Why can't I get this simple calculus problem right?

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Patashu
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Why can't I get this simple calculus problem right?

Postby Patashu » Mon May 26, 2008 8:43 am UTC

The question is as follows:
'Two straight roads meet at an angle of 60 degrees. A starts from the intersection and travels along one road at 40 km/hr. One hour later B starts from the intersection and travels along the other road at 50 km/hr. At what rate is the distance between them changing three hours after A starts?'
The answer the book gives is 43.1 km/hr.

da/dt = 40 and db/dt = 50. Letting t = 0 be the point where A is at the intersection, a = 0 + 40t and b = -50 + 50t.

The first method I tried was to convert the movement of a into x and y co-ordinates using the sine rule such that b is moving along the x axis and thus using pythagoras' theorem to find the formula for distance between the two points and differentiate it. Letting b = x, ax = 20t and ay = 20sqrt(3)t. I got dr/dt = 490t+350/sqrt(49t^2+70t+37). When t = 3 dr/dt = 455/sqrt(43) approximates 69.4 km/hr. No good.

The second method I tried was to use the cosine rule, drawing a and b as sides of a triangle changing with time, the distance between the two points c the third angle of the triangle opposing an angle of 60 degrees. Thus, r = sqrt(a^2+b^2-ab). This yielded r = 10sqrt(21t^2+15t+25), a curiously different equation from my last. Differentiating gave dr/dt = 210t+75/sqrt(21t^2+15t+25). When t = 3, dr/dt = 705/sqrt(259) approximates 43.8 km/hr, close but certainly not the right answer.

As you can see I am perplexed and baffled as to how two different methods are giving me two different equations, none of which is right. What am I doing wrong in each approach?

EDIT: I got my second method to give the correct answer by mastering the daunting task of adding and subtracting numbers correctly. I am still morbidly curious as to what is wrong with my first approach, however.
Last edited by Patashu on Mon May 26, 2008 10:15 am UTC, edited 1 time in total.

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Re: Why can't I get this simple calculus problem right?

Postby jaap » Mon May 26, 2008 9:13 am UTC

Patashu wrote:The first method I tried was to convert the movement of a into x and y co-ordinates using the sine rule such that b is moving along the x axis and thus using pythagoras' theorem to find the formula for distance between the two points and differentiate it. Letting b = x, ax = 20t and ay = 20sqrt(3)t. I got dr/dt = 490t+350/sqrt(49t^2+70t+37). When t = 3 dr/dt = 455/sqrt(43) approximates 69.4 km/hr. No good.


I get r2 = ( bx - ax )2 + ( by - ay )2 = ( 30t - 50 )2 + ( 40t )2 = 100( 21t2 -30t +25 ).

Patashu wrote:The second method I tried was to use the cosine rule, drawing a and b as sides of a triangle changing with time, the distance between the two points c the third angle of the triangle opposing an angle of 60 degrees. Thus, r = sqrt(a^2+b^2-ab). This yielded r = 10sqrt(21t^2+15t+25), a curiously different equation from my last. Differentiating gave dr/dt = 210t+75/sqrt(21t^2+15t+25). When t = 3, dr/dt = 705/sqrt(259) approximates 43.8 km/hr, close but certainly not the right answer.


I get the same answer as my first one. It looks like you're closer this time, but you got the coefficient of the middle term wrong. I think the final answer is then 41.3, not what you said the correct answer was.

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Re: Why can't I get this simple calculus problem right?

Postby Patashu » Mon May 26, 2008 9:28 am UTC

jaap wrote:
Patashu wrote:The second method I tried was to use the cosine rule, drawing a and b as sides of a triangle changing with time, the distance between the two points c the third angle of the triangle opposing an angle of 60 degrees. Thus, r = sqrt(a^2+b^2-ab). This yielded r = 10sqrt(21t^2+15t+25), a curiously different equation from my last. Differentiating gave dr/dt = 210t+75/sqrt(21t^2+15t+25). When t = 3, dr/dt = 705/sqrt(259) approximates 43.8 km/hr, close but certainly not the right answer.


I get the same answer as my first one. It looks like you're closer this time, but you got the coefficient of the middle term wrong. I think the final answer is then 41.3, not what you said the correct answer was.

Aha, you're right; I got the second co-efficient completely wrong somehow. Now I get r = 10sqrt(21t^2-30t+25) therefore dr/dt = 210t-150/sqrt(21t^2-30t+25) which at t = 3 approximately equals 43.1km/h.
Now how did you get 41.3?

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Re: Why can't I get this simple calculus problem right?

Postby daryuu » Mon May 26, 2008 9:31 am UTC

Your method for part one strikes me as a little odd, and your formulas for the position of a (assuming that's what they were) don't make much sense... I'm really not sure where you're getting them...

