Calc Exam 3

The acceleration of the bowling ball is given by a (t) = −32. Let h be the height of the window above the ground. Integrating a (t) gives us the function describing the ball's velocity, v (t). Since the ball is dropped and not thrown, v (0) = 0. v (t) = −32 (0) + C = 0 C = 0, so that v (t) = −32t. Now we can find the function p (t) describing the height of the ball above the ground. We know that p (0) = h, so p (0) = −16 (0)2 + C = h C = h. Then p (t) = −16t 2 + h. Let tg be the time the bowling ball hits the ground. Since it hits the ground with a velocity of −128 ft / sec, we have v (tg ) = −32tg = −128 tg = 4. We also know that at t = tg, p (t) is equal to zero (the ball hits the ground). p (tg ) = −16 (tg )2 + h = 0 −16 (4)2 + h = 0 h = 256 feet.

A bowling ball dropped out of a window hits the ground with a velocity of −128 ft / sec. How high above the ground is the window? Assume that the acceleration of the bowling ball due to gravity is −32 ft / sec2.

Taking the antiderivative of the acceleration function will give you the velocity function. v (t) = 6t 2 + C Plugging in the initial condition (velocity at t = 0 is zero) gives you the correct velocity equation. v (t) = 6t 2 Taking the antiderivative of this equation will give you the position function. p (t) = 2t 3 + C Plugging in the initial conditions will solve for the specific solution. p (t) = 2t 3 Setting the solution equal to 1200 will determine the time when the dragster completes the race. A quick calculation will reveal that time to be t = 8.43 sec. Since the question asks for the velocity at this time, just plug this value into the velocity equation and solve. The answer you get from the velocity equation is 426.4 ft / sec2.

A dragster on its way down a 1200 ft course has an acceleration function a (t) = 12t, where t is the time in seconds and a (t) is measured in ft / sec2. If the dragster starts the race from a standstill at the beginning of the course at t = 0, how fast is the dragster going when it crosses the finish line?

The velocity function v (t) is the antiderivative of the acceleration function. Since the rocket is moving upwards at 3 meters / sec at t = 1, v (1) = 3. We use this fact to solve for C. 8 (1)2 − 2 (1)3 + C = 3 8 − 2 + C = 3 C = −3. The rocket's velocity function is v (t) = 8t 2 − 2t 3 − 3

A model rocket blasts off and experiences an acceleration described by the function a (t) = 16t − 6t 2, where a (t) is in meters / sec2. Find the function which describes the velocity of the rocket if it is moving upwards at 3 meters / sec at t = 1.

The acceleration of the penny is described by a (t) = −32. We can integrate a (t) to get the velocity function for the penny. Since the penny is thrown downwards with an initial velocity of 40 ft / sec, v (0) = −40, so v (0) = −32 (0) + C = −40 C = −40. v (t) = −32t − 40. Integrating again gives us the distance the penny is above the ground as a function of time. We can use the fact that p (0) = 200 to solve for C: p (0) = −16 (0)2 − 40 (0) + C = 200 C = 200. Thus p (t) = −16t 2 − 40t + 200. Since the penny hits the ground when p (t) = 0, we solve p (t) = 0 for t: p (t) = −16t 2 − 40t + 200 = 0 t 2 + 5 2 t − 25 2 = 0. Using the quadratic formula results in t = 5 2 or t = −5. We discard the negative time since it makes no sense in the context of the problem. The penny hits the ground when t = 5/2 2.5 s

A penny is thrown downward from a tower that is 200 feet above ground with an initial velocity of 40 ft / sec. The acceleration of gravity is −32 ft / sec2 . When does the penny hit the ground?

−16t 2 + 400

A stone is dropped from a 400 foot high sea cliff at time t = 0. The acceleration due to gravity is given by a (t) = −32 ft / sec2 Find a formula for the distance the stone is above the water at any time t.

Start by finding the position function. Now solve p (t) = 0. The negative answer is not allowed in the context of this problem, so the stone hits the water in 5 seconds.

A stone is dropped from a 400 foot high sea cliff at time t = 0. The acceleration due to gravity is given by a (t) = −32 ft / sec2. How many seconds after dropping the stone will it hit the water?

The basic formula for vertical motion is y(t)=-16t^2+v0t+y0. The initial velocity is zero since it was dropped and not thrown. y0 = 576 feet

A student drops a penny off the top of a building. If the penny hits the ground 6 seconds later, how tall is the building?

