Mu Individual 2012 National MAO Solution Manual

1.The cross sections are squares. First, we need to find the side length of one square. The side-length of the square will be . Now, we find the lower and upper bounds of the integral by setting it equal to 0. Now, since the cross-section areas are squares, the volume of the figure will be equal to . from -1 to 1. Algebraic simplification leads the answer to be . D.

2. The way that the differential equation is solved is through a multi-layered substitution. . ; , so that means that . Substitute back into the first expression. Cross multiply and use separable equations like in the first one. Integrating both sides we get:

, and . Substitute one last time to get terms of X and Y, and . A.

3. , and . the expression is minimized. This means that Apply the quotient rule: Since the denominator of the quotient rule does not matter since we are setting the numerator equal to 0, we set equal to 0.

, so this means that

. The two roots are and. Since it was given that , , . So the expression evaluates to which simplifies to 2. E.

4. THROWN OUT . Take the derivative of both sides with respect to x, , so this means that  At point . C.

5. . A.


Find for the equation at the point (6,4)










At point (6,4)

; A.


; C.

8. Find . Now this seems like it could be a pain in the butt trying to simplify this with trigonometric identities. There is one identity which makes integrating this expression much easier. . Now, we can substitute into . Now, let’s simplify . Use a u-substitution to get =. The total expression turns into: . From 0 to , this turns into . E.

9. First, use Newton’s formula starting with .


., since and .

. Now apply Newton’s formula again with to find .

 through simplification.

Now that we have in the form , we need to find the number of factors in . 336 breaks down in its prime factorization as such:

.  # of factors= . ; B or E Accepted.

10. So we have given that and . Using the Mean Value Theorem for Derivatives, we can conclude that for some numbers and . Setting up the Mean-Value Theorem for Derivatives up, we can see that:

for some . If we rearrange this so that we can isolate , we come up with , so that means that . , so we can substitute: . Also, we can substitute . So we end up with . Since we were given that , this means that . This means that , and , given our bounds from earlier. So this means if we substitute , we come up with the final inequality: . Now we can deduce that and . . B.

11. (doesn’t this look fun?) Ok first you need to assess the method of integration most appropriate for evaluating this integral. Partial fractions seem the best!

Using that we can figure out that: .

. Factoring we get:

. By this logic:

, , , and . Through common sense we can now obtain the fact that: , that , , and that . Now you can re-substitute into the partial fraction you had before.


The first integral is the definition of the integral of arc tan of x. So now we have:

. Now use u substitution for the other integral.

. That makes that integral  So the final simplified integral is: ; D.

12. This question is a very elementary derivative question:

. The first term is easy. By power rule, this will be the power taken down 2011 times, so it is the same as 2012!.

. For the second and third terms of the function, you have to know that both and have cyclical derivatives in cycles of 4. That means that the 4th derivative is the same as the 8th as the 12th and so on, same goes for the 1st and the 5th and so on. because 2011 is one before the cycle ends, which is equivalent to the third derivative of , which is . Same with . This is equivalent to the third derivative, which is . Now the last term is

For this , know that and . That means it is just . At 0:

0+0+3-1=2; A.

13. First lets start off with the curve . You can use either horizontal slices or cylindrical shells to calculate half of the volume of the curve, and double it to obtain the volume of doughnut

Method 1: Horizontal Slices. We can manipulate the formula before to integrate in terms of one variable. evaluated from 0 to 1, which means , so cubic feet of jelly.

Method 2: Cylindrical Shells. Substitute into the integral to integrate in terms of one variable. Now we use u-substitution to integrate this function: , so the integral is now:  from 0 to 1 from 0 to 1, so , D.


We start of this problem by stating what we know. Chris Kim is 170 feet tall, if we make a right triangle, we can see that this is the hypotenuse. This makes and 80 and 150 feet respectively. If we set up the equation up to:

, and derive it, we get. We know that using our given information. This means that , so this means that . D.

15. = Use L’Hospitals: ==0. A.

16. Set this up as: , so this means that . The volume of the ice-cream cone is given as: . Substitute . Take the derivative with respect to , . Our given information is that . This means that , therefore, . C.

