Find the area of the region bounded by the parabola y = 4x^2, the tangent line to this parabola at (4, 64), and the x-axis.
First, find the equation of the tangent line:
y = 4x^2
y' = 8x
slope at (4,64) = 32
(y-64)/(x-4) = 32
y = 32x - 64
32x-64 crosses the x-axis at x=2
So, we need to break the area up into two parts.
Area between the curve and y=0 on [0,2]
Area between curve and tangent line on [2,4]
Area = Int(4x^2 dx)[0,2] + Int(4x^2 - (32x-64))[2,4]
= (4/3 x^3)[0,2] + (4/3 x^3 - 16x^2 + 64x)[2,4]
= [4/3 * 8] + [4/3 * 64 - 16*16 + 64*4] - [4/3 * 8 - 16*4 + 64*2]
= 64/3
To find the area of the region bounded by the parabola, the tangent line, and the x-axis, we need to break down the problem into simpler steps.
First, let's find the x-coordinate where the tangent line intersects the parabola. The tangent line at (4, 64) has the equation of the form y = mx + c, where m is the slope and c is the y-intercept. Since the point (4,64) lies on the tangent line, we can substitute these values into the equation to find the value of c.
Using the slope-intercept form of a line equation, we have:
64 = 4(4) + c
64 = 16 + c
c = 48
So, the equation of the tangent line is y = 4x + 48.
Next, we find the x-values where the parabola and the tangent line intersect. To do this, we set the equations for the parabola and the tangent line equal to each other and solve for x.
4x^2 = 4x + 48
4x^2 - 4x - 48 = 0
To solve this quadratic equation for x, we can use factoring or the quadratic formula. Factoring might not be straightforward in this case, so let's solve it using the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this equation, a = 4, b = -4, and c = -48. Substituting these values into the formula, we get:
x = (-(-4) ± √((-4)^2 - 4(4)(-48)))/(2(4))
x = (4 ± √(16 + 768))/8
x = (4 ± √784)/8
x = (4 ± 28)/8
Simplifying further, we have two possible solutions for x:
x₁ = (4 + 28)/8 = 32/8 = 4
x₂ = (4 - 28)/8 = -24/8 = -3
Since we are dealing with a real-valued problem, we discard the negative solution. Therefore, the x-coordinate where the parabola and the tangent line intersect is x = 4.
Now, we can integrate to find the area between the parabola and the x-axis in the interval from x = 0 to x = 4. The parabola is defined as y = 4x^2, so the area can be calculated as:
Area = ∫[0,4] (4x^2) dx
Integrating this expression will give us the area desired. We would perform this integration according to the standard rules of integration.
Area = [4/3 * x^3] evaluated from 0 to 4
Area = 4/3 * (4^3) - 4/3 * (0^3)
Area = 4/3 * 64 - 4/3 * 0
Area = 256/3
Therefore, the area of the region bounded by the parabola y = 4x^2, the tangent line at (4, 64), and the x-axis is 256/3 square units.
To find the area of the region bounded by the given parabola, the tangent line, and the x-axis, we need to find the x-coordinates of the points of intersection first.
The given parabola is y = 4x^2.
The tangent line at (4, 64) can be determined using the derivative of the parabola. Taking the derivative of y = 4x^2, we have dy/dx = 8x. Substituting x = 4 into the derivative, we find the slope of the tangent line to be 8(4) = 32.
So, the equation of the tangent line is y - 64 = 32(x - 4).
To find the x-coordinates of the points of intersection, we need to solve the equation of the parabola and the equation of the tangent line simultaneously.
First, let's solve for x in the equation of the tangent line:
y - 64 = 32(x - 4)
y - 64 = 32x - 128
32x - y = 64
Next, we set the equation of the parabola equal to the equation of the tangent line:
4x^2 = 32x - y
Now we can substitute the equation of the tangent line into the equation of the parabola:
4x^2 = 32x - (32x - 64)
4x^2 = 64
Simplifying further, we get:
x^2 = 16
Taking the square root of both sides, we have:
x = ±4
So, the points of intersection are (4, 0) and (-4, 0).
To find the area bounded by the parabola, the tangent line, and the x-axis, we need to integrate the difference between the parabola and the x-axis from -4 to 4.
The area formula is given by: A = ∫[a,b] |f(x) - g(x)| dx, where f(x) is the upper function and g(x) is the lower function.
In this case, the upper function is the parabola y = 4x^2, and the lower function is the x-axis y = 0.
So, the area is given by:
A = ∫[-4,4] |4x^2 - 0| dx
A = ∫[-4,4] 4x^2 dx
We can integrate this expression using the power rule:
A = [4/3 * x^3] from -4 to 4
A = 4/3 * (4^3 - (-4)^3)
A = 4/3 * (64 - (-64))
A = 4/3 * 128
A = 512/3
Therefore, the area of the region bounded by the parabola y = 4x^2, the tangent line at (4, 64), and the x-axis is 512/3 square units.