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Question 6

Page history last edited by Rence 11 years, 6 months ago

A Calculator, or Paul, MAY be used to help answer this question.

Be certain to limit yourself to the four permissible calculator functions.

 

Paul runs to catch the bus along a line in such a way that at time t, 1 ≤ t ≤ 8, his position is given by:

 

(a) Write a formula for the velocity of Paul at time t.

 

(b) At what instant does Paul reach his maximum speed.

 

(c) When is Paul moving to the left?

 

(d) Find the total distance traveled by Paul from t = 1 to t = 8.

 

Quick Find Code For This Problem.

 

Legend: [QUES6LGND]

Solution For Part a.): [QUES6PA]

Solution For Part b.): [QUES6PB]

Solution For Part c.): [QUES6PC]

Solution For Part d.): [QUES6PD]

 

How To Use Quick Find.

 

-To Use Quick Find, copy (Ctrl+C/Command+C) the code (All the stuff INSIDE the square brackets in the code region above) for which ever section you'd like to go to, then, hit in this order;

 

Ctrl+F

Ctrl+V

Enter/Return

 

or if your on a Mac You'd hit

 

Command+F

Command+V

Enter/Return.

 

Then, your webbrowser should bring you directly to the beginning of the solution indicated by the code, making navigation through longish solutions much easier. Enjoy :]

 

Solution

[QUES6LGND]

Legend

Blue: Side Notes or "Paul says"

Green: Terms and Definitions

Red: Final Answers

 

The four permissible (TI-83) calculator functions that we're allowed to use to solve these wiki questions:

  1. Graphing the function: [Y=] [GRAPH]
  2. Determining the roots or y-intercepts: [2nd] [CALC] [2:zero]
  3. Numerically differentiating a function: [2nd] [6:dy/dx] or [MATH] [8:nDeriv]
  4. Numerically integrating a function: [2nd] [7:∫f(x)dx] or [MATH] [9:fnInt]

 


 

[QUES6PA]

PART A

 

Differentiating a position function results in a velocity function. Symbolically, v(t) = s’(t).

 

In this case, we're differentiating an antiderivative. (Paul says: Note the "∫".) Differentiation and antidifferentiation can be seen as inverse processes of each other: The derivative of x2 is 2x; an antiderivative of 2x is x2.

 

Not only are we differentiating an antiderivative, we’re differentiating an accumulation function, a function that measures the accumulating area under a graph.

 

The derivative of an accumulation function is the original function, by The Second Fundamental Theorem of Calculus.

 

 

 

We can see there is a function within a function in s(t). (Paul says: Note there are two variables, t and x.)

 

To differentiate a function within a function, we use The Chain Rule.

 

 

The velocity function or the derivative of the position function is:

 

 


 

[QUES6PB]

 

PART B

 

To determine Paul’s maximum speed, we first need to determine the extrema of v(t).

 

Velocity is speed with direction. Speed doesn't have direction. So, the absolute value of velocity is speed--all values of v(t) must be positive. To have a negative velocity means Paul is going in a direction opposite to what we designated positive. By convention, right is designated positive and left is designated negative.

 

We’re limited to using those four permissible calculator functions, so we can’t use the calculator’s maximum function. We’ll have to use a different approach: finding the roots of v'(t)!

 

 

To differentiate products, we use The Product Rule:

 

 

 

 

Paul says: Wherever v’(t) = 0 (the roots/the zeroes/the y-intercepts), we know v(t) has a local extrema (minimum or maximum)!

 

Since v(t) is a transcendental function, a function that contains an exponential function and a trigonometric function, we can’t apply the algebra we know to solve for the roots of v’(t), so we have to use our calculator.

 

Using the calculator, v’(t) = 0 at t = 1.2340473, 3.7311818, and 6.7002108. This can be seen on the graph of v'(t).

 

The graph of v'(t).

Note: Because the graphing program used has its limitations, the domain from 0 < t < 9 is shown. The domain of v(t), as indicated in the problem, is 1 ≤ t ≤ 8.

 

But, don’t forget all this happens in a closed interval, and, by The Extreme Value Theorem, the endpoints of a function are also considered as local extrema.

 

Paul says: The endpoints are at t = 1 and t = 8, since the domain is 1 ≤ t ≤ 8!

 

  • v(1.2340473) = 0.39375796
  • v(3.7311818) = 4.833368255
  • v(6.7002108) = -5.89643534
  • v(1) = 0.45969769
  • v(8) = 0.10668763

 

Therefore, at t = 6.7002108, Paul is at his maximum speed, which is 5.89643534.

 

Paul says: t = 6.7002108 is a minimum on the graph of v(t). But, in the context of the problem, the question asked is to determine the maximum speed. A negative velocity indicates running in the direction opposite to what we designated as positive. So, 5.90 m/s [left] = -5.90 m/s!

 


 

[QUES6PC]

PART C

 

When the v(t) is negative, Paul is moving to the left. So, we need to know where the roots of v(t) are because the roots indicate a change in sign, positive to negative or negative to positive.

