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AP Calculus AB



This wiki is a collaborative solutions manual built by the student's in Mr. Kuropapwa's AP Calculus class. Don't you just love this wiki math stuff?



 



User edited FrontPage

Fri, 02 Mar 2007 04:43:14 +0000

AP Calculus Wiki Solutions Manual
This wiki is now closed. It is open for viewing only. Note: All solutions have been generated by students and have not been "corrected." Some of them are correct and some are not. It is up to the reader to verify the accuracy of any solution found on these pages.
This is a place for students to help each other learn by writing a collaborative solutions manual! This counts for marks. You must make at least two edits/contributions to the wiki. One must be a significant contribution and the other must be a constructive modification. Here is what you have to do:
A Significant Contribution means ...



User edited SideBar

Fri, 17 Nov 2006 18:10:51 +0000

!! The Questions
|
**Question Number**
|
**Topic**
| **Topic** | |
|%5Bquestion01|Question 1%5D|Area and Volume|
|%5Bquestion02|Question 2%5D|Velocity|
|%5Bquestion03|Question 3%5D|Differential Equations|



User edited question02

Wed, 10 May 2006 23:45:15 +0000

It is by using the Second Fundamental Theorum of Calculus
|If f is a continuous function on some interval (a, b) and an accumulation function A is defined by A(x) = (integral of) f(t) dt, then A’(x) = f(x).|
Therefore, v(t) is equal to the underlying function.
**V(t) = 1- x(cos x) - (ln x)(sin x)**
We find the roots of the function and whether its positively or negatively valued on either side of the root.
We remember our domain and only search for the roots in between.
**v(t) 1 ____+_____|'___-___ 8**
**5.2056**
Thus we know when our particle is moving to the left when it is negatively valued.



User edited question02

Wed, 10 May 2006 23:43:30 +0000

(d) Find the total distance traveled by the particle from ''t'' = 1 to ''t'' = 8.
!!**Solution**
!!!a) Write a formula for the velocity of the particle at time t.
(image)

There are two ways on finding the velocity of this accumulation function. The long way moves anti-differentiating s(t) and then differentiating it once again. With that equation being a bit on the ugly side. We’ll go straight on the short, sweet, and easy way.
It is by using the Second Fundamental Theorum of Calculus
If f is a continuous function on some interval (a, b) and an accumulation function A is defined by A(x) = (integral of) f(t) dt, then A’(x) = f(x).
Therefore, v(t) is equal to the underlying function.
**V(t) = 1- x(cos x) - (ln x)(sin x)**
---
!!!b) At what instant does the particle reach its maximum speed?
Always understand what you are being asked for. Speed is the absolute value of velocity. Therefore does not relate to the direction its going (+/-). Another thing it is asking for the maximum, meaning the endpoints are possibilities.
Graphing the velocity function and tracing for a maximum of the function, we get it has one on where **x=3.7312**. Tracing for a minimum of the function, we get that it has one on where **x=1.2340** and **x= 6.7002**.
We evaluate the points and endpoints using the absolute value of the velocity function, **|V(t)|**.
**|V(1)| = 0.4597 <--Endpoint
|V(1.2340)| = 0.3938
|V(3.7312)|= 4.8334
|V(6.7002)| = 5.8964
|V(8)| = 0.1067 <-- Endpoint
Therefore, the particle reach its maximum speed at the instant when x=6.7002.**
---
!!!c) When is the particle moving to the left?
We look at the velocity function we found in part a.
We find the roots of the function and whether its positively or negatively valued on either side of the root.
We remember our domain and only search for the roots in between.
**v(t) 1 ____+_____|___-___ 8**
**5.2056**
Thus we know when our particle is moving to the left when it is negatively valued.
**Therefore, the particle is moving to the left during (5.2036,8%5D.**
---
!!!d)Find the total distance traveled by the particle from t=1 to t=8.
**Keyword: total distance**
To find total distance we split the integral to the # of roots it has in the domain plus one. Our particular function has one root in the restricted domain. Therefore, we split the calculations into two.
**S(5.2036) - (integral from 5.2036 to 8) v(t) dt
= 10.9923 - (-10.4675)
= 21.4599**



