Related Rates Question- 2 cars leave at 10 am... thanks!

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math

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2 cars leave at 10 am. 1 car travels at 40 mph. other car at 50 mph. the cars are separated by an angle of 120 degrees. How fast does distance change at 12 noon.

here's my attempt:

dA/dt is 40 dB/dt is 50 find dC/dt at what?
since 10 to 12 is 2 hours, A = 80 B= 100 and use law of cosines to find c
c^2 = a^2 + b^2 - 2abcosC c =
im having trouble finding the derivative of law of cosines, specifically the past 2abcosC
is that part (2a)(-sinC)db/dt + (bcosC(2a)(da/dt)

thanks for your help!
 
Let's give this a go. Check me out and make sure I didn't veer off the beaten path.

Law of Cosine is, indeed, the thing to use.

\(\displaystyle c^{2}=a^{2}+b^{2}-2abCos({\theta})\)

From 10 am to 12 noon the vehicles travel for 2 hours. That's 80 miles for A and 100 miles for B.

Differentiate:

\(\displaystyle \L\\2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}-2ab(-1/2)\)

\(\displaystyle \L\\2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}+a\frac{db}{dt}+b\frac{da}{dt}\)

\(\displaystyle \L\\2c\frac{dc}{dt}=(2a+b)\frac{da}{dt}+(a+2b)\frac{db}{dt}\)

We need dc/dt, knowing da/dt=40 and db/dt=50, a=80, b=100.

Use the law of cosines to find c when a=80 and b=100 and theta equals 120 degrees.

We find \(\displaystyle c=20\sqrt{61}\) when a=80 and b=100 and theta=120 deg.

\(\displaystyle \L\\2(20\sqrt{61})\frac{dc}{dt}=(2(80)+100)(40)+(80+2(100))(50)\)

\(\displaystyle \L\\\frac{dc}{dt}=\frac{24400}{40\sqrt{61}}\approx{\fbox{78.1}}\)


**EDIT: Soroban's answer is different than mine, which leads me to believe I may have an error. If anyone spots it let me know.
 
Friendly Note:

Please try to make sense when you write something.

"im having trouble finding the derivative of law of cosines"

I dare you to tell me what that means.

Hint:

You may wish to observe that your angle is fixed.

cos(120º) = -1/2

That should simplify a few things.
 
Hello, math!

Here's my approach to the problem . . .

Two cars leave at 10 am.
One car travels at 40 mph; the other car at 50 mph.
The cars are separated by an angle of 120°.
How fast does distance change at 12 noon?
Click to expand...
Code: 
    A *
       \  *
        \     *     x
     40t \        *
          \           *
           \ 120°         *
            * - - - - - - - - * B
          O         50t

The first car leaves point \(\displaystyle O\) at 40 mph.
In \(\displaystyle t\) hours, it has travelled \(\displaystyle 40t\) miles to point \(\displaystyle A\).

The other car leaves point \(\displaystyle O\) at 50 mph.
In \(\displaystyle t\) hours, it has travelled \(\displaystyle 50t\) miles to point \(\displaystyle B\).

Angle \(\displaystyle AOB \,=\,120^o.\;\;\)Let \(\displaystyle x\,=\,AB\).


Law of Cosines:
. . \(\displaystyle x^2\;=\;(40t)^2\,+\,(50t)^2\,-\,2(40t)(50t)\cos120^o \;=\;6100t^2\)


Differentiate with respect to time: \(\displaystyle \:2x\left(\frac{dx}{dt}\right)\:=\:12,200t\)
. . and we have: \(\displaystyle \L\:\frac{dx}{dt} \;=\;\frac{6100t}{x}\)


When \(\displaystyle t\,=\,2:\;x^2\:=\:6100\cdot2^2\:=\:24,400\;\;\Rightarrow\;\;x\:=\:20\sqrt{61}\)


Therefore: \(\displaystyle \L\:\frac{dx}{dt}\:=\:\frac{6100(2)}{20\sqrt{61}} \:\approx\:\fbox{78.1\text{ mph}}\)

 
galactus and soroban- thank you very much on your thorough replies.

galactus- O is theta

for the derivative of law of cosines, i am confused how you got the last part :
absin(O) (dO/dt)
and also, when you plugged in the values a few steps later, how did you get this part of the derivative equal to zero. how do we know the value of dO/dt?

i'm not trying to say there's a mistake, but just don't get the previous parts. thanks!

any final conclusions on the answer?
 
math said:
galactus and soroban- thank you very much on your thorough replies.

galactus- O is theta

for the derivative of law of cosines, i am confused how you got the last part :
absin(O) (dO/dt)
Click to expand...

That's because the derivative of cos is -sin.

[quote:1mskvjpr]and also, when you plugged in the values a few steps later, how did you get this part of the derivative equal to zero. how do we know the value of dO/dt?
Click to expand...

Because the angle remains constant. There is no change, therefore \(\displaystyle \frac{d{\theta}}{dt}=0\)

i'm not trying to say there's a mistake, but just don't get the previous parts. thanks!

any final conclusions on the answer?[/quote:1mskvjpr]
 
sorry last thing!

for the derivative of law of cosines, i am confused how you got the last part :
absin(O) (dO/dt)

c^2 = a^2 + b^2 - 2abcosO
to find the derivative of this last part, i get that deriv of cos is -sin, but where does the -2 go. did you use product rule. if so, what did you make "f(x)" and "g(x)" for product rule?

when i tried, i made f(x) = 2a and g(x) = bcosO but i think I'm wrong.
 
I changed my post. I had an error. I differentiated incorrectly.
 
Hello, galactus!

I found your error . . . a silly one.
. . (Is there any other kind?)

Law of Cosine is, indeed, the thing to use.

\(\displaystyle \L c^2\:=\:a^2 + b^2 - 2ab\cos120^o \:=\:a^2 + b^2 + ab\)

From 10 am to 12 noon the vehicles travel for 2 hours.
That's 80 miles for A and 100 miles for B.

Differentiate:

\(\displaystyle \L 2c\frac{dc}{dt}\:=\:2a\frac{da}{dt}+2b\frac{db}{dt}-2ab\left(\text{-}\frac{1}{2}\right)\)

\(\displaystyle \L 2c\frac{dc}{dt}\:=\:2a\frac{da}{dt}+2b\frac{db}{dt}+a\frac{db}{dt}+b\frac{da}{dt}\)

\(\displaystyle \L 2c\frac{dc}{dt}\:=\:3\left(a\frac{da}{dt}+b\frac{db}{dt}\right)\) . ← here!
Click to expand...

We have: .\(\displaystyle \L 2c\frac{dc}{dt}\:=\:\underbrace{2a\frac{da}{dt}}_{\downarrow}+\underbrace{2b\frac{db}{dt}}_{\searrow}+\underbrace{a\frac{db}{dt}}_{\downarrow}+\underbrace{b\frac{da}{dt}}_{\swarrow}\)

. . . . . . . . \(\displaystyle \L2c\frac{dc}{dt} \:=\:(2a + b)\frac{da}{dt} \,+ \,(a + 2b)\frac{db}{dt}\)

 
tkhunny- i'm very sorry. thank you for pointing out that the angle is fixed.
 
DUH!!. Thanks Soroban. After all this time and trip up on algebra. :oops:
 
math said:
tkhunny- i'm very sorry. thank you for pointing out that the angle is fixed.
Click to expand...
No worries. I'm just messing with your head.
 
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