17. Set this up as. from 0 to 1. This comes out to . C.

18. CHANGED TO D: This looks like a tough one. A trick with trigonometric substitution should make this a little easier! We have our function , which would be very tough to just do the chain rule straight out. Let’s try and simplify this as much as possible: Let’s simplify . Construct a right triangle with angle . Opposite we have , with hypotenuse 1. We need to find our other opposite leg. We set up the Pythagorean Theorem. Simplify in y in terms of x and we get . Using this same triangle, we need to find the tangent angle , which simplifies to . So we are really trying to find . First let’s find the derivative of the function. Finding the derivative of the inside should be done first if chain rule is going to be the most effective:

Finding the derivative of is easier for me treating as a product rule rather than a quotient ruleBy chain rule, taking the derivative of the entire ,Simplifying algebraically, we get . Using the chain rule, the second derivative is as follows=. So evaluated at is =. E.

19. First find the critical numbers of the function. , so that means that . , the critical numbers are . To determine which is the relative min and which is the relative max, use those values in the second derivative test. Testing , we get -4. This is less than 0, and therefore must be the relative maximum. Testing at , we get 4, which is greater than 0. So this is our relative minimum. So now we know that and . So . B.

20. First let’s find the value of the derivative of the function at the x-value given. . Using the chain rule, . At , =. Now we plug this back into point-slope, . We need to find by plugging into the original function, which gives us. So the tangent line simplifies to. So now we can tell that , , and . So . B.

21. The total profit generated by an item is defined as the revenue minus the manufacturing cost: . =. is going to be the total cost per day; . Set up the equation:

. Determine any critical numbers of , so and . A.

22. . This integral looks like a lot to do. Actually, with a few trigonometric manipulations, this question becomes easily integrable.First use the fact that to rearrange the integral toWe can now use u-substitution for each integral given== from 0 to the square root of pi/2. This evaluates to =. Do the same thing for the second integral: ==. . D.

23. Linda’s answer can be found using the following differential approximation:

at the point = =. So now we can deduce that , , and . Steve’s approximation is a bit easier since it is centered at the origin. At (0,0), this will be . Now we can deduce that , , and . So this makes our expression: evaluate into 10+3+2+0=15. B.

24. For this question, simply set up the equation , . To find , we have, which is equal to at , or . B.

25. First, we need to write the MacLaurin series for . To do this, we use a Taylor series centered at . Our question only calls for the first three terms, so this means the 0th term is included. Our 0th term is . The value of the 0th derivative at 0 is 1. So this means our 0th term is 1. To find the 1st term, take the derivative of the function. Chain rule yields us . The value of the 1st derivative at is . Again, since 1!=1, our 1st term will be . Now we find the 2nd term. By using chain rule again, we find that the second derivative of the function is . The value of second derivative at is . Now we multiply that by , and our term comes out to be . So now we know our series is . Now we can manipulate the series to fit by plugging in for every . So our first three terms centered there are . We integrate this from 0 to 1: = from 0 to 1 evaluates to . Simplifying this algebraically we get , so this means that evaluates to 2+10-3= 9. E.

26. Use the ratio test to find the interval of convergence:

, therefore, the radius of convergence is , D.

27. The differential equation is separable, but leads to the equation  We will use partial decomposition to solve the left side of the differential equation. When , so . Using the same method, we get . Multiply each side by 600.

. The constant of integration can be solved as such, since . Solving this yields , so the new function can be modeled after . So this means that = 600-= 603. C.

28. We should rearrange the integral given to . Now we can use trigonometric substitution, .

Now, we have to change the bounds of integration: When , ,so and when , so that means . We now change the integral to terms of trigonometric functions:

= This turns into . We use trigonometric simplification to simplify the integral to

trigonometric substitution from here will not work. So instead we’re going to rearrange the integral to make a substitution.

. Now we can use a substitution since we have a function in terms of u and its derivative. , so now we change the integral bounds AGAIN to get the new integral:

When , , and when , then , so the integral becomes:

, and at this point we simply integrate by splitting fractions: , from ½ to 1, which simplifies to . C.

29. . Use a u-substitution to rearrange this into. Evaluated from 0 to . So BC-AD= 6-2=4. B.

30.  Use a u-substitution . Use a substitution from 0 to infinity, since arctangent’s bound is , this evaluates out to . A.