User edited question03

Sun, 07 May 2006 21:57:38 +0000

**Solution**
Part a)
(image)
Multiply both sides of equation by 1/√y
**Answer to Part a) y = %5B (-k/2)t + C %5D²**

Part b)
Use **''y(0) = 9''** to solve for **''C''** by substituting it into the equation

Use y(0) = 9 to solve for C by substituting it into the equation
9 = %5B (-k/2) (0) + C %5D²
Square root both sides
(+/-) √9 = √%5B (-k/2) (0) + C %5D²
Must take the positive square root of 9 because their cannot be a negative value in terms of feet
**''3 = C''**
3 = C
y = %5B (-k/2)t + 3 %5D²
Substitute the value of ''**y(20) = 4**'' into the equation to solve for ''**k**''k
4 = %5B (-k/2) (20) + 3 %5D²
Take the square root of both sides
√4 = √ %5B (-10)k + 3 %5D²
2 = (-10)k + 3
Solve for **''k''**
Solve for k
-1 = (-10)k
Multiply by -1/10
1/10 = k
**Answer to Part b) y = %5B (-0.05)t + 3 %5D²**

Part c)
**dy/dt = -(1/10) √ (y)**
Let ''**dy/dt**'' equal ''**-0.1**'' and solve for ''**y**''y
(-0.1) = -(1/10) √y
Multiply by ''**-10**''-10
(-0.1) (-10) = √y
''**1 = √y**''
1 = √y
Square both sides
1² = (√y)²
**''1 = y''**
1 = y
Use the value of ''**y = 1**'' to solve for **''t''**t, while using the equation ''**y(t)**''
y = %5B (-0.05)t + 3 %5D²
1 = %5B (-0.05)t + 3 %5D²
Take the square root of both sides



User edited question03

Sun, 07 May 2006 21:50:08 +0000

**Solution**
(image)
dy/dt = -k√y


Multiply both sides of equation by 1/√y
(1/√y) dy/dt = -k
√(y) = (-k/2)t + C
Square both sides of equation
√(y)² = %5B (-k/2)t + C %5D²
**Answer to Part a)
y = %5B (-k/2)t + C %5D²
**
**Answer to Part a) y = %5B (-k/2)t + C %5D²**

Part b)
Use **''y(0) = 9''** to solve for **''C''** by substituting it into the equation
9 = %5B (-k/2) (0) + C %5D²
Square root both sides
y = %5B (-0.1/2) t + 3 %5D²
y = %5B (-0.05)t + 3 %5D²
**Answer to Part b)
y = %5B (-0.05)t + 3 %5D²
**
**Answer to Part b) y = %5B (-0.05)t + 3 %5D²**

Part c)
**dy/dt = -(1/10) √ (y)**
Let ''**dy/dt**'' equal ''**-0.1**'' and solve for ''**y**''
(-2)(-1/0.05) = t
**t = 40 min.**
**Answer to Part c)
When t = 40 minutes
**
**Answer to Part c) When t = 40 minutes**



User edited question03

Sun, 07 May 2006 21:44:21 +0000

**Solution**
(image)
dy/dt = -k√y
Multiply both sides of equation by 1/√y.
(1/√y)dy/dt = -k
Multiply both sides of equation by dt.
(1/√y)dy = (-k)dt
Antidifferentiate both sides.
2√(y) = -kt + C
Multiply by√(y) = (1-k/2).t + C
Square both sides of equation.√(y)² = %5B(-k/2)t + C%5D²
Answer to Part a) **y = %5B(-k/2)t + C%5D²**

