Consider the curves y = x^2and y = mx, where m is some positive constant. No matter what positive constant m is, the two curves enclose a region in the first quadrant.Without using a calculator, find the positive constant m such that the area of the region
bounded by the curves y = x^2 and y = mx is equal to 8.
I would very much appreciate it if someone could find the answer and explain how you did it.
the anti derivative is (m/2)x^2 - 1/3(x)^3 so if you use the 2nd fundamental theorem of calculus using (mx - x^2) over the interval (0,m) and set that equal to 8 (the interval is 0 to m because the function with the larger area is mx) you should get m^3 / 8 = 8. Solve for m and you're done :)
Well, isn't this a quadratic question! Let's get things rolling, shall we?
First, we need to find the points of intersection between the two curves. So we set the equations equal to each other:
x^2 = mx
Rearranging this equation, we have:
x^2 - mx = 0
Now, let's solve for x by factoring out the common factor x:
x(x - m) = 0
From this equation, we can see that we have two potential solutions for x: x = 0 and x = m.
Now, let's set up and solve the definite integral to find the area between the curves. The area A can be calculated by subtracting the equation of the parabola from the equation of the line (y = mx - x^2) and integrating from x = 0 to x = m:
A = ∫(mx - x^2)dx (from 0 to m)
Integrating, we get:
A = [m(x^2/2) - (x^3)/3] (from 0 to m)
Simplifying further, we have:
A = (m(m^2/2) - (m^3)/3) - (0/2 - 0/3)
Simplifying some more:
A = (m^3/2) - (m^3)/3
To find the value of m that gives an area of 8, we set this equation equal to 8 and solve for m:
(m^3/2) - (m^3)/3 = 8
Combining the fractions on the left side, we have:
(3m^3 - 2m^3)/6 = 8
Simplifying further:
(m^3)/6 = 8
Multiplying both sides by 6:
m^3 = 48
And finally, taking the cube root of both sides, we find:
m = 3.634
So, the positive constant m that gives an area of 8 is approximately 3.634.
I hope that adds up to a satisfying answer for you!
To find the positive constant m such that the area of the region bounded by the curves y = x^2 and y = mx is equal to 8, we need to determine the points where the two curves intersect in the first quadrant.
First, let's set the equations of the two curves equal to each other and solve for x:
x^2 = mx
Rearrange the equation:
x^2 - mx = 0
Factor out x:
x(x - m) = 0
From this equation, we can determine that either x = 0 or x = m.
Now we need to find the corresponding y-values for these x-values.
For y = x^2, substituting x = m gives us:
y = (m)^2 = m^2
For y = mx, substituting x = 0 gives us:
y = 0
So the two curves intersect at the points (0,0) and (m, m^2).
To find the area between the curves, we need to find the definite integral from 0 to m of the difference between the equations of the two curves:
Area = ∫[0, m] (mx - x^2) dx
Integrating, we get:
Area = [1/2 mx^2 - 1/3 x^3] evaluated from 0 to m
Area = 1/2 m(m)^2 - 1/3 (m)^3 - (0) - (0)
Area = 1/2 m^3 - 1/3 m^3
Area = 1/6 m^3
Now we can set this equal to 8 and solve for m:
1/6 m^3 = 8
Multiply both sides by 6:
m^3 = 48
Take the cube root of both sides:
m = ∛(48)
Without using a calculator, we can approximate the cube root of 48 by recognizing that 48 is divisible by 8:
m = ∛(8 x 6) = ∛(2^3 x 6) = 2∛6
Therefore, the positive constant m such that the area of the region bounded by the curves y = x^2 and y = mx is equal to 8 is approximately 2∛6.
To find the positive constant m such that the area of the region bounded by the curves y=x^2 and y=mx is equal to 8, we need to set up the integral to calculate the area of the region and solve for m.
First, let's find the points of intersection between the two curves:
x^2 = mx
Rearrange the equation:
x^2 - mx = 0
Factor out the common factor x:
x(x - m) = 0
From this equation, we get two possible solutions: x = 0 or x = m.
Next, we need to set up the integral to calculate the area between the curves. The area can be obtained by integrating the difference between the two functions, from x = 0 to x = m:
A = ∫[0 to m] (mx - x^2) dx
Now, let's integrate the function. We'll split it into two different integrals:
A = ∫[0 to m] mx dx - ∫[0 to m] x^2 dx
Simplifying the integrals:
A = 1/2 * m * x^2 ∣∣[0 to m] - 1/3 * x^3 ∣∣[0 to m]
A = 1/2 * m * m^2 - 1/3 * m^3
To find m such that A = 8, we can set up the equation:
8 = 1/2 * m * m^2 - 1/3 * m^3
Multiplying through by 6 to eliminate the fractions:
48 = 3m^3 - 2m^3
48 = m^3
Taking the cube root of both sides:
3√(48) = m
Simplifying:
3√(48) = 3√(16 * 3) = 3√(16) * √(3) = 4√(3)
Therefore, the positive constant m such that the area of the region bounded by the curves y = x^2 and y = mx is equal to 8 is 4√(3).