If your solution to stated problem does not match the given solution, you should self-critique per instructions at

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises. 

If x(t) and y(t) are smooth continuous functions, then the parametric equations x = x(t), y = y(t) define a curve in the x-y plane.  We can call this curve C.  Between clock times t = t1 and t = t2, the distance we travel along the curve C is given by the line integral

integral ( sqrt( x ' (t)^2 + y ' (t)^2) dt, as shown in more detail below:

If t is interpreted as time, then the vector x ' (t) i + y ' (t) j is interpreted as the velocity vector v(t).  If we multiply the velocity vector by a short time interval `dt, we get the approximate displacement vector `ds = v(t) `dt.  Summarizing the situation at clock time t:

• At clock time t, we are at the point (x(t), y(t)) of the curve C, our velocity at that instant is v(t) and our displacement between t and t + `dt is approximately equal to `ds = v(t) `dt.= (x ' (t) i + y ' (t) j) * `dt

Our actual points on the curve C are (x(t), y(t)) and (x(t + `dt), y(t + `dt)).

• If we move from point (x(t), y(t)) to point (x(t + `dt), y(t + `dt)) then our vector displacement between these two instants is very close to x ' (t) `dt i + y ' (t) `dt j, and we therefore move a distance very close to sqrt( (x ' (t) `dt)^2 + (y ' (t) `dt)^2)
• We can simplify this expression to sqrt (x ' (t))^2 + ( y ' (t) )^2 ) `dt, abbreviated simply sqrt( x ' ^2 + y ' ^2) `dt.

If we want to know how far we move during a longer interval between t = t1 and t = t2, during which our speed and/or our direction might change significantly, we begin by partitioning the t interval into shorter intervals. 

The typical subinterval is characterized by duration `dt_i and sample point t_hat_i. 

Evaluating x ' and y ' at our sample point, our approximate displacement during the interval is x ' (t_hat_i) `dt_i i + y ' (t_hat_i) `dt_i `j and our distance is sqrt( x ' ^ 2 + y ' ^2) `dt_i, where is is understood that x ' and y ' are evaluated at t = t_hat_i.

If we sum our distances over all subintervals and take the limit as interval size shrinks to 0, we get the integral int( sqrt( x ' ^2 + y ' ^2) dt ), which is called a line integral.  This particular line integral simply evaluates the distance we travel along our path.

Question: 

`q001.  Sketch and describe the path defined by x(t) = cos(t) and y(t) = 2 sin(t), 0 <= t <= pi/2.

This path is our curve C.  Find, but do not evaluate, the integral for the distance traveled along this path. 

Your solution: 

Confidence rating:
 

Given Solution: 

Solution:  The integral is integral( x ' (t)^2 + y ' (t)^2 ) dt, t from 0 to pi/2. 

x ' (t) = sin(t) and y ' (t) = 2 cos(t), so the integral becomes

integral ( sqrt(sin^2(t) + 4 cos^2(t) ) dt, t from 0 to pi / 2.

Self-critique (if necessary):

Self-critique rating:

#$&*

Suppose now that our curve C represents the shape of a thin object (think of a wire) whose density is given by some function f(x, y), in the sense that the average density is multiplied by the length to get the mass.

Again considering the i_th subinterval of our partition:

The length of the subinterval is again sqrt( x ' ^2 + y ' ^2 ) `dt_i, where x ' and y ' are evaluted at the sample point t_hat_i.

The approximate average density on the subinterval will be taken to be f(x(t_hat_i), y(t_hat_i)).

So the mass of the subinterval is approximate

`dm_i = f(x, y) sqrt(x ' ^ 2 + y ' ^2) `dt_i, where again x, y, x ' and y ' are evaluted at sample point t_hat_i.

Summing up our masses and taking the limit as partition mesh approaches zero, we find that the total mass is

integral ( f(x, y) sqrt( x ' ^2 + y ' ^2) dt, t from 0 to pi/2).

Effectively we have simply inserted the density function f(x, y) into the arc length integral.

Question:  `q002.  If the curve in the first question represents a curved wire whose density at point (x, y) is f(x, y) =  4 + x^2 + y^2, then what is the total mass of the wire?

Your solution: 

Confidence rating:
 

Given Solution: 

Our arc length increment is represented by sqrt( x ' ^2 + y ' ^2 ) dt, our density by our function f(x, y), so our mass increment is density * length

• dm = f(x, y) sqrt(x ' ^2 + y ' ^ 2 ) dt.

Our total mass is therefore

integral ( f(x, y) sqrt( x ' ^2 + y ' ^2 ) dt, t from 0 to pi / 2)

= integral ( (4 + x^2 + y^2) sqrt( x ' ^2 + y ' ^2 ) dt, t from 0 to pi / 2)

= integral ( (4 + cos^2(t) + 4 sin^2(t) ) sqrt( sin^2(t) + 4 cos^2(t) ), t from 0 to pi / 2).

Self-critique (if necessary):

Self-critique rating:

#$&*

Now suppose we want to find the work done by a force field F ( x, y ) = F_1(x, y) i + F_2(x, y) j as we move along this path. 

• On our typical subinterval our displacement is x ' (t_hat_i) `dt_i i + y ' (t_hat_i) `dt_i `j and the force is F_1(x, y) i + F_2(x, y) j , where F_1 and F_2 are evaluated at x(t_hat_i), y(t_hat_i)). 
• The work on this interval is the dot product of the force and the displacement:

  `dW_i = F dot `ds_i

  = F(x, y) dot ( x ' (t_hat_i) `dt_i i + y ' (t_hat_i) `dt_i `j )  

  = ( F_1(x, y) i + F_2(x, y) j ) dot (x ' i + y ' j ) `dt_i

  = (F_1 * x ' + F_2 * y ') `dt_i.

  Again all quantities are evaluted at the sample point t_hat_i in our i_th time subinterval, so in detail the above quantity is

  (F_1 (x_hat_i, y_hat_j) * x ' (x_hat_i, y_hat_j)  + F_2 (x_hat_i, y_hat_j) * y ' (x_hat_i, y_hat_j) ) `dt_i

  It's easier to look at the expression

  (F_1 * x ' + F_2 * y ') `dt_i,

  but the preceding line indicates exactly what this expression means

   

• When we sum the contributions over the various subintervals and take limits we obtain an integral which we can evaluate.  Specifically the integral is

  integral ( F_1 * x ' + F_2 * y ' ) dt, t from t1 to t2.

Your solution: 

Confidence rating:
 

Given Solution: 

The path is defined by x(t) = cos(t) and y(t) = 2 sin(t), 0 <= t <= pi/2.

The force function on this path, at clock time t, is therefore

F (x(t), y(t) ) = x(t)^2 y(t) i + x(t) y(t)^2 j = cos^2(t) * 2 sin(t) i + cos(t) * (2 sin(t) )^2 j = 2 cos^2(t) sin(t) i + 4 cos(t) sin^2(t) j

and our displacement increment is

ds = v (t) dt =  (x ' (t) i + y ' (t) j ) dt = (-sin(t) i + 2 cos(t) j ) dt,

so our integral is

integral ( F dot `ds )

= integral ( ( 2 cos^2(t) sin(t) i + 4 cos(t) sin^2(t) j ) dot ((-sin(t) i + 2 cos(t) j ) dt), t from 0 to pi/2)

= integral ( ( -2 cos^2(t) sin^2(t) + 8 cos^2(t) sin^2(t) ) dt, t from 0 to pi/2)

= integral ( (6 cos^2(t) sin^2(t)) dt, t from 0 to pi / 2).

Self-critique (if necessary):

Self-critique rating: