Can someone help me with some "easy" calc?

[magician13134]

I'm really struggling in calculus class, and we have an assignment today that I need to do very well on since I just failed the test over the subject. I just want verification that my answer right...

A policeman is parked 200 ft north of an intersection. He sees a reckless driver who is traveling due east through the intersection. The driver is traveling at 200ft/s. If the police leaves the intersection immediately traveling due south at 400ft/s, find the rate of change of the distance between the police and driver after 2 seconds.

I got -316.228 ft/s...

Thanks

 

[magician13134]

Ok, one more question. My friend and I are debating on this one. We both got the same answer, but through SLIGHTLY different methods, I want to make sure mine is ok.
We were solving a problem and got the same equation for ds/dt, but she simplified a bit BEFORE differentiating, and I didn't.
I had

300=s(50-x)

Which gave me

0=ds/dt(50-x)+s(-dx/dt)

or

ds/dt=[s(dx/dt)]/(50-x)

She simplified the original formula to

300/(50-x)=s

Giving her

[(50-x)(0)-(300)(-dx/dt)]/(50-x)^2

or

[(300)(dx/dt)]/(50-x)^2

Are both of those legal and correct?

 

[JimB]

There is not enough information given to solve this problem, no accellerations for a start.

Also, the problem bears no relation to reality, 400ft/s is equivalent to 272 miles per hour.

JimB

 

[magician13134]

We were supposed to ignore acceleration. Sorry if that was unclear...

 

[MrAl]

Hi there,

Something is definitely wrong here.

For one thing, if the policeman is going south (at any rate) and the
driver is going east (at any rate) and they both leave the intersection
then they are traveling 90 degrees
apart and so their rate of change of distance is going to be positive,
not negative as you indicated.

SO, either i dont understand your problem right or the problem is not
stated correctly, or you got the wrong answer.

Post all of your work to show how you got that answer and we can take it
from there, if you like...

 

[magician13134]

I checked with others at school today, I believe I got it right.

The question was asking for the rate of change of the distance between the two cars. Since the further the police goes, the smaller that gets, it is negative. It was just one differentiation of the quadratic formula... I get it back tomorrow or the next day, so I can let you know for sure if it was right...

 

[MrAl]

A policeman is parked 200 ft north of an intersection. He sees a reckless driver who is traveling due east through the intersection. The driver is traveling at 200ft/s. If the police leaves the intersection immediately traveling due south at 400ft/s, find the rate of change of the distance between the police and driver after 2 seconds.
Thanks

Hi again,

Ok, well is this the way the problem is stated, because i dont see
how a car that is parked 200 feet north of an intersection can
leave that same intersection 'immediately'. Traveling at 400ft/s
(even if that was possible) it would take him 1/2 second just to
reach the intersection, since he is parked 200 feet away from it
and supposedly is traveling south.
Maybe when they say 'due south' they mean he travels south
but also east?

Show us how you got your answer (all the constants) and maybe
we can figure this out. Another correct or incorrect answer to the
problem will not help.
We also need to know if your teacher gave you any special
instructions that apply to problems of this type.

 

[Willbe]

I'm too lazy to do it

Try these guys. . .
**broken link removed**

 

[MrAl]

Hello again,


Well, here is the logic behind my concerns...

The cop starts out at 200 feet north of the intersection.
He has good eyesight (he he) and so he is prompted to start
heading south at 400 ft per sec to catch up with the 'speeder'
he sees, who is traveling at 200 ft per sec east.

The equation for distance between cop and speeder for this problem is:
s(t)=sqrt((400*t-200)^2+(200*t)^2)
because the cop is going at 400*t and the speeder at 200*t
and the cop starts out at -200 feet from the intersection and
they are traveling at right angles to each other.

Now the derivative is:
ds/dt=(1000*t-400)/sqrt(5*t^2-4*t+1)

and solving this for zero (0) we get:
0=(1000*t-400)/sqrt(5*t^2-4*t+1)
and so when ds/dt=0, t equals 0.4 seconds.

What this means is that for any time less than 0.4 seconds the
derivative is negative, and for any time more than 0.4 seconds the
derivative is positive.
Thus, for t=2 seconds, the derivative (rate of change of distance
between cop and speeder) is positive.

This means the problem is not stated correctly or something else
is wrong. For example, perhaps the time t to solve for is 0.2 seconds
rather than 2.0 seconds, which would make the derivative negative
and around -316 ft/sec like the so called 'correct' stated answer to
this problem.

It's easy to make a mistake like this because it only involves leaving
out the decimal point from the .2 to make it incorrectly 2 instead.

This means the original 'correct' statement for the problem probably read like this:

Originally Posted by magician13134 but corrected here...
A policeman is parked 200 ft north of an intersection. He sees a reckless driver who is traveling due east through the intersection. The driver is traveling at 200ft/s. If the police leaves the intersection immediately traveling due south at 400ft/s, find the rate of change of the distance between the police and driver after 0.2 seconds.
Thanks


Note that i changed "2 seconds" to "0.2 seconds".

Answer: -316.2267778 feet per second.


Here is a graph of the distance and the rate of change of distance from t=0 to t=2 seconds...

**broken link removed**

 

[BaCaRdi]

Looks perfect to me

-BaC
Hate when those tests are wrong.lolol

Hello again,


Well, here is the logic behind my concerns...

The cop starts out at 200 feet north of the intersection.
He has good eyesight (he he) and so he is prompted to start
heading south at 400 ft per sec to catch up with the 'speeder'
he sees, who is traveling at 200 ft per sec east.