Use y(0) = 9 to solve for C by substituting it into the equation.
9 = %5B(-k/2)(0) + C%5D²
Square root both sides.(+/-)√9 = √%5B(-k/2)(0) + C%5D²
Must take the positive square root of 9 because their cannot be a negative value in terms of feet.
3 = C
y = %5B(-k/2)t +3%5D²
Substitute the value of y(20) = 4 into the equation to solve for k.
4 = %5B(-k/2)(20) +3%5D²Take the square root of both sides.
√4 = √%5B-10k +3%5D²
2 = -10k +3
Solve for **k**.
-1 = -10k
Multiply by -1/10.
1/10 = k
k = 0.1
y = %5B-(0.1/2)t +3%5D²
y = %5B-0.05t +3%5D²
Answer to Part b) **y = %5B-0.05t +3%5D²**
Let dy/dt =equal -(1/10)√(y)0.1 and solve for y.
-0.1 =Multiply by -(1/10)√y.
-0.1(-10) = √y
1 = √y
Square both sides.1² = √y²
1 = y
Use the value of y = 1 to solve for t, while using the equation y(t).
y1 = %5B-0.05t +3%5D²
Take the square roo√1 = √%5B-0.05t of both sides.+3%5D²
1-2 = -0.05t +3
Multiply by (-2)(-1/0.05). = t
t = 40 min.
Answer to Part c) **When t = 40 minutes**
Multiply both sides of equation by 1/√y
(1/√y) dy/dt = -k
Multiply both sides of equation by dt
(1/√y) dy = (-k) dt
Antidifferentiate both sides
2√(y) = (-k)t + C
Multiply by (1/2)
√(y) = (-k/2)t + C
Square both sides of equation
√(y)² = %5B (-k/2)t + C %5D²
**Answer to Part a)
y = %5B (-k/2)t + C %5D²
**
Use **''y(0) = 9''** to solve for **''C''** by substituting it into the equation
9 = %5B (-k/2) (0) + C %5D²
Square root both sides
(+/-) √9 = √%5B (-k/2) (0) + C %5D²
Must take the positive square root of 9 because their cannot be a negative value in terms of feet
**''3 = C''**
y = %5B (-k/2)t + 3 %5D²
Substitute the value of ''**y(20) = 4**'' into the equation to solve for ''**k**''
4 = %5B (-k/2) (20) + 3 %5D²
Take the square root of both sides
√4 = √ %5B (-10)k + 3 %5D²
2 = (-10)k + 3
Solve for **''k''**
-1 = (-10)k
Multiply by -1/10
1/10 = k
k = 0.1
y = %5B (-0.1/2) t + 3 %5D²
y = %5B (-0.05)t + 3 %5D²
**Answer to Part b)
y = %5B (-0.05)t + 3 %5D²
**
**dy/dt = -(1/10) √ (y)**
Let ''**dy/dt**'' equal ''**-0.1**'' and solve for ''**y**''
(-0.1) = -(1/10) √y
Multiply by ''**-10**''
(-0.1) (-10) = √y
''**1 = √y**''
Square both sides
1² = (√y)²
**''1 = y''**
Use the value of ''**y = 1**'' to solve for **''t''**, while using the equation ''**y(t)**''
y = %5B (-0.05)t + 3 %5D²
1 = %5B (-0.05)t + 3 %5D²
Take the square root of both sides
√1 = √ %5B (-0.05)t + 3 %5D²
1 = (-0.05)t + 3
-2 = (-0.05)t
Multiply by (-1/0.05)
(-2)(-1/0.05) = t
**t = 40 min.**
**Answer to Part c)
When t = 40 minutes
**



