#$&* course PHY 121 7/22 10 018. Vectors
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Given Solution: The leg from (7, 13) to the x axis drops from the point (7, 13) to the point (7,0) and so has length 13. The second leg runs from (7,0) to the origin, a distance of 7 units. The legs therefore have lengths a = 13 and b = 7, so that the hypotenuse c satisfies c^2 = a^2 + b^2 and we have c = `sqrt(a^2 + b^2) = `sqrt( 13^2 + 7^2 ) = `sqrt( 216) = 14.7, approximately. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. Sketch a right triangle on a set of coordinate axes whose leg along the x axis has length 12 and whose hypotenuse has length 15. What must be the length of the second leg of the triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a^2 + b^2 = c^2 b^2 = c^2 - a^2 b^2 = 15^2 - 12^2 b^2 = 225 - 144 b^2 = 81 b = 9 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Let c stand for the length of the hypotenuse and a for the length of the known side, with b standing for the length of the unknown side. Then since c^2 = a^2 + b^2, b^2 = c^2 - a^2 and b = `sqrt(c^2 - a^2) = `sqrt( 15^2 - 12^2) = `sqrt(225 - 144) = `sqrt(81) = 9. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The leg on the x axis will be slightly less than 10, so I'll estimate 8. The leg on the y axis will be smaller because it is across from the 37 degree angle. I'll estimate that to be 6. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The line will run closer to the x axis then to the y axis, since the line is directed at an angle below 45 degrees. It won't run a whole lot closer but it will run significantly closer, which will make the x coordinate greater than the y coordinate. Since the line itself has length 10, it will run less than 10 units along either the x or the y axis. It turns out that the x coordinate is very close to 8 and the y coordinate is very close to 6. Your estimates should have been reasonably close to these values. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK I am currently in shock! ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. Using your calculator you can calculate sin(37 deg) and cos(37 deg). First be sure your calculator is in degree mode (this is the default mode for most calculators so if you don't know what mode your calculator is in, it is probably in degree mode). Then using the sin and cos buttons on your calculator you can find sin(37 deg) and cos(37 deg). What are these values and what should therefore be the x and y coordinates of the line directed from the origin at 37 degrees from the x axis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sin (37) = .601815023 cos (37) = .7986355 I am not sure what length to use. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: sin(37 deg) should give you a result very close but not exactly equal to .6. cos(37 deg) should give you a result very close but not exactly equal to .8. Since the x coordinate is L cos(`theta), then for L = 10 and `theta = 37 deg we have x coordinate 10 * cos(37 deg) = 10 * .8 = 8 (your result should be slightly different than this approximate value). Since the y coordinate is L sin(`theta), then for L = 10 and `theta = 37 deg we have y coordinate 10 * sin(37 deg) = 10 * .6 = 6 (your result should be slightly different than this approximate value). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn't know to back to the previous problem to get this answer. When I plug in the 10, I do get the correct answer. ------------------------------------------------ Self-critique rating: 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6^2 + 8^2 = c^2 36 + 64 = c^2 100 = c^2 c = 10 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If c stands for the hypotenuse of the triangle, then if a = 8 and b = 6 are its legs we have c = `sqrt(a^2 + b^2) = `sqrt(8^2 + 6^2) = `sqrt(100) = 10. The same will hold for the more precise values you obtained in the preceding problem. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q006. A vector drawn on a coordinate system is generally depicted as a line segment of a specified length directed from the origin at a specified angle with the positive x axis. The vector is traditionally drawn with an arrow on the end away from the origin. In the preceding series of problems the line segment has length 10 and was directed at 37 degrees from the positive x axis. That line segment could have been drawn with an arrow on its end, pointing away from the origin. The components of a vector consist of a vector called the x component drawn from the origin along the x axis from the origin to x coordinate L cos(`theta), and a vector called the y component drawn from the origin along the y axis to y coordinate L sin (`theta). What are the x and y components of a vector directed at an angle of 120 degrees, as measured counterclockwise from the positive x axis, and having length 30 units? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 30 (cos)(120) = -15 y = 30 (sin)(120) = 25.98 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The x component of this vector is vector along the x axis, from the origin to x = 30 cos(120 degrees) = -15. The y component is a vector along the y axis, from the origin to y = 30 sin(120 degrees) = 26, approx.. Note that the x component is to the left and the y component upward. This is appropriate since the 120 degree angle, has measured counterclockwise from the positive x axis, takes the vector into the second quadrant, which directs it to the left and upward. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 8.7, y = 5: magnitude 10.034; angle 29.886 degrees x = -2.5, y = 4.3: magnitude 5; angle -60 + 180 = 120 degrees x = 10, y = - 17.3: magnitude approx 20; angle -60 + 360 = 300 degrees confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A vector with x component 8.7 and y component 5 makes an angle of arctan (5/8.7) = 30 degrees with the x axis. Since the x component is positive, this angle need not be modified. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt(8.7^2 + 5^2) = 10. A vector with x component -2.5 and y component 4.3 makes an angle of arctan (4.3 / - 2.5) + 180 deg = -60 deg + 180 deg = 120 degrees with the x axis. Since the x component is negative, we had to add the 180 degrees. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt((-2.5)^2 + 4.3^2) = 5. A vector with x component 10 and y component -17.3 makes an angle of arctan (-17.3 / 10) = -60 degrees with the x axis. Since the angle is negative, we add 360 deg to get 300 deg. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt(10^2 + (-17.3)^2) = 20. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q008. Find the angle and length of a vector whose x component is 8 and whose y component is -4. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: angle: tan-1 (-4/8) = -27 + 360 = 333 degrees magnitude: 8^2 + (-4)^2 = c^2 64 + 16 = 80; `sqrt 80 = 8.94 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!