AREA, DISPLACEMENT, DISTANCE

Calculus II Project

Objective:

To learn how to use integrals to compute areas, displacements, and distances. You can use Maple to graph functions, to solve equations, and to evaluate integrals.

Solved Example 1:

Compute the area enclosed by the curve y = 1-x-x^2 and the x -axis.

Solution:

First let's plot the curve to get an idea of the region whose area we need to compute.

> plot(1-x-x^2,x=-2.5..1);

[Maple Plot]

To compute the area we have to find the two x -intercepts. We'll use Maple to solve the equation y = 0 .

> xint:=[solve(1-x-x^2,x)];

xint := [-1/2-1/2*sqrt(5), -1/2+1/2*sqrt(5)]

Since y is positive for all x between the two intercepts, the area in question is the definite integral.

> area:=Int(1-x-x^2,x=xint[2]..xint[1]);

area := Int(1-x-x^2,x = -1/2+1/2*sqrt(5) .. -1/2-1/...

> area:=int(1-x-x^2,x=xint[2]..xint[1]);

area := -sqrt(5)-1/2*(-1/2-1/2*sqrt(5))^2+1/2*(-1/2...

> simplify(%);

-5/6*sqrt(5)

The area is 5/6*sqrt(5) .

Solved Example 2:

A particle is moving along a straight line. The velocity function is v(t) = t^4-5*t^3-t+15 in meters per second.

a) Find the displacement in the time period from t = 1 to t = 5.2.

b) Find the distance traveled in the time period from t = 1 to t = 5.2.

Solution:

> v:=t->t^4-5*t^3-t+15;

v := proc (t) options operator, arrow; t^4-5*t^3-t+...

a) The displacement is the integral:

> Int(v(t),t=1..5.2);

Int(t^4-5*t^3-t+15,t = 1 .. 5.2)

> int(v(t),t=1..5.2);

-102.5139360

The displacement is -102.5 meters.

b) The distance traveled by the particle is int(abs(v(t)),t = 1 .. 5.2) . To evaluate this integral we have to look at the graph of v ( t ).

> plot(v(t),t=0..7,y=-80..50);

[Maple Plot]

The distance is A[1]-A[2]+A[3] . We have to find the x -intercepts of the function.

> solve(v(t),t);

RootOf(_Z^4-5*_Z^3-_Z+15,index = 1), RootOf(_Z^4-5*...
RootOf(_Z^4-5*_Z^3-_Z+15,index = 1), RootOf(_Z^4-5*...

Perhaps fsolve will work better.

> fsolve(v(t),t);

1.576928399, 4.915065288

1.576928399, 4.915065288

It does. Let's name the two solutions.

> xinter:=[fsolve(v(t),t)];

xinter := [1.576928399, 4.915065288]

> distance:=Int(v(t),t=1..xinter[1])-Int(v(t),t=xinter[1]..xinter[2])+Int(v(t),t=xinter[2]..5.2);

distance := Int(t^4-5*t^3-t+15,t = 1 .. 1.576928399...

> distance:=int(v(t),t=1..xinter[1])-int(v(t),t=xinter[1]..xinter[2])+int(v(t),t=xinter[2]..5.2);

distance := 119.0828944

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Assignment

Problem 1

Find the area enclosed by the curve y = x^2/sqrt(x+5)-2 and the x -axis.

Problem 2

A particle is moving along a straight line. The velocity function is v(t) = 4*t^5-25*t^4-10*t^2+300*t in meters per second.

a) Find the displacement in the time period from t = 1 to t = 5.

b) Find the distance traveled in the time period from t = 1 to t = 5.

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MSIP Grant #P120A80089-98: "Three Urban Calculus Reform Programs: Adopting the Best," 1998-2001; MSEIP Grant #P120AA010031: "Four Colleges: Calculus + Enhancements", 2001-2004