Rates of Change Questions
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A man 6 ft tall walks at a rate of 5 ft/sec toward a streetlight that is 16 ft above the ground.
a) At what rate is the tip of his shadow moving? (Ans: -8)
b) At what rate is the length of his shadow changing? (Ans: -3)
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A highway patrol plane flies 3 mi above a level, straight road at a steady 120 mi/h. The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 mi, the line-of-sight distance is decreasing at the rate of 160 mi/h. Find the car's speed along the highway. (Ans: 80 mi/h)
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Many thanks for helping me out! =D
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Hi,
I will address Q1, which is a nice question on rates of change involving geometry.
Make a diagram (try and draw it), and one sees similar triangles.
Let the man be of x feet from the streetlight. Let the tip of the shadow be of y feet from the streetlight. We wish to find dy/dt.
By similar triangles, 6/16 = (y - x) / y (*), which simplifies to y = 8/5 x.
We are given that dx/dt = -5. Now dy/dt = (8/5) (dx/dt) = -8 ft/s.
In the next part, we want to find d/dt (y - x), where y - x is the length of the shadow.
From expression (*) we obtain y - x = 3/8 y.
Now d/dt (y - x) = (3/8) (dy/dt) = -3 ft/s.
Thanks!
Cheers,
Wen Shih -
Thanks to wee_ws for helping me out before. :) But I have some more questions to ask about rates of change. Hopefully anyone can help me out? 1) A girl flies a kite at a height of 300 ft, the wind carrying the kite horizontally away from her at a rate of 25 ft/sec. How fast must she let out the string when the kite is 500 ft away from her? (Ans: 20 ft/sec) _____________________________________________________________ 2) Boat A is 100km due North of Boat B. Boat A is travelling at East at a constant speed of 50km/h while Boat B is travelling North-East at a constant speed of 70km/h. After how many hours will Boat A & Boat B be the nearest to each other? What is the nearest distance between Boat A & Boat B? Thanks again for helping me out! =D
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Originally posted by -Wanderer-:
Thanks to wee_ws for helping me out before.
But I have some more questions to ask about rates of change. Hopefully anyone can help me out?
1) A girl flies a kite at a height of 300 ft, the wind carrying the kite horizontally away from her at a rate of 25 ft/sec. How fast must she let out the string when the kite is 500 ft away from her? (Ans: 20 ft/sec)
2) Boat A is 100km due North of Boat B. Boat A is travelling at East at a constant speed of 50km/h while Boat B is travelling North-East at a constant speed of 70km/h. After how many hours will Boat A & Boat B be the nearest to each other? What is the nearest distance between Boat A & Boat B?
Thanks again for helping me out! =D
1)
sqrt(500^2 - 300^2) = 400ft = current horizontal distance from kite
1sec later, it will be 425ft.
Distance from kite now = sqrt(425^2 + 300^2) = 520.216m
Thus, speed required = (520.216 - 500)ft / 1sec = 20 ft/sec
2)
Will get back to you if time permits.
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Originally posted by wee_ws:
Hi,
I will address Q1, which is a nice question on rates of change involving geometry.
Make a diagram (try and draw it), and one sees similar triangles.
Let the man be of x feet from the streetlight. Let the tip of the shadow be of y feet from the streetlight. We wish to find dy/dt.
By similar triangles, 6/16 = (y - x) / y (*), which simplifies to y = 8/5 x.
We are given that dx/dt = -5. Now dy/dt = (8/5) (dx/dt) = -8 ft/s.
In the next part, we want to find d/dt (y - x), where y - x is the length of the shadow.
From expression (*) we obtain y - x = 3/8 y.
Now d/dt (y - x) = (3/8) (dy/dt) = -3 ft/s.
Thanks!
Cheers,
Wen Shihurmm maybe it is due to my lack of understanding of the question, but i cant seem to get the diagram out... can someone please help me. thanks!
just a question off topic... does concentration of acid affects the acidity of the acid? if yes or no, pls explain. thanks! -
Originally posted by ItchyArmpit:
just a question off topic... does concentration of acid affects the acidity of the acid? if yes or no, pls explain. thanks!Yup..acidity is affected by the amount of H+ ions present as compared to OH- ions
So more moles of acid = more H+ = higher acidity
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Originally posted by -Wanderer-:
Thanks to wee_ws for helping me out before.