 

Using the calculator function to determine roots on the graph, we find v(t) = 0 at t = 5.2035697 and t = 7.987021 in 1 ≤ t ≤ 8.

 

The graph of v(t).

Note: Because the graphing program used has its limitations, the domain from 0 < t < 9 is shown. The domain of v(t), as indicated in the problem, is 1 ≤ t ≤ 8.

 

Looking at the graph, we can deduce that v(t) is negative at 5.2035697 < t < 7.987021.

 

Or, instead of using a graph, we could use a line analysis.

 

 

  • By plugging a number to the left of 5.2035697, we obtain a positive number.
  • By plugging a number to the right of 5.2035697 or to the left of 7.987021, we obtain a negative number.
  • By plugging a number to the right of 7.987021, we obtain a positive number.

 

We see that v(t) is negative (Paul is moving to the left) at 5.2035697 < t < 7.987021.

 

 


 

[QUES6PD]

PART D

 

Integrate, from t = 1 to t = 8, the absolute value of the velocity function to get distance.

 

The graph of v(t) integrated from t = 1 to t = 8.

Note: Because the graphing program used has its limitations, the domain from 0 < t < 9 is shown. The domain of v(t), as indicated in the problem, is 1 ≤ t ≤ 8.

 

Paul says: If we don’t find the absolute value, the positive areas and the negative areas will cancel each other, and our final answer would be the displacement!

 

Distance can be defined as how much ground Paul walked, ignoring his direction.

 

Displacement can be defined as the change in position, where his direction is indicated.

 

Using the function numerical integrator (fnInt) in our calculator, the integral of the absolute value of the velocity function at 1 ≤ t ≤ 8 is 21.46124296.

 

Paul says: NEVER PUT THIS ON THE AP EXAM!

 

Paul says: PUT THIS ON THE EXAM INSTEAD!

 

Therefore, Paul ran 21.46124296 units in total.

 

 

Comments (10)

Benchmen said

at 6:57 am on Apr 14, 2009

What is this question the same as Question 4?

Darren Kuropatwa said

at 8:27 am on Apr 14, 2009

Oops ... Thanks for letting me know. I'll fix that.

Kristina said

at 5:11 pm on Apr 14, 2009

I lol'd. Nice one Mr. K, reminds me of that DEV question me and Joyce did last year that was based on Paul as well XD

Benchmen said

at 5:51 pm on Apr 14, 2009

Nice Question Mr K

Darren Kuropatwa said

at 6:59 pm on Apr 14, 2009

Glad you all like it. I hope Paul does too. ;-)

zeph said

at 7:50 pm on Apr 14, 2009

This is a test to see my comment works. :)

zeph said

at 8:11 pm on Apr 14, 2009

test

zeph said

at 8:25 pm on Apr 14, 2009

Unfortunately, I can’t edit, but I can comment. This is my ROUGH DRAFT as of yet.

PART A

This is an accumulation function. Differentiating this accumulation function would give us the argument of the accumulation function by The Fundamental Theorem of Calculus (since derivatives and antiderivatives are like inverses of each other). Algebraically speaking, F’(x) = f(x), where F(x) is the accumulation function.

v(t) = 1 – tcost – lntsint

PART B

Since we’re not allowed to use our calculators to determine the maximum velocity in the interval 1 ≤ t ≤ 8, we’ll have to use a differentiate approach. We’ll use The First Derivative Test!

By using The First Derivative Test, we can determine the parent function’s roots, then use a line analysis to determine where the parent function is increasing or decreasing. Remember, wherever the first derivative is negative or positive, the parent function is decreasing or increasing, respectively, which means that the roots of the first derivative are local extrema.

We’re allowed to use our calculator to determine roots. So, using the calculator, v’ = 0 at 1.2340473 and 3.7311818. Using a line analysis (that I’ll insert later after the EDIT button is fixed), we find that 1.2340473 is a local minimum and 3.7311818 is a local maximum.

But, don’t forget that this is a closed interval and, by The Extreme Value Theorem, the endpoints of a function are considered local extrema. However, v(1) = 0.45969769 and v(8) = 0.10668763.

Therefore, at t = 3.7311818, Paul is at his maximum velocity.

PART C

When the velocity function is negative, Paul is moving to the left. Graph v, determine the roots, and then deduce on what interval the velocity is negative. I’ll elaborate further when I have an image loaded up.

Paul is moving to the left at 5.2035697 < t < 7.987021.


zeph said

at 8:25 pm on Apr 14, 2009

PART D

Integrate the absolute value of the velocity function to get distance. We’re finding the absolute value of the function because, if we don’t find the absolute value, we’ll end up finding displacement. Using the fnInt function in our calculator, the absolute value of the velocity function at 1 ≤ t ≤ 8 is 21.46124296.

Fernando Noveron said

at 2:34 am on Dec 2, 2009

Thnks

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