Hint:
Spoiler:
Try letting b move along the x-axis and thinking of a as being at the 60 degree position of a circle of radius (straight-line distance a has traveled), and getting the distance between them as a function of t.


That method seemed fairly straightforward to me, and I got the right answer (letting t=0 represent the beginning of b's movement and finding the derivative at t=2)...

As for your second method, it should work except that you're assuming (as I did above) that t=0 signifies the beginning of b's movement (don't forget to shift the t used in a's formula), and you applied the law of cosines wrong... Just those two corrections and I got the right answer.

Edit: ninja'd by two people? I must be really slow...

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Re: Why can't I get this simple calculus problem right?

Postby Patashu » Mon May 26, 2008 9:38 am UTC

daryuu wrote:Your method for part one strikes me as a little odd, and your formulas for the position of a (assuming that's what they were) don't make much sense... I'm really not sure where you're getting them...

Okay, I'll explain what I was thinking of.
I let b be travelling along the x axis and wanted to restate a as x and y parts on this set of axiis.

So, I drew a triangle with a for its hypotenuse and an angle of 60 degrees at its base. The far side is ay = asin60 = 20sqrt(3)t by the sine rule and the near side is ax = asin30 = 20t by the sine rule.
Now the distance between a and b should be sqrt((b-ax)^2+ay^2)).
I just went through and corrected the math without changing the method and it's still way off, so I'd like to know what is wrong with this approach.

Hint:
Spoiler:
Try letting b move along the x-axis and thinking of a as being at the 60 degree position of a circle of radius (straight-line distance a has traveled), and getting the distance between them as a function of t.


That method seemed fairly straightforward to me, and I got the right answer (letting t=0 represent the beginning of b's movement and finding the derivative at t=2)...

I can't figure out what you're doing here. Tell me more?

As for your second method, it should work except that you're assuming (as I did above) that t=0 signifies the beginning of b's movement (don't forget to shift the t used in a's formula), and you applied the law of cosines wrong... Just those two corrections and I got the right answer.

No, I had t=0 be the beginning of a's movement, just like the previous (now failed) solution I tried.
And if that's not how you apply the law of cosines I'm a monkey's uncle.

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Re: Why can't I get this simple calculus problem right?

Postby mostly_true_comics » Mon May 26, 2008 9:53 am UTC

I'm highly drugged and highly sleepy, so there's something about this problem statement I'm not getting. After both A and B have begun their journeys, we have two individuals, each with constant velocity, neither of which are undergoing acceleration of any sort.

The rate of chance of distance, therefore, shouldn't be time-dependent. Everybody's velocity with respect to time is constant, so their relative velocities are going to constant. Imagine astronauts or spaceships instead of cars.

What am I missing?

(ps)
Unless these roads aren't actually straight but are instead embedded the surface of a sphere. But this is clearly listed as an introductory proble, nad hte first first thing you learn in introductory physics is that the earth is a flat, airless surface devoid of friction.
"Because Truth is stranger than fiction, but for best results, try an alloy of 90% truth, 10% confabulation, and 2% butterscotch ripple."

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Re: Why can't I get this simple calculus problem right?

Postby Patashu » Mon May 26, 2008 9:57 am UTC

mostly_true_comics wrote:I'm highly drugged and highly sleepy, so there's something about this problem statement I'm not getting. After both A and B have begun their journeys, we have two individuals, each with constant velocity, neither of which are undergoing acceleration of any sort.

The rate of chance of distance, therefore, shouldn't be time-dependent. Everybody's velocity with respect to time is constant, so their relative velocities are going to constant. Imagine astronauts or spaceships instead of cars.

What am I missing?


The question is asking for the rate of change of distance between A and B, not the rate of change of velocity between A and B.

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Re: Why can't I get this simple calculus problem right?

Postby daryuu » Mon May 26, 2008 9:58 am UTC

Patashu wrote:And if that's not how you apply the law of cosines I'm a monkey's uncle.


No, I just meant that the law of cosines is c^2 = a^2 + b^2 - 2ab*cos(theta), but you used c^2 = a^2 + b^2 - ab...

Patashu wrote:
Hint:
Spoiler:
Try letting b move along the x-axis and thinking of a as being at the 60 degree position of a circle of radius (straight-line distance a has traveled), and getting the distance between them as a function of t.


That method seemed fairly straightforward to me, and I got the right answer (letting t=0 represent the beginning of b's movement and finding the derivative at t=2)...

I can't figure out what you're doing here. Tell me more?


A better explanation might be to think of the position of a at time t as a multiple of the unit vector in the direction of a relative to b... If that makes any more sense... I don't want to just give you a formula or anything, but hopefully this is a bit easier to interpret.

Patashu wrote:No, I had t=0 be the beginning of a's movement, just like the previous (now failed) solution I tried.


