## Problem 3: Position and Velocity

2011-2012 An object is dropped from the second-highest floor of the Sears Tower, 1542 feet off of the ground. (The top floor was unavailable, occupied by crews taping for the new ABC special “Behind the Final Behind the Rose Final Special, the Most Dramatic Behind the Special Behind the Rose Ever.”) (a) Construct the position … Continue reading "Problem 3: Position and Velocity"

An object is dropped from the second-highest floor of the Sears Tower, 1542 feet off of the ground.

(a) Construct the position and velocity equations for the object in terms of t, where t represents the number of seconds that have elapsed since the object was released.

(b) Calculate the average velocity of the object over the interval t = 2 and t = 3 seconds.

(c) Compute the velocity of the object 1, 2, and 3 seconds after it is released.

(d) How many seconds does it take the object to hit the ground? Report your answer accurate to the thousandths place.

(e) If the object were to hit a six-foot-tall man squarely on the top of the head as he (unluckily) passed beneath, how fast would the object be moving at the moment of impact? Report your answer accurate to the thousandths place.

### Solution:

(a) The position function for a projectile is s(t) = –16t2 + v0t + h0, where v0  represents the initial velocity of the object (in this case 0) and h0 represents the initial height of the object (in this case 1,542 feet). Note that this position equation represents the height in feet of the object t seconds after it is released. Thus, the position equation is s(t) = –16t2 + 1,542. The vecocity equation v(t) is the derivative of the position equation: v(t) = –32t.

## How to Find Average Velocity

(b) Average velocity is the slope of the secant line, rather than the slope of the tangent line. Finding average velocity is easy. Plug t = 2 and t = 3 into the position equation to calculate the height of the object at the boundaries of the indicated interval to generate two ordered pair: (2, 1478) and (3, 1398). Apply the slope formula from basic algebra to calculate the slope of the line passing through those points.

(c) Substitute t = 1, 2, and 3 into v(t).

(d) The object hits the ground when its position is s(t) = 0. Set the position equation equal to zero and solve for t.

(e) The problem asks you to calculate the velocity of the object when it is exactly six feet off of the ground, when s(t) = 6. Apply the same technique you completed in part (d), but instead of calculating the time t when the object’s position is 0, calculate the time t when its position is 6.

Now calculate the velocity of the object at that time: v(9.79795897113) = –32(9.79795897113) = –313.535 ft/sec.

## Problem 14: Power Rule for Integration

Looking to figure out how to sole the power rule for integration for your calculus homework? We have the answers at calculus-help.com.

2019-2020

Find the antiderivatives of each with respect to x, without using u-substitution (i.e., without changing variables):

(a)

(b) , assuming a, b, c, d, and e are constants

## Integral Power Rule Solution:

(a) For the integral power rule solution, start by expanding the numerator, squaring (x – 1) and mutliplying by (2x + 1). Rewrite the denominator as a term with a negative exponent, and then distribute it through the quantity.

Now you can antidifferentiate by adding 1 to each power individually and multiplying the coefficient by the reciprocal of the resulting exponent. In other words, for the first term, add 1 to 3/2 to get 5/2. Then, multiply 2 by the reciprocal of 5/2, which is 2/5.

Don’t forget that a general antiderivative (i.e., an indefinite integral) must always contain “+ C.”

(b) Follow the same procedure, beginning by writing the radical term with a fractional exponent. Then,  distribute and find the antiderivtaive. The variables make it marginally more difficult, especially when you add 1 to the weird variable powers.

To apply the power rule for integration to this type of function, you have to remember an important rule from algebra.

If we can write the function using exponents then we most likely can apply the power rule.

Let’s solve this problem:

∫ √x+4 dx

Before even using any calculus, you can rewrite the function using the above rule with exponents. So, you have exponents and can apply the power  rule.

It becomes:

∫√x + 4 dx = ∫x^1/2 +4 dx

Then, you can apply the power rule.

It then becomes:

x^1/2 +4 dx= (x^(1/2+1)) / ( 1/2 + 1) + 4x + C

When simplified, you get the final answer.

That is:

(x^3/2 / 3/2) + 4x + C

Which equals:

(2/3 * x^2/3) + 4x + C

The final answer can usually be written with exponents, like we did here. Or, it can be written with roots. However, your teacher or professor may have a preference, so make sure you ask!

## Finding the Power Rule for Integration

As with anything else related to calculus, perfecting the power rule for integration will take a lot of practice if it doesn’t come to naturally.

Make sure you look up practice problems online! You can browse our other calculus problems here. Best of luck!

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## 2019-2020 Derivative Table

Two functions, f(x) and g(x), are continuous and differentiable for all real numbers. Some values of the functions and their derivatives are given in the following table.

