Write the integral in one variable to find the volume of the solid obtained by rotating the first‐quadrant region bounded by y = 0.5x2 and y = x about the line x = 5.
Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = -x^3 and y = -x.
A stone is thrown upward with a speed of 30 feet per second from the edge of a cliff 200 feet above the ground. What is the speed of the stone when it hits the ground? Use acceleration due to gravity as -32 feet per second squared and approximate your answer to 3 decimal places.
I have solved the first one! I'm still confused about the second and third :(
for the integral, there are two ways.
shells of thickness dx:
v = ∫[0,2] 2πrh dx
where r = 5-x and h=x-x^2/2
v = ∫[0,2] 2π(5-x)(x-x^2/2) dx = 16π/3
discs (washers) of thickness dy:
v = ∫[0,2] π(R^2-r^2) dy
where R=5-y and r=5-√(2y)
v = ∫[0,2] π((5-y)^2-(5-√(2y))^2) dy = 16π/3
For #2, the curves intersect at (-1,1) and (1,-1) so each of the 4 subintervals has width 2/4 = 1/2
That means tne midpoints of the subintervals are at -3/4, -1/4, 1/4, 3/4
You might as well take advantage of the symmetry, and just take twice the sum on the right, so that is
2[(1/2)f(1/4)+(1/2)f(3/4)]
= f(1/4)+f(3/4)
where f(x) is the distance between the curves: -x^3-x
For #3, you know that g = -32 ft/s^2, so every second, v decreases by 32 ft/s:
v(t) = 30-32t
so, how long does it take for the stone to hit the ground?
h(t) = 200+30t-16t^2
solve for t, and then use that to find v.
To find the volume of the solid obtained by rotating the first-quadrant region bounded by y = 0.5x^2 and y = x about the line x = 5, you need to use the method of cylindrical shells.
First, let's find the limits of integration. To do this, set y = 0.5x^2 and y = x equal to each other:
0.5x^2 = x
Solving for x, we get x = 2.
Now, the volume of the solid is given by the integral:
V = ∫[a,b] 2πx(y_2 - y_1)dx
where a = 0 and b = 2, y_1 = 0, and y_2 = x.
So the integral becomes:
V = ∫[0,2] 2πx(x - 0)dx
V = 2π ∫[0,2] x^2 dx
To find the antiderivative of x^2, we add 1 to the exponent and divide by the new exponent:
V = 2π (1/3)x^3 ∣[0,2]
Substituting the limits:
V = 2π (1/3)(2^3) - 2π (1/3)(0^3)
V = 2π (8/3)
Finally, simplify the expression:
V = (16/3)π.
Therefore, the volume of the solid obtained by rotating the first-quadrant region bounded by y = 0.5x^2 and y = x about the line x = 5 is (16/3)π cubic units.
To approximate the area of the region bounded by y = -x^3 and y = -x using the mid-point rule with n = 4, you need to divide the region into four equal subintervals and use the midpoint of each subinterval as the representative x-value to calculate the area of a rectangle. Then, sum up the areas of all the rectangles to get an approximation of the total area.
First, find the width of each subinterval:
Δx = [(b - a) / n]
where a = -1 and b = 1, and n = 4.
Δx = [(1 - (-1)) / 4]
Δx = 0.5
Now, calculate the representative x-values for each subinterval:
x_1 = -1 + (0.5 / 2) = -0.75
x_2 = -0.5
x_3 = -0.25
x_4 = 0.25
Next, calculate the height of each rectangle using the corresponding representative x-value and the function value at that x-value:
h_1 = -(-0.75)^3 = -0.421875
h_2 = -(-0.5)^3 = -0.125
h_3 = -(-0.25)^3 = -0.015625
h_4 = -(0.25)^3 = -0.015625
Now, calculate the area of each rectangle:
A_1 = 0.5 * 0.421875 = 0.2109375
A_2 = 0.5 * 0.125 = 0.0625
A_3 = 0.5 * 0.015625 = 0.0078125
A_4 = 0.5 * 0.015625 = 0.0078125
Finally, sum up the areas of all the rectangles to get the approximate area:
A ≈ A_1 + A_2 + A_3 + A_4
A ≈ 0.2109375 + 0.0625 + 0.0078125 + 0.0078125
A ≈ 0.2880625.
Therefore, using the mid-point rule with n = 4, the approximate area of the region bounded by y = -x^3 and y = -x is approximately 0.2880625 square units.
To find the speed of the stone when it hits the ground, you can use the following kinematic equation:
v^2 - u^2 = 2as
where v is the final velocity (the speed when it hits the ground), u is the initial velocity (30 feet per second), a is the acceleration due to gravity (-32 feet per second squared), and s is the displacement (200 feet).