Yes, but you're forgetting to take into account that a has been moving for one more unit of time than b. I adjusted a's time up because adjusting b's down would give an incorrect solution for t<1, but since that doesn't really matter for your problem either way is fine...

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Re: Why can't I get this simple calculus problem right?

Postby Patashu » Mon May 26, 2008 10:09 am UTC

daryuu wrote:
Patashu wrote:And if that's not how you apply the law of cosines I'm a monkey's uncle.


No, I just meant that the law of cosines is c^2 = a^2 + b^2 - 2ab*cos(theta), but you used c^2 = a^2 + b^2 - ab...

Cos60 is 1/2, which cancels out the 2. You'll notice I left both out.

Patashu wrote:
Hint:
Spoiler:
Try letting b move along the x-axis and thinking of a as being at the 60 degree position of a circle of radius (straight-line distance a has traveled), and getting the distance between them as a function of t.


That method seemed fairly straightforward to me, and I got the right answer (letting t=0 represent the beginning of b's movement and finding the derivative at t=2)...

I can't figure out what you're doing here. Tell me more?


A better explanation might be to think of the position of a at time t as a multiple of the unit vector in the direction of a relative to b... If that makes any more sense... I don't want to just give you a formula or anything, but hopefully this is a bit easier to interpret.

I sort of grasp it but not how phrasing it this way is supposed to help.

Patashu wrote:No, I had t=0 be the beginning of a's movement, just like the previous (now failed) solution I tried.


Yes, but you're forgetting to take into account that a has been moving for one more unit of time than b. I adjusted a's time up because adjusting b's down would give an incorrect solution for t<1, but since that doesn't really matter for your problem either way is fine...
[/quote]
With a = 0+40t and b = -50+50t I correctly get b having just started at t = 1 and a's head start of 40 km.

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Re: Why can't I get this simple calculus problem right?

Postby daryuu » Mon May 26, 2008 10:43 am UTC

Patashu wrote:Cos60 is 1/2, which cancels out the 2. You'll notice I left both out.


Oh, sorry... I never remember sines/cosines of specific angles and I didn't bother to calculate that (just plugged it into my calculator along with everything else). So, yeah, my bad.

Patashu wrote:I sort of grasp it but not how phrasing it this way is supposed to help.


When I looked back at it, the circle concept seemed unnecessary. I was involving an related geometric structure, so I thought it would be more direct to just say it in terms of vectors.

Patashu wrote:With a = 0+40t and b = -50+50t I correctly get b having just started at t = 1 and a's head start of 40 km.


I'm going to level with you: I am very sleepy right now. I wasn't thinking about the equation you derived when I said that; I was just going off of what I had just written in my text editor... Anyway, I can see your edit that you can get the correct answer with method two, so there's not much point in discussing it any further (in other words, I'd like to look like less of an idiot now, if at all possible)...

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Re: Why can't I get this simple calculus problem right?

Postby jaap » Mon May 26, 2008 11:52 am UTC

Patashu wrote:
jaap wrote:I get the same answer as my first one. It looks like you're closer this time, but you got the coefficient of the middle term wrong. I think the final answer is then 41.3, not what you said the correct answer was.

Aha, you're right; I got the second co-efficient completely wrong somehow. Now I get r = 10sqrt(21t^2-30t+25) therefore dr/dt = 210t-150/sqrt(21t^2-30t+25) which at t = 3 approximately equals 43.1km/h.
Now how did you get 41.3?

I don't know. Now that I plug in the numbers again, I also get 43.1. I'm better at algebra than arithmetic.

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Re: Why can't I get this simple calculus problem right?

Postby Cosmologicon » Mon May 26, 2008 3:07 pm UTC

Patashu wrote:da/dt = 40 and db/dt = 50. Letting t = 0 be the point where A is at the intersection, a = 0 + 40t and b = -50 + 50t.

The second method I tried was to use the cosine rule, drawing a and b as sides of a triangle changing with time, the distance between the two points c the third angle of the triangle opposing an angle of 60 degrees. Thus, r = sqrt(a^2+b^2-ab). This yielded r = 10sqrt(21t^2+15t+25), a curiously different equation from my last. Differentiating gave dr/dt = 210t+75/sqrt(21t^2+15t+25). When t = 3, dr/dt = 705/sqrt(259) approximates 43.8 km/hr, close but certainly not the right answer.

This is just a personal preference thing, but I like to use implicit differentiation to avoid as much algebra as possible. I wouldn't have solved for r(t). I would have implicitly differentiated r2 = a2 + b2 - ab to get:

2 r dr/dt = 2 a da/dt + 2 b db/dt - a db/dt + b da/dt

Then I'd plug in a, b, r, da/dt, and db/dt. I don't know if it's any faster or easier, but I like not having to deal with any quadratics.


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