User edited question01

Sun, 07 May 2006 18:46:23 +0000

**Solution**
a)
y=4-x^(2/3) and y=0
4-x^(2/3)=0 (find the intersection between the graph and x- axis)
x=8
4-x^(2/3)=0 (find the intersection between the graph and x-axis)
x=8 Is where the two lines intersect and enclose the Area of region R.
(Σ takes the place of the integral sign and the numbers after it are the (a,b) of the integral.)
Σ(0,8)4-x^(2/3) dx
=(4x-(3/5)x^(5/3))(0,8) (the area of R is the integral of the intergral of the function minus 0)
=12.8
=(4x-(3/5)x^(5/3))(0,8) (the area of R is the integral of the function **4-x^(2/3)** from **(0,8)**)
A= %5B4(**8**) -(3/5)(**8**^(5/3))%5D-%5B4(**0**) -(3/5)(**0**^(5/3))%5D (Substitute **8** and **0** into the function **4-x^(2/3)** to solve for the Area of region R)
A= %5B12.8%5D -%5B0%5D
**A=12.8**
b)
V=πΣ(0,8)(4-x^(2/3))^2dx
=91.8951 (the volume of the solid πr^2 and radius is the function,and just use calculator to calculate the value is 91.8951)
c) x=K divides the region into two regions
so,
V=πΣ(0,k) (4-x^(2/3))^2 dx=91.8951/2=45.94755
Σ(0,k)(4-x^(2/3))^2=14.6286
(16x-(24/5)x^(5/3)+ (9/13)x^(13/9))(0,K)=14.6286 (Antidifferentiate it)
16K-(24/5)k^(5/3)+(9/13)k^(13/9)=14.6286 (substutition)
k=1.333
V=πΣ(0,k) (4-x^(2/3))^2 dx=91.8951/2 (The volume is divided by 2 in order to solve for the x value of **k** which divides the volume generated by rotating region R into two equal volumes.)
%5BπΣ(0,k)(4-x^(2/3))^2%5D/π = 45.94755/π (Divide both sides by π)
Σ(0,k)**(4-x^(2/3))^2**=14.6286 **Antidifferentiate** the integral)
(16x-(24/5)x^(5/3)+ (9/13)x^(13/9))(0,K)=14.6286 (The next step involves ''substitution'')
''16K-(24/5)**k**^(5/3)+(9/13)**k**^(13/9)''=14.6286 (Solve for **k**)
**k=1.333**



User edited question06

Mon, 01 May 2006 23:44:15 +0000

Σ(10,20)(f'(x))= 25-12.5
Σ(10,20)(f'(x))= 12.5
Because the integral from 10 to 20 is positive, the relative maximum occurs when x=20.
**Answer to Part a) The relative maximum occurs when x=20.**



User edited question06

Mon, 01 May 2006 23:43:20 +0000

Because A between 0 and 10 is positive,when x=10 the value of **f** is **greater** than when x=0.
Now that x=0 has been eliminated as a choice for maximum, the other contender needs to be compared to x=10.
Take the integral from 10 to 20, and if it is positive, then x=20 is the max, if negative, x=10 is the max.
Σ(10,15) is a triangle.
A = b(h)/2
A = 5(-5)/2
A = -12.5
Σ(15,20) is also a triangle.
A = b(h)/2
A = 5(10)/2
A = 25
Σ(10,20)(f'(x))= 25-12.5
Σ(10,20)(f'(x))= 12.5
Because the integral from 10 to 20 is positive, the relative maximum occurs when x=20.



User edited question06

Mon, 01 May 2006 23:34:48 +0000

**Solution**
The relative maximum depends on the where the graph of the derivative of **f** equals zero. Due to the fact that the function **f**, exists from 0 to 20 including both endpoints. Between 0 and 20, the maximum will be visible on **f'** because it will be above the x axis before it **crosses** the x axis and exists below the x axis.
There are three x coordinates that need to be considered for this problem. The obvious x=10, and the endpoints, x=0 and x=20. No one knows if x=10 is greater than x=0, but if you take the integral from 0 to 10, you will find that out.
Σ(this takes the place of the integral sign)
Σ(0,10) Can be found by finding the area of a triangle.
A = b(h)/2
A = 10(10)/2
A = 50
Because A between 0 and 10 is positive,when x=10 the value of **f** is **greater** than when x=0.
Now that x=0 has been eliminated as a choice for maximum, the other contender needs to be compared to x=10.
Take the integral from 10 to 20, and if it is positive, then x=20 is the max, if negative, x=10 is the max.



