Find c > 0 such that the area of the region enclosed by the parabolas
y=x^2-c^2 and y=c^2-x^2 is 13.
notice that one parabola opens up, the other down.
their intersection .....
x^2 - c^2 = c^2 - x^2
2x^2 = 2c^2
x^2 = c^2
x = ± c ---- y = 0 in both
they intersect at (-c,0) and (c,0)
using symmetry
area = 2∫(2c^2 - 2x^2) dx from 0 to c
= 2(2c^2 x - (2/3)x^3) | from 0 to c
= 2( 2c^3 - (2/3)c^2 - 0)
= 2(4/3 c^3)
so (8/3)c^3 = 13
8c^3 = 39
c = ∛39 /2 = appr 1.7
check my algebra and arithmetic, was expecting a "nicer" answer.
To find the value of c we need to determine the points at which the two parabolas intersect. We can do this by equating the two equations and solving for x.
The equations of the parabolas are:
y = x^2 - c^2 ...(1)
y = c^2 - x^2 ...(2)
Setting equation (1) equal to equation (2), we have:
x^2 - c^2 = c^2 - x^2
Re-arranging the equation, we get:
2x^2 = 2c^2
x^2 = c^2
Taking the square root of both sides, we have:
x = ± c
Now we have two values of x at which the parabolas intersect, x = c and x = -c.
To find the corresponding y-values, we substitute these values of x into either equation (1) or (2). Let's use equation (1):
For x = c:
y = c^2 - c^2 = 0
For x = -c:
y = c^2 - (-c^2) = 2c^2
So, the coordinates of the points of intersection are (c, 0) and (-c, 2c^2).
To calculate the area of the region enclosed by the parabolas, we need to find the definite integral of the difference between the two equations over the interval [-c, c]:
Area = ∫ [x^2 - c^2] - [c^2 - x^2] dx with the limits of integration from -c to c
Expanding and simplifying the integral, we get:
Area = ∫ [2x^2 - 2c^2] dx
Evaluating the integral, we have:
Area = [ (2/3)x^3 - (2c^2)x ] evaluated from -c to c
Area = (2/3)(c^3 - (-c)^3) - (2c^2)(c - (-c))
Area = (2/3)(2c^3) - (2c^2)(2c)
Area = (4/3)c^3 - 4c^3
Area = -10c^3/3
Now we can set this equal to 13 and solve for c:
-10c^3/3 = 13
Multiplying both sides by -3/10, we get:
c^3 = -39/10
Taking the cube root of both sides, we have:
c = -∛(39/10)
Therefore, c is approximately equal to -1.253.