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Question 6

This version was saved 14 years, 11 months ago View current version     Page history
Saved by zeph
on April 18, 2009 at 6:58:35 pm
 

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.

 

Solution

 

Legend

Blue: Side Notes or "Paul says"

Green: 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]

 


 

PART A

 

Velocity is defined as the change in position. Derivative is defined as the rate of change. Therefore, velocity is the derivative of position. Symbolically, v(t) = s’(t). We differentiate the position function that is given to determine the velocity function.

 

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.) The outer function, which we’ll call f(x), is , and the inner function, which we’ll call g(x), is .

 

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

 

 

The derivative of the position function is:

 

 


 

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.) 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)!

 

Acceleration is defined as the change in velocity. Derivative is defined as a rate of change. Acceleration is the derivative of velocity. Symbolically, a(t) = 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 velocity, 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 velocity. 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!

 


 

PART C

 

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 saw this in part B.

 

When the velocity function 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 and v(t) is negative at 5.2035697 < t < 7.987021.

 

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.

 

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.

 

 


 

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.

 

 

 

Therefore, Paul ran 21.46124296 units in total.

 

 

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