Rate of increase of a side of a triangle

[oddjobmj]

Hello,

As I believe is normal policy, please don't give me the answer to the problem :) I would, however, greatly appreciate a bump in the right direction.

Homework Statement

Triangle abc: side a=12, side b=15, angle C is increasing at 2 degress/minute. How fast is the length of side c increasing when angle C, theta, is 60 degrees?

Homework Equations

Law of Cosines: b^2=c^2+a^2-2ac*cos(theta)

The Attempt at a Solution

I understand that if I simplify the law of cosines by taking the square root of both sides and then take the derivative of both sides with respect to time(t) I can plug in my a, b, and 60 degree angle into find c' (rate of increase of the length of side c). What I'm having trouble doing is understanding how to deal with taking the derivative of sin(theta) with respect to t. Should I plug in 60* before taking the derivative? Here is what I have so far:

c=[(a^2)+(b^2)-2ab(cos(theta))]^1/2

c'=1/2[2a(db/dt)+2b(da/dt)-2((da/dt)b(cos(theta))+(db/dt)a(cos(theta))-(sin(theta))ab))]

Unfortunately the computer I'm using won't load an equation editor so if this is unreadable I can edit it when I get home. Either way, I'm just getting caught up on how to deal with cos of 60 degrees.

Thank you for your time,
Odd

 

[cragar]

take the derivative of <br /> sin(\theta)
and the multiply it by \frac{d\theta}{dt}<br />
chain rule

 

[oddjobmj]

Thank you for the rapid response :) Having taken the next step in the problem I feel like things became too simple. Since sides a and b (12 and 15) are constant... when I plug in the values for a' and a or b' and b the a and b primes will be 0 which cuts out most of the equation. Here is what I'm getting:

s' = 1/2[-2(-sin(theta)(dtheta/dt)ab)]^-1/2

The issue is that it's giving me something around 0.02 m/minute. However, to check the validity of this result, if you plug 60* into the equation for the law of cosines and then find the difference between that and the result for 62* it's around 0.39 m. I can't figure out where I've gone wrong.

 

[cragar]

I think we can just do
<br /> c^2=a^2+b^2-2abcos(x)<br />
then take the derivative
<br /> 2cc&#039;=2absin(x)\frac{dx}{dt}<br />
then we solve for c' and then plug in all of our values.

 

[oddjobmj]

Yeah, that's what I'm getting. The problem is that equation suggests that side c is increasing at a rate of 22.68 m/minute which is far from accurate.

According to the law of cosines with side a being 12 and side b being 15 and angle C being 60 side c should be sqrt(189). (Result makes sense)

c'=(2*12*15*sin(60)*2)/2*sqrt(189)

 

[oddjobmj]

Is anyone able to help me understand where the discrepancy is? I'm still not able to figure this one out.

Thanks,
Odd

EDIT: I've searched google, these forums, and even tried using wolfram alpha to calculate the derivative of the law of cosines. I can't find any information on this topic. It seems so simple but it's giving me the wrong answer.

 

[HallsofIvy]

You said you got c c&#039;= ab sin(\theta), which is correct, but then you say "that equation suggests that side c is increasing at a rate of 22.68 m/minute which is far from accurate."

No, it does not. Setting a= 12, b= 15, c= 18 and \theta= 60 degrees does not give anything of the sort. Check your arithmetic again!

 

[cragar]

are the units correct. we might need to use radians and not degrees
when i take your number 22.68m/min and i multiply it by pi/180
i get .396 so out dx/dt might need to be in radians

 

[HallsofIvy]

are the units correct. we might need to use radians and not degrees
when i take your number 22.68m/min and i multiply it by pi/180
i get .396 so out dx/dt might need to be in radians

No. c is not an angle so is in neither degrees nor radians. The only angle involved is inside the trig function so, as long as the trig function is calculated properly, it doesn't matter if it is in degrees or radians. sin(60^o)= sin(\pi/3 rad)= \sqrt{3}/2

So this is simply a matter of calculating
\frac{ab}{c}sin(60)= \frac{(15)(12)}{18}\frac{\sqrt{3}}{2}

 

[cragar]

what about the \frac{d\theta}{dt}<br />

 

[oddjobmj]

HallsofIvy, I'm failing to see where you're getting your equation.

Here is what I have(derivative of law of cosines):

2cc'=2absin(theta)*d(theta)/dt

Because it's a derivative with respect to time I have to use the chain rule and add the theta prime on the end. So, with that considered it could make sense to convert to radians which actually gives the exact result I'm getting if I take the difference of the lengths; one at 60* and the other at 62*.

If you plug in our values for the variables into the equation above we get:

2*12*15*sin(60)*(2*/minute)=2*c*c' and c is about 13 meters according to the law of cosines; not sure where we got 18:

144+225-360cos(60)
cos(60)=1/2
c^2=189
c=sqrt(189)

Cragar, I think I'm going to tack on the conversion to radians at the end of the equation there and solve for c'. I'm just not sure why this conversion is necessary.

HallsofIvy, I am definitely still interested in understanding your logic.

Thank you both for your help!

 

[sunnycal]

It is late answer for the OP, but I am leaving it here as the page comes up in Google search and the answer may benefit others. There are two things that the OP (oddjob) need to change.

The most important thing is that when plugging the rate of change 2 deg/min, this should be changed to radian (which becomes 90/Pi). This should take care of the discrepancy in the answer.

The other two minor things are..
Derivative of sqrt(x) will be [1/2sqrt(x)]. Based on this, the derivative of c from law of cosine above is incorrect. It may be an error in tying, but I just wanted to point out.

Do not substitute the value for rate of change until a derivative has been taken. In this case it may not matter, but in many other cases it may matter.