User edited question06

Mon, 01 May 2006 23:24:03 +0000

**Solution**
The relative maximum depends on the where the graph of the derivative of **f** equals zero.
The relative maximum depends on the where the graph of the derivative of **f** equals zero. Due to the fact that the function **f**, exists from 0 to 20 including both endpoints. Between 0 and 20, the maximum will be visible on **f'** because it will be above the x axis before it **crosses** the x axis and exists below the x axis.



User edited question06

Mon, 01 May 2006 23:19:24 +0000

**Solution**
The relative maximum depends on the where the graph of the derivative of **f** equals zero.



User edited FrontPage

Sat, 29 Apr 2006 15:12:01 +0000

You may have another idea for a significant contribution to the wiki. I'm open to your suggestions. Talk to me and make a suggestion. If we both agree that it would fall under this category then go ahead. Be creative with this; I'd really love to hear your ideas.
!!! A Constructive Modification means ...
you edit someone elses work, not your own, or only onw part of Question #6. You might correct we part of Question #6 (for example). You might correct a significant error or several small errors. Maybe you want to reorganize a page or the navigation from this home page. Maybe you want to edit someone else's entry, not for content, but for the way it's written such as by adding some meaningful colour or graphics. The main idea here is to move this project forward in some constructive way. Again, be creative with this.
! How this works....



User edited question04

Thu, 27 Apr 2006 01:35:45 +0000

**Solution**
ok... im not sure if i can do this but you guys.... this one's mine already. :D - ARA
a. g'(0) = %5B(x^2 + 1) * f ' (x)%5D + %5B2x * f(x)%5D
= %5B(0 + 1) * 1%5D + %5B0 * 3%5D
g'(0) = 1
b. g'(1) = %5B(1^2 + 1) * f ' (1)%5D + %5B2(1) * f(1)%5D
= 2*2 + 2* f(1)
**note: We are not sure about the value of f(1), BUT we can say that it is in fact a positive number since f(0) is equal to 3 and the graph of f'(x) is positive from that point onwards, therefore f(1) will definitely be greater than 3. **
- So, g'(1) is increasing at x = 1 because 4 + 2 * (+no.) is equal to a positive value.
c. g''(0) = ''%5B(x^2 + 1) * f''(x) + 2x * f ' (x)%5D + %5B2x * f ' (x) + 2 * f(x)%5D
= %5B1 * f''(0) + 0%5D + %5B0 + 2 * f(0)%5D
looking at the graph, I have estimated that f''(0) is quite close to .9 rather than 1. So,
= .9 + 6
g''(0) = 6.9
d. g''''(1) = %5B(1+1) * f''(1) + 2(1) * f ' (1)%5D + 2(1) * f '( 1) + 2 * f(1)%5D
**note: As we can see, f''(1) is hard to determine but because f'(1) is increasing, we can say that f''''(1) is a positive number.**
So, summing up the products, we get (if not exactly) a positive value. Therefore, g is concave up at x = 1 because its second derivative g'''' is positively valued at that point.



