# Complex area between curves problem.

1. Apr 16, 2010

### bur7ama1989

1. The problem statement, all variables and given/known data

I am working on a Calculus I project and the last question has me stunned. I would really appreciate some assistance in this. Maybe a push in the right direction.
So here is the problem word for word. There is also an attachment containing a picture of the problem.

Suppose that the cubic function "y = a(x^3) + b(x^2) + cx + d" and the parabola "y = k(x^2) + mx + n" intersect at "x=A" and "x=B" and that the curves are tangent at B (that is, the derivatives are equal at "x=B").
Show that the area between the curves is equal to

Area = ((absolute(a))/12)((B-A)^4)

2. Relevant equations

I made the first equation into f(x):

f(x)=a(x^3)+b(x^2)+cx+d

and the second equation into g(x):

g(x)=k(x^2)+mx+n

I found the derivatives:

f'(x)=3a(x^2)+2bx+c

g'(x)=2kx+m

I equated them

3a(x^2)+2bx+c=2kx+m

I now have a system of three equations with 9 variables and the answer shows only 3.

3. The attempt at a solution

I played with these equations until my neck hurt.

1)

f(A)=g(A)

2)

g(B)=g(B)

After those two steps I began solving for one variable at a time and trying to remove them from the question. Needless to say I have failed. Please help me out.

#### Attached Files:

• ###### Calc.bmp
File size:
96.9 KB
Views:
79
2. Apr 17, 2010

### lanedance

i think you're overestimating your unknowns as some of the relations are a little hidden...

so you have
$$f(x) = ax^3 + bx^2 + cx + d$$
$$g(x) = kx^2 + mx + n$$

with area given by the integral of h(x) = f(x) - g(x) from A to B
$$\h(x) = f(x)-g(x) = ax^3 + (b-k)x^2 + (c-m)x + (d-n)$$

as you don't know any of the constants anyway, simplify the function by considering new constants r,s,t
$$\h(x) = ax^3 + rx^2 + sx + t$$

now you know
$$h(A) = 0$$
$$h(B) = 0$$
$$h'(B) = 0$$

h has 4 unknowns with 3 equations...

3. Apr 17, 2010

### bur7ama1989

Ok, you were tremendously helpful in moving me along, but i seem to be stuck once more. I used substitution rules and was able to get rid of two of the four variables. I ran out of equations to use to get rid of the last unknown. Any thoughts as to where I may be going wrong.

4. Apr 17, 2010

### lanedance

show your work, so i can see where you're at

5. Apr 17, 2010

### lanedance

keep in mind the answer is written in terms of a

6. Apr 17, 2010

### bur7ama1989

So i have the three equations:
(1) 3a(B^2)+2rB+s=0
(2) a(A^3)+r(A^2)+sA+t=0
(3) a(B^3)+r(B^2)+sB+t=0

I equated (2) and (3) to remove "t" from the equation and my result was:

(4) a((A^3)-(B^3))+r((A^2)-(B^2))+s(A-B)=0

I solved for "s" in equation (1):

(1) s=-3a(B^2)-2rB

I substituted "s" into the new equation (4) and my result was:

a((A^3)-(B^3))+r((A^2)-(B^2))+(-3a(B^2)-2rB)(A-B)=0

Now I don't know what to do next.

7. Apr 17, 2010

### Dick

Don't lose sight of what you are actually trying to do. Write down an expression for the integral 'I' as well. You want to use your other equations to eliminate the r, s and t from the expression for the integral.

8. Apr 18, 2010

### bur7ama1989

I created an equation for the integral I. Each time I solved for an unknown I plugged it into the equation of the integral so as to eliminate the unknowns in it as well. My result was an answer too long to type and impossible to simplify further. I am going to try this using Maple as that is what the Prof. ultimately wants us to hand in, but I don't know if my neck can take one more try at doing this problem with a pencil and paper. Interesting, but dreadfully long.

Thank you and lanedance for your help, I will post what happens using Maple.

9. Apr 18, 2010

### lanedance

ok i had a bit more of a think about this, and think solving for the constant might not be the quickest way, should help to re-write the cubic equation in term of its roots... .

using the boundary conditions h(A) = h(B) = 0, we know A & B are both real roots, so it must also have a 3rd real root, call it n, then we can write the equation as:
$$h(x) = ax^3 + rx^2 +sx + t = a(x-A)(x-B)(x-n)$$

To use that last, differentiate with product rule and have a look what happens when you set h'(B) = 0

10. Apr 18, 2010

### lanedance

so hopefully that helps the simplification, the integral will still be complicated, but should hopefully yield the (B-A)^4

11. Apr 18, 2010

### Dick

I don't think you need to simplify solving it by hand. This seems to be a Matlab assignment. That makes more sense, it's pretty sadist to do it by hand. I did it using Maxima and it solves with a CAS just fine. So the setup is good. The only issue that may require some thought is why you should write |a| instead of a or -a. The CAS won't give you that automatically.

12. Apr 19, 2010

### bur7ama1989

I did it using Maple. Much quicker and easier but it really frustrated me that I couldn't do it by hand. I'm sure I made an error somewhere and with finals coming up i don't have the leisure time to be go over it again. As for the absolute value "a" i thought i knew why, but when i thought about it deeper I realized i didn't really know. I assumed it had something to do with the orientation of the cubic function and its position in the integral. I am not sure if that is right though and I don't really know how it applies. The area between two curves is given as the integral from one point to another of the upper function minus the lower function. So I thought that in order for the cubic function to be on top it would have ta have a positive leading coefficient. When i thought about it i realized that may not necessarily be the case.

I attached my work from Maple.

#### Attached Files:

• ###### Calc.gif
File size:
21.5 KB
Views:
55
13. Apr 19, 2010

### Dick

You've got the right idea, if the cubic is the larger function then you integrate ax^3+rx^2+sx+t. If it's smaller you integrate -ax^3+rx^2+sx+t. So the 'a' will flip sign in the answer. But you know the answer should be positive in either case. Hence |a|.

14. Apr 19, 2010

### bur7ama1989

Excellent. Thank you for your help. I really appreciate it. I am a new member on this site and I will try and contribute as much as possible as a sign of my gratitude.

15. Apr 19, 2010

### Dick

Thanks in advance for the help offer! It's rewarding to help someone as motivated as you are.

16. Apr 20, 2010

### lanedance

ditto -

also did you see that in post #9, with a bit of thought you can write the cubic directly as:
$$h(x) = a(x-A)(x-B)^2$$

without having to do the substitution

17. Apr 20, 2010

### Dick

Now that's a good idea!