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course Phy 241
08/06 6:14 pm
018. Vectors
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Question: `q001. Note that this assignment contains 8 questions.
. The Pythagorean Theorem: the hypotenuse of a right triangle has a length c such that c^2 = a^2 + b^2, where a and b are the lengths of the legs of the triangle. Sketch a right triangle on a set of coordinate axes by first locating the point (7, 13). Then sketch a straight line from the origin of the coordinate system to this point to form the hypotenuse of the triangle. Continue by sketching a line straight down from (7, 13) to the x axis to form one leg of the triangle, then form the other leg by continuing straight along the x axis back to the origin. How long are these two legs? How long therefore is the hypotenuse?
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Your solution:
a^2 + b^2 = c^2
7^2 + 13^2 = c^2
sqrt(218) = sqrt(c^2)
14.76 = c
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Question: `q002. Sketch a right triangle on a set of coordinate axes whose leg along the x axis has length 12 and whose hypotenuse has length 15. What must be the length of the second leg of the triangle?
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Your solution:
a^2 +b^2 = c^2
12^2 + b^2 = 15^2
144 + b^2 = 225
144 -144 + b^2 = 225 - 144
sqrt(b^2) = sqrt(81)
b = 9
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Question: `q003. If a line of length L is directed from the origin of an x-y coordinate system at an angle `theta with the positive x axis, then the x and y coordinates of the point where the line ends will be y = L * sin(`theta) and x = L * cos(`theta).
Sketch a line of length 10, directed from the origin at an angle of 37 degrees with the positive x axis. Without doing any calculations estimate the x and y coordinates of this point. Give your results and explain how you obtained your estimates.
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Your solution:
L = 10, theta = 37 deg
y = 10*sin(37 deg) x = 10*cos(37 deg)
y = 6.01 x = 7.98
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Question: `q004. Using your calculator you can calculate sin(37 deg) and cos(37 deg). First be sure your calculator is in degree mode (this is the default mode for most calculators so if you don't know what mode your calculator is in, it is probably in degree mode). Then using the sin and cos buttons on your calculator you can find sin(37 deg) and cos(37 deg). What are these values and what should therefore be the x and y coordinates of the line directed from the origin at 37 degrees from the x axis?
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Your solution:
L = 10, theta = 37 deg
y = 10*sin(37 deg) x = 10*cos(37 deg)
y = 6.01 x = 7.98
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Question: `q005. Show that the x and y coordinates you obtained in the preceding problem in fact give the legs of a triangle whose hypotenuse is 10.
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Your solution:
a^2 + b^2 = c^2
6^2 + 8^2 = c^2
36 + 64 = c^2
sqrt(100) = sqrt(c^2)
10 = c
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Question: `q006. A vector drawn on a coordinate system is generally depicted as a line segment of a specified length directed from the origin at a specified angle with the positive x axis. The vector is traditionally drawn with an arrow on the end away from the origin. In the preceding series of problems the line segment has length 10 and was directed at 37 degrees from the positive x axis. That line segment could have been drawn with an arrow on its end, pointing away from the origin.
The components of a vector consist of a vector called the x component drawn from the origin along the x axis from the origin to x coordinate L cos(`theta), and a vector called the y component drawn from the origin along the y axis to y coordinate L sin(`theta).
What are the x and y components of a vector directed at an angle of 120 degrees, as measured counterclockwise from the positive x axis, and having length 30 units?
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Your solution:
L = 30, theta = 120 deg
y = 30*sin(120 deg) x = 30*cos(120deg)
y = 25.98 x = -15
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Question: `q007. The angle of a vector as measured counterclockwise from the positive x axis is easily determined if the components of the vector are known. The angle is simply arctan( y component / x component ) provided the x component is positive. If the x component is negative the angle is arctan( y component / x component ) + 180 deg. If the x component is positive and the y component negative, arctan( y component / x component ) will be a negative angle, and in this case we generally add 360 degrees in order to obtain an angle between 0 and 360 degrees.
The arctan, or inverse tangent tan^-1, is usually on a calculator button marked tan^-1.
Find the angle and length of each of the following vectors as measured counterclockwise from the positive x axis:
A vector with x component 8.7 and y component 5.
A vector with x component -2.5 and y component 4.3.
A vector with x component 10 and y component -17.3.
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Your solution:
Arctan(5/8.7)
= 29.88 deg
Arctan(4.3/-2.5)
= - 59.82 deg
Arctan(17.3/10)
= 59.97 deg
L = `sqrt(8.7^2 + 5^2)
L = 10
L = `sqrt((-2.5)^2 + 4.3^2)
L = 5
L = `sqrt(10^2 + (-17.3)^2)
L = 20.
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If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: `q008. Find the angle and length of a vector whose x component is 8 and whose y component is -4.
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Your solution:
C =sqrt((8 mi)^2 + (-4 mi)^2)
C = sqrt(64 mi^2 + 16 mi^2)
C =sqrt(80 mi^2)
C =8.94 mi
Theta = Arctan (-4 mi/8 mi)
Theta = -26.57 deg + 360 deg
Theta = 333.43 deg
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&#Your work looks good. Let me know if you have any questions. &#