The jet will cover the most distance if it accelerates at its constant rate of acceleration up to its maximum speed and then cruises at that speed for the rest of the trip. Since the jet goes through two distinct stages, this is really a two part problem. Stage 1) The jet is accelerating at its constant rate of 150 feet / sec2. Stage 2) The jet is cruising at its maximum velocity of 1200 feet / sec. The correct answer is: 19,200 feet

A supersonic jet accelerates at a constant rate of 150 feet / sec2 until it reaches its maximum velocity of 1200 feet / sec. What is the maximum distance the jet can travel in 20 seconds if it starts from rest?

The width of each rectangle if 1 unit. The sum of all areas is found by multiplying the heights by the width and adding them all together.

Approximate the area under the curve y = −x 2 + 16 between 0 and 4 by finding the area of the circumscribed rectangles using 4 subdivisions.

To solve for the area you will need to find the limits of integration. The left point of integration occurs at the y‑axis, or when x = 0. The right point of integration is at the x-value that satisfies the equation e x = 4. Solving for x results in x = ln 4

Consider the function y = e x. What is the area bound between this curve and the line e x = 4 in the first quadrant?

Remember that the derivative of cos x is − sin x. So the derivative of − cos x is sin x. Remember that integrating is the opposite of differentiating. Therefore the integral of sine is negative cosine. -cosx+C

Evaluate the integral ∫sinxdx

2x^2-2/x+C

Evaluate ∫(4x^3+2)/(x^2) dx

∫dx= ∫1dx=(x/1)+C=x+C

Evaluate ∫dx

You need to find an equation that when you take the derivative of it gives you zero. Remember that the derivative of a constant is zero, so the antiderivative of zero could be any constant. The correct answer is: F (x) = π

Find an antiderivative for f (x) = 0.

The derivative of tan x is sec2 x, so the derivative of tan 2x is (sec2 2x) ⋅ 2 = 2 sec2 2x. Let's multiply tan 2x by 1/2 to cancel out the 2. The derivative of 1/2 tan 2x is 1/2 sec^2 2x ⋅ d /dx (2x)=1/2 sec^2 2x ⋅ 2 = sec^2 2x. 1/2 tan 2x

Find an antiderivative of the function f (x) = sec2 2x.

Think about the power rule for differentiation in reverse. We know that the antiderivative of x 2 will be x 3 multiplied by some constant. The derivative of x 3 is 3x 2, so consider (1/3) x 3.

Find an antiderivative of the function f (x) = x 2

sketch the curves, the region fails the vertical line test but passes the horizontal line test. the curve is y-easy. set up the integral with respect to y

Find the area of the region bound between the curves x =1−y 4 and y =x/5+1.

By taking the limit as Δx approaches zero, you can integrate the height of the shape along the width.

How is the concept of the limit important in finding the area of exotic shapes?

As long as there is an x-term in the numerator, the limit will produce an indeterminate form, the number of derivatives it takes to remove all the x-terms from a polynomial is equal to the power of the x-term to remove the power b from the exponent, you would need to take the derivative b times

How many times is L'Hopital's rule used to solve for lim x-> infinity x^b/e^x, where b is a positive integer?

The region passes the vertical line test and fails the horizontal line test, so the area is x‑easy but not y‑easy. The region fails the vertical line test and fails the horizontal line test, so the area is neither y‑easy nor x‑easy.

Is the area bound by these curves x‑easy, y‑easy, both, or neither?

Try drawing a vertical line that passes through the same curve more than once. You can't do it. Now try drawing a horizontal line that passes through the same curve more than once. You can't do it. The region passes both the vertical line test and the horizontal line test, so the area is x‑easy and y‑easy.

Is the area bound by these curves x‑easy, y‑easy, both, or neither?

Let a1 (t), v1 (t), and p1 (t) be the functions describing the acceleration, velocity, and position of the first ball, and let a2 (t), v2 (t), and p2 (t) be the corresponding functions for the second ball. Then Since the first ball is thrown up with a velocity of 50 feet / sec at t = 0, v1 (0) = 50 so that v1 (0) = −32 (0) + C = 50 C = 50, or v1 (t) = −32t + 50. Now The first ball is thrown up at t = 0, so p1 (0) = 0: p1 (0) = 16 (0)2 + 50 (0) + C = 0 C = 0. We can conclude that p1 (t) = −16t 2 + 50t. Now for the second ball. Since the second ball is thrown up with a velocity of 20 feet / sec at t = 1, v2 (1) = 20 so that v2 (1) = −32 (1) + C = 20 C = 52, or v2 (t) = −32t + 52. Now The second ball is thrown up at t = 1, so p2 (1) = 0: p2 (1) = −16 (1)2 + 52 (1) + C = 0 C = −36. We can conclude that p2 (t) = −16t 2 + 52t − 36. So do the two balls cross? If they ever cross, their positions will be the same at the same instant, so we solve the equation p1 (t) = p2 (t) −16t 2 + 50t = −16t 2 + 52t − 36 2t = 36 t = 18 But at t = 18, the positions of the balls are p1 (18) = −4284 feet. The balls have hit the ground long before t = 18, so the two balls do not cross.