Based on that glorious table, calculate the following:

(a)

(b)

(c)

(d)

### Solution:

(a) Take the derivative of each function separately (the derivative of a sum is equal to sum of its derivatives) and plug in 4 to each to get your answer. Reference the chart for the values of f ‘(4) andg(4).

(b) This time you have to use the Product Rule, because f(x) and g(x) are multiplied. Once again, after you apply the derivative rule, just nab the needed function and derivative values from the chart.

(c) This time it’s the Quotient Rule that has to be applied.

(d) How about a big, warm welcome for the Chain Rule! Remember, you apply the Chain Rule when one function is composed with (inside of) another. To differentiate, take the derivative of the outer functionf(x) while leaving g(x) alone inside f(x). Then multiply by the derivative of g(x).

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## Problem 15: A Grizzly Motion Problem

2011-2012 The motion of a grizzly bear stalking its prey, walking left and right of a fixed point in feet per second, can be modeled by the motion of a particle moving left and right along the x-axis, according to the following acceleration equation: Assume that the origin corresponds to the fixed point, and that … Continue reading "Problem 15: A Grizzly Motion Problem"

2011-2012

The motion of a grizzly bear stalking its prey, walking left and right of a fixed point in feet per second, can be modeled by the motion of a particle moving left and right along the x-axis, according to the following acceleration equation:

Assume that the origin corresponds to the fixed point, and that a positive value for position means that the bear is located to the right of the fixed point as we watch said bear from a safe location.

If the bear’s velocity is 1 ft/sec when t = 0, answer the following questions:

(a) Identify the velocity equation that represents the bear’s motion.

(b) Determine how fast the bear was traveling at t = 7 seconds.

(c) In what direction is the bear traveling at t = 5 seconds?

(d) How far does the bear walk during the first 10 seconds?

Note: You can (and should) use a graphing calculator for part (d).

### Solution:

(a) You are given the acceleration equation. Recall that velocity is the antiderivative of acceleration, so integrate a(t) and use the fact that v(0) = 1  to identify the velocity equation.

(b) Evaluate v(7).

Speed is the absolute value of velocity, so the bear is traveling at a speed of 1.579 ft/sec when t = 7.

(c) Evaluate v(5).

Because v(5) is negative, the bear is traveling left at t = 5.

(d) You must split the interval [0,10] into segments based on the t-intercepts of v(t). Those values of t are the times at which the bear changes direction; you must measure how far the bear traveled forward and backward separately.

The velocity equation has only one t-intercept on [0,10]: t = 1.17012095. To calculate the total distance traveled, compute the area between v(t) and the t-axis on the intervals [0,1.17012095] and [1.17012095,10] independently. When the bear is traveling left, this integral is a negative value, but you are asked to find the total distance traveled, not the final position of the bear. Therefore, both integrals must be positive values, so take the absolute value of the second definite integral.

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## Problem 13: Polar Derivatives

2011-2012 Find all angles on the interval  at which the tangent line to the graph of the polar equation  is horizontal. Solution: Express the polar equations parametrically (in terms ofx and y) and calculate the slope of the polar equation. The tangent lines to the polar graph are horizontal when the numerator of this derivative is … Continue reading "Problem 13: Polar Derivatives"

## 2011-2012

Find all angles on the interval  at which the tangent line to the graph of the polar equation  is horizontal.

### Solution:

Express the polar equations parametrically (in terms ofx and y) and calculate the slope of the polar equation.

The tangent lines to the polar graph are horizontal when the numerator of this derivative is equal to 0. In other words, at .

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## Problem 12: Super Related Rates

2011-2012 Have you ever loved something so deeply, so meaningfully, so completely, so profoundly that it would really irk you if you dropped that thing into a bubbling vat of acid? I have, and so that you may learn from my tragedy, I will share a horrific tale from my past. Once, on a whim, … Continue reading "Problem 12: Super Related Rates"

2011-2012

Have you ever loved something so deeply, so meaningfully, so completely, so profoundly that it would really irk you if you dropped that thing into a bubbling vat of acid? I have, and so that you may learn from my tragedy, I will share a horrific tale from my past.

Once, on a whim, I spent an entire summer trying to carve a perfect cube from a piece of driftwood on the beach. Don’t ask why; this is what all math teachers do during summer break, and if teachers tell you otherwise, they are lying. Look at their hands carefully—they are probably whittling as they lie to you!

After months of hard work and risking my fish-belly white skin to the sun, I finally found the perfect piece of driftwood. In just a few days, up to my knees in whittled-away wood chips, I beheld the most beautiful hand-carved wooden cube, each edge precisely one inch long. To say it was a thing of beauty is an understatement. In fact, we became fast friends, and spent much time laughing together about the absurdity of life. (I was later to learn that I had a severe case of sun poisoning and may have hallucinated some of this.)