Plugging in the known values into the equation, we get:
v^2 - (30^2) = 2(-32)(200)
v^2 - 900 = -12800
v^2 = -12800 + 900
v^2 = -11900
Taking the square root of both sides to solve for v:
v = sqrt(-11900)
Since the acceleration due to gravity is negative (opposite direction as upward motion), we take the negative square root of -11900:
v ≈ -sqrt(11900) ≈ -109.108 (approximated to 3 decimal places).
Therefore, the speed of the stone when it hits the ground is approximately 109.108 feet per second (rounded to 3 decimal places).
To find the volume of the solid obtained by rotating the first-quadrant region bounded by y = 0.5x^2 and y = x about the line x = 5, we can use the method of cylindrical shells.
First, let's sketch the region bounded by the given curves in the first quadrant of the xy-plane:
```
| y = x
| /
| /
|-----/-----
| /
| /
| /
|/__________
x = 5
```
To determine the interval of integration, we need to find the x-coordinate of the point where the curves y = 0.5x^2 and y = x intersect. Set the two equations equal to each other:
0.5x^2 = x
Simplifying, we have:
0.5x^2 - x = 0
Factoring out an x, we obtain:
x(0.5x - 1) = 0
So, either x = 0 or 0.5x - 1 = 0.
From the second equation, we find:
0.5x = 1
x = 2
Therefore, the region of interest is bounded by x = 0 and x = 2.
Now, let's derive the formula for the volume of the solid using the cylindrical shell method. Consider a horizontal strip of thickness ∆x located at x within the region of interest. The height of this strip is given by the difference between the two curves:
h(x) = x - 0.5x^2
The circumference of the cylindrical shell at x is given by the distance around the line of rotation, which is 2π times the distance from x to the axis of rotation (x = 5):
C(x) = 2π(5 - x)
The volume of each cylindrical shell is then given by:
∆V = C(x) * h(x) * ∆x
To find the total volume, we need to sum up the volumes of all the cylindrical shells. This requires integrating the expression for ∆V from x = 0 to x = 2:
V = ∫[0,2] C(x) * h(x) dx
Substituting the expressions for C(x) and h(x), we have:
V = ∫[0,2] 2π(5 - x)(x - 0.5x^2) dx
To approximate the area of the region bounded by y = -x^3 and y = -x using the mid-point rule with n = 4, we can divide the interval [a, b] into n subintervals of equal width ∆x = (b - a) / n. In this case, the interval is not explicitly given, so let's find it first.
To determine the interval, we need to find the x-coordinate of the points where the curves y = -x^3 and y = -x intersect. Set the two equations equal to each other:
-x^3 = -x
Dividing by -x, we have:
x^2 = 1
Taking the square root of both sides and considering both positive and negative solutions, we find:
x = 1 or x = -1
Therefore, the region of interest is bounded by x = -1 and x = 1.
Using the mid-point rule, the approximation for the area can be computed as:
A ≈ ∆x * [f(x1) + f(x2) + ... + f(xn)]
For n = 4, we have 4 subintervals, meaning n+1 = 5 equally spaced points within the interval [-1, 1]. Let's label these points x0, x1, x2, x3, x4.
∆x = (1 - (-1)) / 4 = 2/4 = 0.5
The midpoints of the subintervals are given by:
x1 = -1 + ∆x/2
x2 = x1 + ∆x
x3 = x2 + ∆x
x4 = x3 + ∆x
Once we have the values of x1, x2, x3, and x4, we can evaluate the function f(x) = -x^3 or f(x) = -x at these points and compute the approximation for the area using the formula.
To find the speed of the stone when it hits the ground, we need to determine the time it takes for the stone to reach the ground. We can use the kinematic equation:
y = y0 + v0t + 0.5at^2
where y is the position (y = 0 at the ground), y0 is the initial position (200 feet above the ground), v0 is the initial velocity (30 ft/s), a is the acceleration due to gravity (-32 ft/s^2), and t is the time.
Setting y = 0 and solving for t, we have:
0 = 200 + 30t - 16t^2
Rearranging, we get a quadratic equation:
16t^2 - 30t - 200 = 0
We can solve this equation using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our case, a = 16, b = -30, and c = -200. Plugging in these values into the quadratic formula will give us the time it takes for the stone to hit the ground.
Once we have the time, we can evaluate the velocity of the stone at that time using the formula:
v = v0 + at
Plugging in the values of v0, a, and t will give us the speed of the stone when it hits the ground.