is 4 m2?
SOLUTION: The area is 4 m2when x2/2=4x2= 8 so x= 22. At that time dA/dt =
(22)2=8 m2/s.
2. A swimming pool is 50m long and 20m wide. Its depth decreases linearly along the length from 3m to
1m. It is initially empty and filled with water at 1 m3/min.
(a) How fast is the water level rising 250 minutes after the filling begins?
SOLUTION: First of all, we need to talk about the dimensions of this. The inclined portion of
the pool has height 2m and length 50m. As a height his filled in, it will have a length bthat is
proportional to 50/2, so b= 25h.
16-1
pf3

Partial preview of the text

Download Calculus I - Fall 2014: Solved Problems on Changing Areas, Volumes, and Distances and more Study notes Calculus in PDF only on Docsity!

Math 180: Calculus I Fall 2014

October 23

TA: Brian Powers

  1. The hypotenuse of an isosceles right triangle is decreasing in length at a rate of 4 m/s. (a) At what rate is the area of the triangle changing when the legs are 5m long? SOLUTION: First we should define the variables. Let x be the length of the legs, and h the length of the hypotenuse.

Since the legs of the isosceles right triangle are in ratio of 1 :

2 to the hypotenuse, the rate of change of the legs will be proportional, i.e. x = h/

  1. Since dhdt = −4, dx dt

dh dt

Because the area A = x^2 /2, dA dt

dA dx

dx dt

= x(− 2

2 x m/s

When x = 5, dA/dt = − 2

2 m^2 /s. (b) At what rate are the length of the legs of the triangle changing? SOLUTION: The legs are decreasing at a constant rate of 2

2 m/s. (c) At what rate is the area of the triangle changing when the area is 4 m^2? SOLUTION: The area is 4 m^2 when x^2 /2 = 4 ⇒ x^2 = 8 so x = 2

  1. At that time dA/dt = −(

2)^2 = −8 m^2 /s.

  1. A swimming pool is 50m long and 20m wide. Its depth decreases linearly along the length from 3m to 1m. It is initially empty and filled with water at 1 m^3 /min. (a) How fast is the water level rising 250 minutes after the filling begins? SOLUTION: First of all, we need to talk about the dimensions of this. The inclined portion of the pool has height 2m and length 50m. As a height h is filled in, it will have a length b that is proportional to 50/2, so b = 25h.

16-2 TA Discussion 16: October 23

The volume when height h ≤ 2 has been filled up is then V (h) = 20 12 (25h)h = 250h^2. For 2 < h ≤ 3, V (h) = 250(2)^2 + 20(50)(h − 2) = 1000h − 1000.

V (h) =

250 h^2 if h ≤ 2 1000 h − 1000 if 2 < h ≤ 3

After 250 minutes, the volume will be 250 m^3 , which means h = 1. We have

dV dt

dV dt

dV dh

dh dt

= 2 · 250 h dh dt

= 500 when h = 1

Which means that 1 = 500

dh dt So dhdt = 1/500 m/min, or 2mm/min (b) How long will it take to fill the pool? SOLUTION: This is easier. We just have to calculate the total volume. The cross section of the pool has area 1 · 50 + 12 · 50 · 2 = 100m^2 , so the volume is 100 · 20 = 2000 m^3 , so it takes 2000 minutes to fill the pool.

  1. An inverted conical water tank with height of 12ft and radius of 6ft is drained through a hole in the vertex at a rate of 2 ft^3 /sec. What is the rate of change of the water depth when the water depth is 3ft? SOLUTION: Let h be the height of the water in the cone. Because the volume of water is a cone that is similar to the large cone, the radius is going to be. 5 h.

So the volume of water is V (h) =

π(. 5 h)^2 h =

π 12

h^3

Furthermore, we have

−2 =

dV dt

dV dh

dh dt

π 4

h^2

dh dt We can solve for dh/dt = − 8 /(πh^2 ), so when h = 3, dh/dt = − 8 /(9π) ft/sec.

  1. A hot-air balloon is 150 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly underneath it going 58.67 ft/s. If the balloon rises vertically at a rate of 10 ft/s, what is teh rate of change of the distance beween the motorcycle and the balloon 10 seconds later? SOLUTION: Let h be the height of the ballon at time t, and b be the distance the motorcycle has traveled by time t. So h = 150 + 10t, b = 58. 67 t