One day I took my little wooden pal to my favorite acid factory. These were the days of the go-go 80s, when you could just walk around giant open vats of hydrochloric acid and take photos, as long as you either wore a hard hat or styled your hair to look like a hard hat. Intent to give my wooden cube the best possible vantage point to see the massive, bubbling vats of acid, I held him precariously over the safety railings.

What happened next? I think you already know. The cube tumbled from my red, blistered fingers into the acid, and the cube vanished from sight. Moments later, it skyrocketed out of the acid, its edges expanding at an absurd 2 ft/sec. I watched with equal parts pride and horror as the cube also gained super crime-fighting powers (including, but not limited to, the power to fly, fill out tax forms correctly, and chew through solid rock). It even retained its perfect cubic shape as it grew to a monstrous size.

I lost consciousness almost instantly, but two moments are seared into my memory, when its still-growing form had an edge length of 12 ft and (later) when the edge length reached an incredible 210 ft.

(a) At what rate did the surface area of the cube change when its edges were 12 feet long? How about 210 feet long?

(b) At what rate did the volume of the cube change when its edges were 12 feet long? Once again, how about later, when its sides measured precisely 210 feet long?

(c) What was the name of the Super Villain whose life ambition was to fight against my beautiful, yet terrifying, cube?

### Solution:

(a) First, you need the equation for surface area of a cube, and you’ll take the derivative of it with respect to time. The surface area, s, of a cube is equal to 6e2, where e is the length of an edge. Basically, it’s a collection of six squares, each with an area of e2.

According to the problem, de/dt = 2 ft/sec, and you need to calculate ds/dt. Take the derivative of the surface area formula with respect to time.

Notice that I plugged in 2 for de/dt. Now you can calculate the two answers required in part (a) by plugging in each, separately, for e:

• When e = 12, ds/st = 24(12) = 288 ft2/sec.
• When e = 210, ds/dt = 5,040 ft2/sec.

Don’t forget units. It’s unnecessary to carry them all the way through, but at least insert them at the end. You use ft2/sec because area is expressed in square units (square feet, in this case, instead of feet).

(b) Apply a technique similar to part (a), using the formula for the volume V of a cube: V = e3.

Substitute e = 12 and e = 210 into this formula to get answers of 864 ft3/sec and 264,600 ft3/sec, respectively.

(c) The Super Villian’s name was Butter Fingers, but we will also accept Cameron, the Devourer of Worldsas a correct answer.

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## Problem 11: Chain Rule

2011-2012 Chain Rule Problems Calculate the derivative of tan2(2x –1) with respect tox using the chain rule, and then verify your answer using a second differentiation technique. Solution for Chain Rule Practice Problems: Note that tan2(2x –1) = [tan (2x – 1)]2. To find the solution for chain rule problems, complete these steps: Apply the power rule, … Continue reading "Problem 11: Chain Rule"

## 2011-2012 Chain Rule Problems

Calculate the derivative of tan2(2x –1) with respect tox using the chain rule, and then verify your answer using a second differentiation technique.

## Solution for Chain Rule Practice Problems:

Note that tan2(2x –1) = [tan (2x – 1)]2. To find the solution for chain rule problems, complete these steps:

1. Apply the power rule, changing the exponent of 2 into the coefficient of tan (2x – 1), and then subtracting 1 from the square.
2. Multiply by the expression tan (2x – 1), which was originally raised to the second power.
3. Take the derivative of tan (2x – 1) with respect to x.
4. Multiply by the derivative of 2x – 1, the expression that is plugged into tangent.

## Derivative of tan2x

These four steps are implemented in the solution below for derivative of tan2x.

To verify the derivative, apply the product rule, noting that  tan2(2x –1) = tan (2x –1) · tan (2x – 1). Evev the product rule will require the chain rule, when you differentiate each factor (2x – 1), as demonstrated below.

Both techniques for these chain rule practice problems result in the same derivative.

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## Problem 10: Diving into Rates of Change

2011-2012 For the second year in a row, one toy looks to dominate the market once again: My First Cliff Diving Kit. It is all the rage, because it comes with everything you need to be an effective cliff diver: swim trunks, neck brace, legal documents for naming next of kin, and very detailed one-paragraph … Continue reading "Problem 10: Diving into Rates of Change"

2011-2012

For the second year in a row, one toy looks to dominate the market once again: My First Cliff Diving Kit. It is all the rage, because it comes with everything you need to be an effective cliff diver: swim trunks, neck brace, legal documents for naming next of kin, and very detailed one-paragraph instruction sheet about how to cliff dive safely. Below, you’ll find a chart representing the sales figures for the each month it has been available. Use that chart to answer the questions that follow.

(a) Assuming that you can draw a continuous graph connecting all the data points, write the equation of line L, the tangent line to the sales graph when x = May 2011.

(b) What is the average rate of sales between November 2010 and September 2011?

(c) Draw a rough sketch of the graph representing the rate of sales between Oct 2010 and Oct 2011.