User edited question01

Wed, 26 Apr 2006 23:39:18 +0000

(c) The vertical line ''x'' = ''k'' divides the region R into two regions so that when these regions are rotated about the x-axis, they generate solids with equal volumes. Find the value of ''k''.
**Solution**
a)
y=4-x^(2/3) and y=0
4-x^(2/3)=0 (find the intersection between the graph and x- axis)
x=8
(Σ takes the place of the integral sign and the numbers after it are the (a,b) of the integral.)
Σ(0,8)4-x^(2/3) dx
=(4x-(3/5)x^(5/3))(0,8) (the area of R is the integral of the intergral of the function minus 0)
=12.8
b)
V=πΣ(0,8)(4-x^(2/3))^2dx
=91.8951 (the volume of the solid πr^2 and radius is the function,and just use calculator to calculate the value is 91.8951)
c) x=K divides the region into two regions
so,
V=πΣ(0,k) (4-x^(2/3))^2 dx=91.8951/2
πΣ(0,k)(4-x^(2/3))^2=45.94755
Σ(0,k)(4-x^(2/3))^2=14.6286
(16x-(24/5)x^(5/3)+ (9/13)x^(13/9))(0,K)=14.6286 (Antidifferentiate it)
16K-(24/5)k^(5/3)+(9/13)k^(13/9)=14.6286 (substutition)
k=1.333



User edited question03

Wed, 26 Apr 2006 23:03:14 +0000

-2 = -0.05t
Multiply by (-1/0.05).
(-2)(-1/0.05) = t
t =
t = 40 min.
Answer to Part c) **When t = 40 minutes**



User edited question03

Wed, 26 Apr 2006 23:01:51 +0000

Use the value of y = 1 to solve for t, while using the equation y(t).
y = %5B-0.05t +3%5D²
1 = %5B-0.05t +3%5D²
Take the square root of both sides.
√1 = √%5B-0.05t +3%5D²
1 = -0.05t +3
-2 = -0.05t
Multiply by (-1/0.05).
(-2)(-1/0.05) = t
t =



User edited question03

Wed, 26 Apr 2006 22:59:09 +0000

1 = y
Use the value of y = 1 to solve for t, while using the equation y(t).
y = %5B-0.05t +3%5D²
1 = %5B-0.05t +3%5D²



User edited question03

Wed, 26 Apr 2006 22:58:23 +0000

Answer to Part b) **y = %5B-0.05t +3%5D²**
dy/dt = -(1/10)√(y)
Let dy/dt equal -0.1 and solve for y.
-0.1 = -(1/10)√y
Multiply by -10.
-0.1(-10) = √y
1 = √y
Square both sides.
1² = √y²
1 = y
Use the value of y = 1 to solve for t, while using the equation y(t).
y = %5B-0.05t +3%5D²



User edited question03

Wed, 26 Apr 2006 22:53:51 +0000

y = %5B-(0.1/2)t +3%5D²
y = %5B-0.05t +3%5D²
Answer to Part b) **y = %5B-0.05t +3%5D²**
dy/dt = -(1/10)√(y)
Let dy/dt equal 0.1 and solve for y.



User edited question03

Wed, 26 Apr 2006 22:50:26 +0000

k = 0.1
y = %5B-(0.1/2)t +3%5D²
y = %5B-0.05t +3%5D²
Answer to Part b) **y = %5B-0.05t +3%5D²**



User edited question03

Wed, 26 Apr 2006 22:48:12 +0000

Multiply by -1/10.
1/10 = k
k = 0.1
y = %5B-(0.1/2)t +3%5D²
y = %5B-0.05t +3%5D²



User edited question03

Wed, 26 Apr 2006 22:45:37 +0000

3 = C
y = %5B(-k/2)t +3%5D²
Substitute the value of y(20) = 4 into the equation to solve for k.
4 = %5B(-k/2)(20) +3%5D²
Take the square root of both sides.
√4 = √%5B-10k +3%5D²
2 = -10k +3
Solve for **k**.
-1 = -10k
Multiply by -1/10.
1/10 = k
k = 0.1



User edited question03

Wed, 26 Apr 2006 22:39:37 +0000

9 = %5B(-k/2)(0) + C%5D²
Square root both sides.
(+/-)√9 = √%5B(-k/2)(0) + C%5D²
Must take the positive square root of 9 because their cannot be a negative value in terms of feet.
3 = C
y = %5B(-k/2)t +3%5D²
Substitute the value of y(20) = 4 into the equation to solve for k.