course Phy 231
I have been having trouble with problem 1.79. I spent a few hours on it, alone and with one of the calculus tutors at PVCC. I think I have the calculations done correctly, so it is probably a problem with technique. **As I typed this up to submit, I found an error and corrected it. My answer seems to make sense now, but the magnitudes are still close to but not quite the ones given in the back of the text (B = 255 meters and C = 70 meters) My solution is B = 263 meters and C = 73.55 meters. I thought that there could only be one solution to this problem, because of the properties of triangles.
You are canoeing on a lake. Starting at your camp on the shore, you travel 240 m in the direction 32 degrees south of east to reach a store. On the return trip you travel distance B in the direction 48 degrees north of west, distance C in the direction 62 degrees south of west, and you are back at your camp. Use vector methods to calculate the distances B and C.The first thing I did was draw a picture. I have axes labelled with compass directions, and drew each of the three vectors with their tails at the origin.
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I used the equations Ax = A cos (theta) and Ay = A sin (theta) to calculate the components of each vector.
Vector A has a magnitude of 240 meters, a direction of -32 degrees, an x component of 203.53 meters and a y component of -127.18 meters
Vector B has a magnitude of B meters, a direction of 132 degrees (180 degrees - 48 degrees), an x component of -.6691 * B meters, and a y component of .7431 * B meters.
Vector C has a magnitude of C meters, a direction of 242 degrees (180 degrees + 62 degrees), an x component of -.4695 * C meters, and a y component of -.8829 * C meters.
Since the starting and ending points of the trip are the same, I wrote down this relationship between the components: Ax + Bx + Cx = 0 and Ay + By + Cy = 0. Then I solved each of them for the C component to get:
Ax + Bx = -Cx and Ay + By = -Cy.
Into the first equation, I substituted the components I calculated earlier:
203.53 - .6691 * B = .3746 * C
Then solved for C:
(203.35 - .6691 * B) / .3746 = C
I substitued this value for C into the other equation:
.7431 * B - 127.18 = (.9272 / .3746) *(203.35 - .6691 * B)
The right-hand side of the equation is simplified:
.7431 * B - 127.18 = 503.77 - 1.6562* B
Similar terms are grouped together:
.7431 * B + 1.6562 * B = 503.77 + 127.18
And simplified:
2.3993 * B = 630.95
Then solved for B = 263 meters.
The components of vector B are calculated:
Bx = 263 cos (132) = -176.0 meters and By = 263 sin (132) = 195.45 meters.
The x component of B is substituted into the equation Ax + Bx = -Cx to find the magnitude of C:
203.53 - 176.0 = .3746 * C , which is solved for C = 73.55
The components of vector C are calculated:
Cx = 73.55 cos (248) = -27.55 meters and Cy = 73.55 sin (248) = -68.19 meters.
All of the components are combined into my original component equations to check for accuracy:
Ax + Bx + Cx = 0 => 203.35 - 176.0 -27.55 = -.2 should be zero, but seems acceptable due to rounding.
Ay + By + Cy = 0 => -127.18 + 195.45 - 68.19 = .08 again, seems like inconsistency due to rounding.
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Your procedure is excellent, and well explained. Despite the sensitivity of some of these calculations to roundoff errors, using 4-significant-figure values there should not be an error of magnitude .2. Let me know if you have questions. However most of your angles are accurate to only 3 significant figures, and .2 is in the third significant figure of your distances, so roundoff error should explain the small discrepancy.