### Solution:

(a) To write the equation of a line, you need two things: a point and a slope. You’ll need to assign some x-values, because the independent variables are months and not numbers. Assign October 2010 an x-value of 0 and October 2011 an x-value of 12. This produces the graph below.

Therefore, the point representing May 2011 is the coordinate (7,1559). To estimate the tangent there, calculate the slope between that point and the point immediately preceding it, (6,1509).

Therefore, a good approximation for the tangent line would be y – 1559 = 50(x – 7), according to point-slope form. You could also approximate the derivative by using the point immediately after May 2011, and you’ll get a slope of 128 rather than 50. That seems like a huge difference, but since you don’t have any intermediate points, who’s to know which is the better approximation?

(b) Calculate the average rate of change using the slope of the secant line connecting the points: (1,634) and (11,2445).

Therefore, the kits are selling better overall on the entire time period when compared with how they were selling in May 2011, no matter which derivative approximation from part (a) you use.

(c) Use the same process as part (a) to approximate each derivative, and you’ll get something that looks roughly like this the graph below.

Notice that the graph dips below the x-axis on roughly the interval (2,4) because the data seems to be decreasing there. the faster the data points increase, the greater the height of the blue derivative graph.

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## Problem 9: The Graph of a Derivative

2011-2012 Derivative Graph In the below graph, two functions are pictured, f(x) and its derivative, but I can’t seem to tell which is which. According to those graphs, which is greater: Solution: You might have noticed that the red function has an even degree whereas the blue has an odd degree. Why? An even-degreed function’s ends will … Continue reading "Problem 9: The Graph of a Derivative"

## 2011-2012 Derivative Graph

In the below graph, two functions are pictured, f(x) and its derivative, but I can’t seem to tell which is which.

According to those graphs, which is greater:

### Solution:

You might have noticed that the red function has an even degree whereas the blue has an odd degree. Why? An even-degreed function’s ends will either both point up or both point down (here they both point up at the edges of the graph), whereas a function with an odd degree has ends that go in opposite directions (like the blue graph which goes down to the left but up to the right). Even though this is interesting, it is not enough to answer the question.

You may also say that the red graph is definitely a lesser degree because it has a fewer number of x-intercepts (4) than the blue graph (5), but this, too, is not sufficient information to answer the question correctly. It is true that the blue graph is f(x) and the red is f ‘(x), but for different reasons.

Here are two acceptable justifications for identifying the blue graph as f(x):

• Whenever the blue graph has a relative max or min (hilltop or valley bottom, respectively), the red graph has an x-intercept; remember that the derivative of a function is either 0 or undefined when the original function has a relative extrema point
• Whenever the blue graph is increasing, the red graph is positive (i.e., above the x-axis), and when the blue graph is decreasing, the red is negative

Now that we’ve got that straight, let’s get down to answering the question. You have no idea what the exact function values are, but it turns out that you don’t really need to know them! Since the red graph is decreasing at x = –1/4, then its derivative must be negative there (for the same reason that the second bullet above is true).

What about f (–1)? Because the red graph represents f ‘(x), it’s easy to see that the red graph is just above the x-axis when x = –1. Therefore, f ‘(–1) > 0.

Back to the original question, which asks you to compare two values. You now know that one of those values is positive and the other is negative, so the positive value must be greater.

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2011-2012 A man wants to build a rectangular enclosure for his herd. He only has $900 to spend on the fence and wants the largest size for his money. He plans to build the pen along the river on his property, so he does not have to put a fence on that side. The side … Continue reading "Problem 8: Optimizing a Dirt Farm" ## 2011-2012 A man wants to build a rectangular enclosure for his herd. He only has$900 to spend on the fence and wants the largest size for his money. He plans to build the pen along the river on his property, so he does not have to put a fence on that side. The side of the fence parallel to the fence will cost $5 per foot to build, whereas the sides perpendicular to the river will cost$3 per foot. What dimensions should he choose?

### Solution:

Start by drawing a picture of the situation.

According to the problem, the farmer is attempting to maximize the size of the rectangular plot. Therefore, the primary equation will be area: A = l ∙ w. However, to take the derivative (maximizing the function), you need to eliminate a variable, either l or w; for that, you’ll need a secondary equation.

You know he wants to spend $900. Furthermore, each foot of the 2 fences he’ll use for the width of the yard will cost$3. Each foot of the fence parallel to the river will cost \$5.

Solve the cost equation for one of its variables; I’ll solve for l.

Now plug this in for l in the primary area equation.

Now that you’ve got one variable, maximize by finding the derivative and setting it equal to 0.

To find the corresponding length of the optimized field, plug 75 in for w into the modified secondary equation.

So the optimized field has a fence parallel to the river measuring 90 feet and two other fences connecting it to the river that measure 75 feet each.

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