The equation for distance between cop and speeder for this problem is:
s(t)=sqrt((400*t-200)^2+(200*t)^2)
because the cop is going at 400*t and the speeder at 200*t
and the cop starts out at -200 feet from the intersection and
they are traveling at right angles to each other.

Now the derivative is:
ds/dt=(1000*t-400)/sqrt(5*t^2-4*t+1)

and solving this for zero (0) we get:
0=(1000*t-400)/sqrt(5*t^2-4*t+1)
and so when ds/dt=0, t equals 0.4 seconds.

What this means is that for any time less than 0.4 seconds the
derivative is negative, and for any time more than 0.4 seconds the
derivative is positive.
Thus, for t=2 seconds, the derivative (rate of change of distance
between cop and speeder) is positive.

This means the problem is not stated correctly or something else
is wrong. For example, perhaps the time t to solve for is 0.2 seconds
rather than 2.0 seconds, which would make the derivative negative
and around -316 ft/sec like the so called 'correct' stated answer to
this problem.

It's easy to make a mistake like this because it only involves leaving
out the decimal point from the .2 to make it incorrectly 2 instead.

This means the original 'correct' statement for the problem probably read like this:

Originally Posted by magician13134 but corrected here...
A policeman is parked 200 ft north of an intersection. He sees a reckless driver who is traveling due east through the intersection. The driver is traveling at 200ft/s. If the police leaves the intersection immediately traveling due south at 400ft/s, find the rate of change of the distance between the police and driver after 0.2 seconds.
Thanks


Note that i changed "2 seconds" to "0.2 seconds".

Answer: -316.2267778 feet per second.


Here is a graph of the distance and the rate of change of distance from t=0 to t=2 seconds...

**broken link removed**

 

[danielsmusic]

Is this not just simple pythagoras' theorem?

 

[MrAl]

Is this not just simple pythagoras' theorem?

Hi daniel,

You can approximate it using a technique like that but you need
to get a few things straightened out first.

One is the cop car starts out 200 feet NORTH of the intersection,
not right at the center of it.

Second, in order to approximate like that you need to take a
smaller increment in time. Say, 0.001 second instead of 1 full
second.

Try these two corrections and see if you can come up with
the same answer.

 

[danielsmusic]

You can offset the answer quite easily.

sqr ( (800-200)*(800-200) + 400*400 ) = 721.1102551 feet apart. at 2 seconds.

sqr ( (800-200+(400*0.001))*(800-200+(400*0.001)) + (400 + (200*0.001))*(400 + (200*0.001))) = 721.5540174 feet apart after 2.001 seconds

721.5540174 - 721.1102551 = 0.4437623 feet more distance after 0.001 seconds

0.4437623 / 0.001 = 443.7623 feet per second

 

[MrAl]

Hi again Daniel,

Ok, you got the answer more accurate now, but notice that the
policeman is moving away from the suspect at t=2 seconds?
Since it doesnt make much sense for him to do that, and the
other answers were coming out to around -316 ft/sec (which makes
sense because then he is catching up to the speeder) we had
figured that there was a mistake in the statement of the problem
in that the 2 seconds (two seconds) was really 0.2 seconds (two tenths
of a second) instead and this would make more sense.

The approximation you are using though is still the time derivative,
and is actually the definition of the derivative and is often written
like this:

f'(x)=lim[h->0]((f(x+h)-f(x))/h)

or for h very small is approximated like this:

f'(x)=(f(x+h)-f(x))/h

But for numerical computations it is often approximated
like this because it gives more accurate results:

f'(x)=((f(x+h)-f(x-h))/(h+h))

where again h is kept small like 0.001 or something like that.

The exact derivative for a quadratic isnt hard to find
however:

for
f(x)=2*x^2+15*x+3

the exact derivative is:
f'(x)=4*x+15

This is found by taking the derivative of each term one
by one:

f'(2*x^2)=2*2*x^(2-1)=4*x
f'(15x)=1*15*x^(1-1)=15*x^0=15
f'(3)=0 (derivative of a constant is zero)

and then putting them back together:

f'(2*x^2+15*x+3)=4*x+15

Since this isnt too hard to do we used the exact derivative, but
it does help to check the answer with the approximation.

If you like, try this again with t=2 seconds and t=0.2 seconds and
see what you come up with.

 

[danielsmusic]

Hey MrAl

You've completely lost me, as I am completely ignorant when it comes to calculus. The only math I know is mostly electrical.

0.2 seconds would make a lot more sense. Although I still get a different answer.

sqr (((0.2x400)+200)²+(0.2x200)²) = 282.8427125
sqr (((0.201x400)+200)²+(0.201x200)²) = 283.2670118

283.2670118-282.8427125 = 0.4243993457

0.4243993457 / 0.001 = 424.2993457 feet per second (they are getting closer)

 

[MrAl]

Hi again Daniel,


Well i dont know you seem to have done pretty well in coming up with
the idea in the first place! I think you almost got it too.

Remember the cop car is 200 feet north of the intersection, so this
has to be negative.

Try this:

sqr (((0.2x400)-200)²+(0.2x200)²) = ?
sqr (((0.201x400)-200)²+(0.201x200)²) = ?


and see if you can come up with the same answer, or close.


Also, since the time is so small now (0.2) it helps somewhat to
decrease the increment (0.001) to 0.0001 as this increases accuracy
a bit. If you do it both ways you can get a good idea how this works.

Let me know how it works out...