SOLUTION: Hi again, I asked this question some time ago, but had never got an answer, if anyone can help me I would be most grateful. a) Find the area of the region bounded by the parabola:

Algebra ->  Graphs -> SOLUTION: Hi again, I asked this question some time ago, but had never got an answer, if anyone can help me I would be most grateful. a) Find the area of the region bounded by the parabola:      Log On


   



Question 214343: Hi again, I asked this question some time ago, but had never got an answer, if anyone can help me I would be most grateful.
a) Find the area of the region bounded by the parabola:
f(x) 2x^2, g(x) = 3x^2 + 5 and the line y = 8.
b) Find dy/dx given that y = l+x(^2)e^(y)
Thanks again, Nick.

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
a) Find the area of the region bounded by the parabola:
f(x) 2x^2, g(x) = 3x^2 + 5 and the line y = 8.
I think you have not typed the problem correctly. As stated, you have two parabolas and they do not intersect each other. So it not clear what the region is. See the graph below which include all three functions:
graph%28400%2C+400%2C+-5%2C+5%2C+-1%2C+11%2C+2x%5E2%2C+3x%5E2+%2B+5%2C+8%29

b) Find dy/dx given that y = l+x(^2)e^(y)
If we call u+=+1, v+=++x%5E2 and w+=+e%5Ey, then u' = 0, v' = 2x and, using the chain rule, w' = y'*e%5Ey. Substituting u, v and w into the original equation we get:
y = u + v*w
Using basic rules of differentiation, we get:
y' = u' + v*w' + w*v'
Replacing u', v, v', w and w' (from above) we get:
y' = 0 + x%5E2*(y'*e%5Ey%29 + e%5Ey%2A%282x%29
Now we use Algebra to simplify and solve for y':
y' = x%5E2%2Ae%5Ey*y' + 2x%2Ae%5Ey
Subtract x%5E2%2Ae%5Ey*y' from each side:
y' - x%5E2%2Ae%5Ey*y' = 2x%2Ae%5Ey
Factoring out y' on the left:
y'*%281+-x%5E2%2Ae%5Ey%29+=+2x%2Ae%5Ey
Dividing both sides by %281+-x%5E2%2Ae%5Ey%29:
y' = %282x%2Ae%5Ey%29%2F%281+-x%5E2%2Ae%5Ey%29