John throws a baseball upwards at t = 0 with an initial velocity of 50 feet / sec. One second later, he throws another baseball upwards with an initial velocity of 20 feet / sec. Do the two balls ever cross each other before they hit the ground? Assume that the effect of gravity gives a constant acceleration of −32 feet / sec2 to each ball.

Let's take the derivative of F (x) + 5: d /dx [ F (x) + 5] = d /dx [ F (x)] + d/dx [5]. F (x) is an antiderivative of f (x), so the derivative of F (x) is f (x). d/dx [ F (x) + 5] = f (x) + d/dx [5] = f (x) + 0 = f (x). We can conclude that F (x) + 5 is also an antiderivative of f (x).

Suppose you are told that an antiderivative of the function f (x) is F (x). Which of the following is also an antiderivative of f (x)?

Let t = 0 be the instant that the ship's engines fail. We can use a (t) to find the functions describing the ship's position and velocity. Since the ship is moving downwards at a velocity of −30 feet / sec the instant the engines fail, we have v (0) = −30: v (0) = −10 (0) + C = −30 C = −30. Thus v (t) = −10t − 30. Integrating again gives us the function describing the position of the ship. At t = 0 the ship is 80 feet above the Martian surface, so p (0) = 80: p (0) = −5 (0)2 − 30 (0) + C = 80 C = 80 Therefore p (t) = −5t 2 − 30t + 80. We can find the time that the ship hits the ground by solving the equation p (t) = 0: −5t 2 − 30t + 80 = 0 t 2 + 6t − 16 = 0 (t + 8) (t − 2) = 0 t = −8 or t = 2. The negative answer doesn't make sense in the context of the problem, so we throw it out. We can conclude that the ship hits the ground at t = 2. At this instant, the ship is moving at a velocity of v (2) = −10 (2) − 30 = −50 feet / sec. The ship doesn't survive!

The Martian lander "Land-o-matic" is designed to withstand a maximum impact velocity of 45 feet / sec. On one particular mission to Mars, the Land-o-matic is descending towards the Martian surface at a constant rate of −30 feet / sec when its engines abruptly fail. If the ship is 80 feet from the surface of the Mars at this instant, will it survive the impact when it hits the ground? Assume that Martian gravity gives the ship an acceleration of a (t) = −10 feet / sec2 .

Let's suppose that a stone is dropped from a height of 10 feet at t = 0 under the force of Dr. Experimento's new gravity. Then Since the stone is dropped at t = 0, and not thrown, v (0) = 0 so that v (0) = −G (0) + C = 0 C = 0. Integrating again gives us the function describing the position of the stone. The stone starts out at a height of 10 feet, so p (0) = 10: p (0) = − G 2 (0)2 + C = 10 C = 10. Therefore p (t) = − G 2 t 2 + 10. We want it to take 1 minute, or 60 seconds, for the stone to hit the ground. In other words, we want the G that satisfies p (60) = 0: p (60) = − G 2 (60)2 + 10 = 0. Solving this equation for G results in G ≈ .0056 feet / sec2.

The mad scientist Dr. Experimento has decided to change the strength of gravity so that it will take 1 minute for a stone dropped from rest to fall 10 feet. Assume that the force of gravity gives the stone a constant acceleration a (t) = −G. Find the value of G in feet / sec2 .

t = 3.5 or t = .36.

The position of a helicopter moving only in the vertical direction is given by p (t) = 4t 3 − 23t 2 + 15t. At what times is the helicopter motionless?

Since u-substitution undoes the chain rule, before you can solve a problem using this technique you have to find the pieces of the integral that the chain rule produced. These pieces are the derivative of other pieces. The integral must be made up of an expression and its derivative.

To determine if an integral is a good candidate for u-substitution:

Let this constant acceleration be A. Then the acceleration of the baseball is given by a (t) = A. We can integrate this to get a function describing the baseball's velocity. v(t)=a(t)dt=Adt =At+C Let's suppose that t = 0 corresponds to when the baseball is moving at 6 feet / sec. Then A (0) + C = 6, or C = 6. Thus v (t) = At + 6. What else do we know? We also know that in 2 seconds, the baseball is moving at 116 feet / sec. Plug in t = 2 and solve for A: v (2) = A (2) + 6 = 116 2A = 110 A = 55 feet / sec2.