But I have some more questions to ask about rates of change. Hopefully anyone can help me out?
1) A girl flies a kite at a height of 300 ft, the wind carrying the kite horizontally away from her at a rate of 25 ft/sec. How fast must she let out the string when the kite is 500 ft away from her? (Ans: 20 ft/sec)
Hi,
Am happy to guide you along :)
Let L be the distance between girl and kite at any time t.
Then L^2 = 300^2 + (25 t)^2 = 300^2 + 625 t^2, by Pythagoras' Theorem --- (1).
Our objective is to find dL/dt when L = 500.
When L = 500, we use Pythagoras' Theorem to find the horizontal distance covered by the kite due to the wind, which is 400 ft.
Now 400/25 = 16 s is the time taken for the kite to be 500 ft from the girl.
Differentiating (1) with respect to t, we obtain
2L dL/dt = 1250 t.
Substituting L = 500 and t = 16 lead us to the desired (exact) value of dL/dt.
Thanks!
Cheers,
Wen Shih -
Originally posted by ItchyArmpit:
urmm maybe it is due to my lack of understanding of the question, but i cant seem to get the diagram out... can someone please help me. thanks!
Hi,
1. Draw a vertical line to represent streetlight of 16 ft.
2. Draw another (shorter) vertical line to represent man of 6 ft, and of a horizontal distance x ft away from streetlight.
3. Draw a line joining the tops of streetlight and man (because light travels in a straight line). Extend this line until it touches the horizontal ground, which is the tip of the shadow.
Now you should see two similar triangles.
Do practise more to improve on your ability to represent mathematical information diagrammatically :)
Thanks!
Cheers,
Wen Shih -
Originally posted by -Wanderer-:
Thanks to wee_ws for helping me out before.
But I have some more questions to ask about rates of change. Hopefully anyone can help me out?
1) A girl flies a kite at a height of 300 ft, the wind carrying the kite horizontally away from her at a rate of 25 ft/sec. How fast must she let out the string when the kite is 500 ft away from her? (Ans: 20 ft/sec)
2) Boat A is 100km due North of Boat B. Boat A is travelling at East at a constant speed of 50km/h while Boat B is travelling North-East at a constant speed of 70km/h. After how many hours will Boat A & Boat B be the nearest to each other? What is the nearest distance between Boat A & Boat B?
Thanks again for helping me out! =D
Back. Sorry for the half baked answer. Dr Wee is still the best
2)
horizontal component of B = vertical component = 49.4975km/h
Horizontal distance of A from B = (50-49.4975)(t) km = 0.5025t km, where t = time in hour
Vertical distance = (100-49.4975t) km, note the placement of 't' here.
Distance at any time = sqrt [(0.5025t)^2 + (100-49.4975t)^2] km ------(1)
To cut the long story short, since unfortunately I have a PBL to deal with, you just need to differentiate this twice, equate it to zero, and you will get a few 't'. Take the one which makes the most sense, and that will be the needed time.
Sub it into (1) to get the distance.
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Originally posted by Forbiddensinner:
Horizontal distance of A from B = (50-49.4975)(t) km = 0.5025t km, where t = time in hour
Hi. I was just wondering why is it (50 - 35sqrt2)? I understand how to get 35sqrt2 and the placement of t, but why 50 minus the 35sqrt2?
Thanks to all who helped me so far. :)
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Originally posted by -Wanderer-:
Hi. I was just wondering why is it (50 - 35sqrt2)? I understand how to get 35sqrt2 and the placement of t, but why 50 minus the 35sqrt2?
Thanks to all who helped me so far. :)
1 hour A will have moved 50km east.
1hour B will have moved 35sqrt2km east.
Thus difference between their horizontal distance = (50 - 35sqrt2)(t) km, where t is in hours.