I've been struggling with this problem for days and I still can't figure out an answer, and none of my peers are able to help either.
"The minute hand on a watch is 8 mm long and the hour hand is 4 mm long. How fast is the distance between the tips of the hands changing at one o'clock?"
I've set A = 4, B = 6.928 (based on Pythagorean theorem), and C = 8. I've tried to find dB/dt after finding the derivative of A^2 + B^2 = C^2, but since the hands on the clock aren't changing dA/dt and dC/dt are both 0 and thus that doesn't help anything.
Then I tried following a similar example in the book, and I'm trying to find dθ/dt (according to the book, this is what I should be looking for), θ being 30. So I get B/4 = tanθ, and B = 4tanθ (again, following example in the book).
But after differentiating it, I STILL have two variables, dB/dt and dθ/dt, and I can't solve for either. I'm trying everything I can think of and nothing is working. Please help!! :(
"The minute hand on a watch is 8 mm long and the hour hand is 4 mm long. How fast is the distance between the tips of the hands changing at one o'clock?"
I've set A = 4, B = 6.928 (based on Pythagorean theorem), and C = 8. I've tried to find dB/dt after finding the derivative of A^2 + B^2 = C^2, but since the hands on the clock aren't changing dA/dt and dC/dt are both 0 and thus that doesn't help anything.
Then I tried following a similar example in the book, and I'm trying to find dθ/dt (according to the book, this is what I should be looking for), θ being 30. So I get B/4 = tanθ, and B = 4tanθ (again, following example in the book).
But after differentiating it, I STILL have two variables, dB/dt and dθ/dt, and I can't solve for either. I'm trying everything I can think of and nothing is working. Please help!! :(
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|C= 8
|
|./
|/A=4
You cannot use pythagorean, since it is not a right triangle.
Divide the circle into minutes: 1 minute= 1/60 of the circle = 6 degrees or 2pi/60= pi/30 rads
The minute hand (C) is moving at (pi/30)rads per min
dC/dt= pi/30 rads/min
The hour hand (A) is moving at 5 minutes per hour, or (1/12) circle per hour,
The speed of the tip of the hour hand is dA/dt= (pi/6)/hour = (pi/360) rads/min
-----
Let R= radian measure between the hands
The minute hand is moving faster, so dR/dt = dC/dt- dA/dt= pi/30-pi/360= 11pi/360 rads/min
-----
To find The distance, B, between A and C :
Law of cosines:
B^2= 4^2+ 8^2 -2(4)(8)cosR
B= sqrt[ 80-64cosR]
dB/dt= (1/2)[80-64cosR]^(-1/2) * (64sinR)* dR/dt
At 1:00, R= 30 degrees, Or pi/6 rads
CosR= sqr(3)/2 and sinR= 1/2
So dB/dt= (1/2)[ 80-32sqrt3]^(-1/2) * 32* (11pi/360)
= [(176pi)/360]* [80-32sqrt3]^(-1/2)
= 0.309...
(mm per min)
Doublecheck the arithmetic:)
Hoping this helps!
|
|./
|/A=4
You cannot use pythagorean, since it is not a right triangle.
Divide the circle into minutes: 1 minute= 1/60 of the circle = 6 degrees or 2pi/60= pi/30 rads
The minute hand (C) is moving at (pi/30)rads per min
dC/dt= pi/30 rads/min
The hour hand (A) is moving at 5 minutes per hour, or (1/12) circle per hour,
The speed of the tip of the hour hand is dA/dt= (pi/6)/hour = (pi/360) rads/min
-----
Let R= radian measure between the hands
The minute hand is moving faster, so dR/dt = dC/dt- dA/dt= pi/30-pi/360= 11pi/360 rads/min
-----
To find The distance, B, between A and C :
Law of cosines:
B^2= 4^2+ 8^2 -2(4)(8)cosR
B= sqrt[ 80-64cosR]
dB/dt= (1/2)[80-64cosR]^(-1/2) * (64sinR)* dR/dt
At 1:00, R= 30 degrees, Or pi/6 rads
CosR= sqr(3)/2 and sinR= 1/2
So dB/dt= (1/2)[ 80-32sqrt3]^(-1/2) * 32* (11pi/360)
= [(176pi)/360]* [80-32sqrt3]^(-1/2)
= 0.309...
(mm per min)
Doublecheck the arithmetic:)
Hoping this helps!
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You cannot apply Pythagoras theorem
You must work with cosine law
find AB in function of the angle and then differentiate
the variable is only the angle the hands form around one 'o clock
As the minute hand rotates by 360° in one hour, it rotates by 0.1° in 1 second
the hour hand rotates by 30° in one hour then it rotates by 0.0083° in 1 second
the angle between the two hands is
a = 30°- 0.09166°t =
the distance of the tips is
AB = √(4^2 + 8^2 - 2·4·8·COS(30°- 0.09166°t))
AB = √(80 - 64 cos(30°- 0.09166°t))
my advice is to convert in radians
my guess is 0.34 mm/sec
calculated in a very basic way :)
http://imageshack.us/a/img23/4194/clockn…
You must work with cosine law
find AB in function of the angle and then differentiate
the variable is only the angle the hands form around one 'o clock
As the minute hand rotates by 360° in one hour, it rotates by 0.1° in 1 second
the hour hand rotates by 30° in one hour then it rotates by 0.0083° in 1 second
the angle between the two hands is
a = 30°- 0.09166°t =
the distance of the tips is
AB = √(4^2 + 8^2 - 2·4·8·COS(30°- 0.09166°t))
AB = √(80 - 64 cos(30°- 0.09166°t))
my advice is to convert in radians
my guess is 0.34 mm/sec
calculated in a very basic way :)
http://imageshack.us/a/img23/4194/clockn…