What constant acceleration is needed to accelerate a baseball from 6 feet / sec to 116 feet / sec in 2 seconds?

From a graph of the functions, you can see that between x = 1 and x = 2, the function f (x) = 5 sin x is on top and the function g (x) = e x − 3 is on the bottom. The shaded area is therefore given by the integral

What is the area between the graphs of the functions f (x) = 5 sin x and g (x) = e x − 3, between x = 1 and x = 2?

The area of the given region is bound to the left by x = 0 and to the right by x = 3. So these two values will be the limits of integration. Since you are looking for an area bound between a curve and the x‑axis, then you just need to make sure that the region is above the axis. Since it is, just integrate the curve along the limits of integration. A=9

What is the area of the region bound between the curve y = x 2, the line x = 3, and the x‑axis?

if you let u=x^2-4, then du=2xdx. the integral becomes ∫u^6du, which can be solved

What is the best choice for a u-substitution for the integral ∫2x(x^2-4)^6 dx

since d(cot x)= -csc^2xdx, the substitution of u=cotx would give you -∫e^udu, which is easier to solve

What is the best choice for a u-substitution for the integral ∫e^cotx(csc^2(xdx))

Since d (sin 2x) = 2 cos 2x dx, then the only factor you don't have in the integral would be a constant. Constants can be multiplied into the expression, so this is a good choice for u.

What is the best choice for a u-substitution for the integral ∫sin^3(2xcos2xdx)

Since dx is supposed to relate to a small change in the x-direction but points at a height (or a y-value), then those two pieces of the graph must be mislabeled. Switching the dx and the f (x) would create a correctly labeled graph

What is wrong with the labeling of this area problem?

The terms along the x‑axis should all be constants or the dx width. Terms along the y‑axis should all be the function evaluated at those constants. So the b and the f (b) are switched.

What is wrong with the labeling of this area problem?

Since all of your equations and expressions are in terms of the y-variable, it logically follows that your boundaries should be in terms of y-values as well. So when you integrate in the y-direction, use the y-value of the boundary point instead of the x-value!

When integrating with respect to y, what is different about the boundary points (the boundaries of integration)?

Increasing the number of partitions or rectangles generally generates a smaller error in the approximation.

When using rectangular partitions to approximate the area under a curve, how does increasing the number of partitions (or rectangles) used to approximate an area generally affect the approximation?

Integrating with respect to y instead of x is like setting the graph on its side and looking at it from a new perspective. This technique lets us more easily evaluate integrals where the regions are better defined by a side-minus-side equation than a top-minus-bottom one. Integrating with respect to y is like integrating with respect to x, but instead of the internal rectangles being vertical, they are horizontal.

Which of the following best explains the concept of integrating with respect to y ?

To solve this expression, you will need to use a technique called integration by parts. There is no simple u-substitution that will work. The correct answer is: u = x 2

Which of the following expressions creates a working u-substitution that solves the following integral? ∫(x^3)sinx^2(dx)

∫xsin xdx, since there is no candidate for us that has its derivative as part of the integral. use integration by parts to solve

Which of the following integrals is good candidate for integration by substitution?

Anything that divides by zero, zero over zero, or infinity over infinity is indeterminate

Which of the following limits does not produce an indeterminate form?

Integration can be graphically interpreted as the summation of an infinite number of rectangles with infinitely small widths. Since you can change the height of the rectangles when you integrate, integration proves to be a valuable tool for finding the areas of exotic shapes. The correct answer is: Integrating the height of the function with respect to the width sums up all the tiny areas created by dividing the area into an infinite number of rectangles.

Why does the definite integral of f (x) dx evaluated from a to b equal the area bound between a curve and the x‑axis?

We use a (t) to find the velocity function, v (t), and the position function, p (t), for the ball. In this case, the position function p (t) measures the height of the ball above the ground. Since the ball was thrown upwards with an initial velocity of 50 ft / sec, v (0) = 50: v (0) = −32 (0) + C = 50 C = 50. Therefore v (t) = −32t + 50. Let's integrate again. Since the ball is thrown from the ground at t = 0, we have p (0) = 0: p (0) = −16 (0)2 + 50 (0) + C = 0 C = 0. Therefore p (t) = −16t 2 + 50t. To find the time that the ball hits the ground, we solve the equation p (t) = 0. The correct answer is: −50 ft / sec

a drag vertically upwards from the ground with an initial velocity of 50 ft / sec. The ball is accelerated by gravity so that its acceleration is given by the function a (t) = −32 ft / sec2 . At what velocity does the ball hit the ground?