#$&* course Mth 277 Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
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Given Solution: Solution: The integral is integral( x ' (t)^2 + y ' (t)^2 ) dt, t from 0 to pi/2. x ' (t) = sin(t) and y ' (t) = 2 cos(t), so the integral becomes integral ( sqrt(sin^2(t) + 4 cos^2(t) ) dt, t from 0 to pi / 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK #$&* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x, y) is f(x, y) = 4 + x^2 + y^2 int((4+cos^2(t)+4sin^2(t))sqrt(sin^2(t)+4cos^2(t))0 to pi/2)=8-10=-2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Our arc length increment is represented by sqrt( x ' ^2 + y ' ^2 ) dt, our density by our function f(x, y), so our mass increment is density * length dm = f(x, y) sqrt(x ' ^2 + y ' ^ 2 ) dt. Our total mass is therefore integral ( f(x, y) sqrt( x ' ^2 + y ' ^2 ) dt, t from 0 to pi / 2) = integral ( (4 + x^2 + y^2) sqrt( x ' ^2 + y ' ^2 ) dt, t from 0 to pi / 2) = integral ( (4 + cos^2(t) + 4 sin^2(t) ) sqrt( sin^2(t) + 4 cos^2(t) ), t from 0 to pi / 2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK #$&* 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. ********************************************* Question: Find, but don't evaluate, the integral for the work done by the force function F(x, y) = x^2 y i + x y^2 j over the path C given by x(t) = cos(t) and y(t) = 2 sin(t), 0 <= t <= pi/2 (this is the same curve given in the first question of this set).. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x, y) = x^2 y i + x y^2 j x(t) = cos(t) and y(t) = 2 sin(t), 0 <= t <= pi/2 x(t) = cos(t) and y(t) = 2 sin(t), 0 <= t <= pi/2 x'(t)=-sin(t) and y'(t)=2cos(t) int(sqrt(sin^2(t) + 4 cos^2(t))dt int((2cos^2(t)sin(t)i+4cos(t)sin^2(t)j)dot((-sin(t)i+2cos(t)j)dt),t from 0 to pi/2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating:OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: