Can someone help me solve these two questions?
Thanks
1) Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = -x^3 and y = -x.
2) A rock is thrown upward with a speed of 48 feet per second from the edge of a cliff 400 feet above the ground. What is the speed of the rock when it hits the ground? Use acceleration due to gravity as -32 feet per second squared.
Of course! Let's tackle each question one by one.
1) To approximate the area of the region bounded by the curves y = -x^3 and y = -x using the midpoint rule, you'll need to follow these steps:
Step 1: Determine the interval over which you'll approximate the area. In this case, we don't have specific limits mentioned, so let's assume the range is from x = a to x = b.
Step 2: Divide the interval [a, b] into smaller subintervals. The number of subintervals, denoted as n, is given as 4 in this question. Each subinterval will have a width of (b - a) / n.
Step 3: Calculate the x-coordinates of the midpoints of each subinterval. To do this, you can start with a + (width / 2) for the first midpoint and continue to add the width until you reach b.
Step 4: Evaluate the corresponding y-coordinates of each midpoint by substituting the x-coordinate into the functions y = -x^3 and y = -x. This will give you the height (y-value) at each midpoint.
Step 5: Calculate the area of each rectangle by multiplying the width of the subinterval (which is the same for all rectangles) by the corresponding height (y-value) at each midpoint.
Step 6: Sum up the areas of all the rectangles to get an approximation of the total area.
2) To determine the speed of the rock when it hits the ground, we can apply the following steps:
Step 1: Identify the known quantities. We have:
- Initial velocity (u) = 48 feet per second (thrown upward)
- Acceleration due to gravity (a) = -32 feet per second squared (negative value indicates downward acceleration)
- Initial height (s) = 400 feet (height from which the rock is thrown)
Step 2: Use the kinematic equation to find the time it takes for the rock to hit the ground. The equation relating distance, initial velocity, acceleration, and time is:
s = ut + (1/2)at^2
Since the rock starts at an initial height of 400 feet and hits the ground (zero height), we can set s = 0 and solve for t.
0 = 48t + (1/2)(-32)t^2
Step 3: Solve the quadratic equation to find t. You can rearrange the equation to:
16t^2 - 48t = 0
Factor out 't' to get:
t(16t - 48) = 0
This equation has two solutions:
- t = 0 (discard this solution as it represents the time at which the rock is thrown)
- 16t - 48 = 0
16t = 48
t = 3
Step 4: Calculate the final velocity (v) when the rock hits the ground using the formula:
v = u + at
Substitute the given values:
v = 48 + (-32)(3)
Simplify:
v = 48 - 96
v = -48
The speed of the rock when it hits the ground is 48 feet per second in the downward direction.