Question

Suppose the minute hand on a clock is 15 inches long and the hour hand is 10 inches long. How fast is the distance between the tips of the hands changing at 3:35 PM?

Answer 1

It's a rates of change problem, and I can't seem to set it up.

Answer 2

Hmmm, first model the equations in terms of \(\theta\).

Answer 3

Hmmm, first model the equations in terms of θ.

Answer 4

Consider that \(x=r\sin\theta,y=r\sin\theta\)

Answer 5

@Opcode

Answer 6

@theonechewbaca Make sense?

Answer 7

is that the formula for the sector that x is in?

Answer 8

\(x=r\cos\theta,y=r\sin\theta\)

Answer 9

yeah that seems to be formula for the sector of the circle, so where would I go from there? I have the lengths of each of the hands (x and y), and also the position they are at.

Answer 10

\[
\frac{dx}{dt} =\frac{d}{dt} r\cos\theta = -r\sin\theta \frac{d\theta}{dt}
\]

Answer 11

This is just the chain rule.

Answer 12

i understand

Answer 13

We know \[
z^2 = x^2+y^2
\]Differentiating with respect to \(t\) gets us: \[
2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}
\]

Answer 14

In this case \(z\) is our distance, and they're asking for \(dz/dt\)

Answer 15

Makes sense, and I know what x and y are equal to

Answer 16

So, I've told you how to find \(dx/dt, dy/dt, x, y,z\) You should be able to do the rest yourself.

Answer 17

Again \(x=r\cos\theta, y=r\sin\theta\) Where \(r\) is the radius (length of hand) and \(\theta\) is the angle

Answer 18

and I find the theta by plugging it into radian form?

Answer 19

So in this case, if its 3:35, the hour hand would be at like -pi/6, and the minute hand would be at -2pi/3?

Answer 20

Now this is 20 minutes away from 15 minutes.

Answer 21

right, but at 35 minutes the hour hand would be between 3 and 4 right?

Answer 22

So the angle is \(-20/60 \times 2\pi\)

Answer 23

Hmmm, that's true...

Answer 24

Umm, after 35 minutes, you have 35/60 hours, and (35/60)/12 rotations.

Answer 25

and at 35 minutes, theta would just be 4pi/3 cause its at the 240 degrees

Answer 26

*5pi/3

Answer 27

Ummm, \(-2\pi/3 = 4\pi/3\)

Answer 28

so now that we've figured out the theta angles for each situation, how do i plug that into the equation to figure out z? I'm thinking that i should put in x=rcos(theta) into the derivative of pythagoreans that we figured out earlier. Do the same for y, and then plug in dx/dt and dy/dt into the same equation and solve for z? But then how do I figure out dz/dt?

Answer 29

You find z once you have found x and y.

Answer 30

yeah plug that into the x^2 + y^2 = z^2 equation?

Answer 31

Then \(dz/dt\) will remain... that is the thing you're trying to find.

Answer 32

awesome! thanks for all the help, im gonna crash right now but im gonna work on it when im up in the morning. I'll tag you in my final post, and if you dont mind id like you to let me know if i got it right or not

Answer 33

how am i supposed to figure out d(theta)/dt? This is what I have so far,
\[2rcos(\theta)\times(-rsin(\theta))\times [d(\theta)\div(dt)]+2rsin(\theta)\times rcos(\theta) \times (d(\theta)/dt)\]All over 2r. Now all I need to figure out is what d(theta)/dt could be

Answer 34

@wio

Answer 35

Remember \(d\theta/dt\) is just how face the hand moves round the clock!

Answer 36

For example, for the minute hand it does 1 revelation for every 60 seconds, so it is going to be \(-2\pi/60\)

Answer 37

The angles are negative because clockwise is backwards.

Answer 38

@wio, But wait, there are two hands in this equation and only one r, shouldn't there be a R and a r, (hour hand and minute hand, respectively) in the equation?

Answer 39

Yes, there are two Rs

Answer 40

Alright let's look at this problem from another angle, using the law of cosines, we could say that the length of the distance between the two hands is equal to \[c^2 = m^2 + h^2 - 2mhcos(\theta)\]

Answer 41

If we plug in the values for the minute hands and the hour hands, we could get a number, then take the derivative of the whole equation right?

Answer 42

We don't need to change anything just because the Rs are different.

Answer 43

I'm not understanding, could you write out the equation?

Answer 44

I'm really busy right now. Just do whatever you think will work.