Hello! I need help with this problem, its been irritating me for days! Any help would be greatly appreciated!
Here’s the question..
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 tip of the hands changing at 10:00 (Hint use the law of cosines).
So far this is what I have..
Min hand= a = 8
Hr hand= b = 4
Distance between these two points = c and I am trying to find dc/dt = ?
If you use the law of cosines (c^2=a^2 + b^2 -2ab cos (c) ) and set this up than it would be c^2=8^2 + 4^2 – 2 (8) (4) cos (c). Simplify and you end this c^2= 80-64cos(c). Also at 10:00 the angle is 300 degrees. What am I missing and where should I go from here? I not sure what the next step would be to get the answer. Again, any help would be greatly appreciated!
This problem is discussed in many ways. You might want to start here, and use your numbers:
http://math.stackexchange.com/questions/867482/the-rate-of-change-of-the-distance-between-the-tips-of-the-hands-of-a-clock
Okay this is what I got...
Minute hand = a = 8
Hour hand = b = 4
Distance between a---b = c
And we are looking for dc/dt = ?
The law of cosine states c^2 = a^2 + b^2 2ab cos (c)
So…
C^2 = 8^2 +4^2 – 2(8)(4) cos (c)
Take derivative
2c * dc/dt = 0 + 0 + 2(8)(4) sin (c) * dc/dt
2c * dc/dt = 2 * 8 * 4 * sin (5pi/6) * 11pi/360
To find side “c” use the Pythagorean theorem = c^2 = 8^2 + 4^2 which equals square root of 80.
So…
2 * sqrt(80) * dc/dt = 2 * 8* 4* (1/2) * 11pi/360
2 * sqrt(80)* dc/dt= 44pi/45
Simplify…get dc/dt on its own..
Dc/dt = 11pi sqrt(5)/450 meters per minute
Also, just got the news that we can not use the Pythagorean theorem b/c its not a right triangle that is why we are using the law of cosines...UGH!! If there is any other way to go about this please PLEASE HELP!
To find the distance between the tip of the minute hand and the hour hand, we can use the Law of Cosines. You've correctly set up the equation as c^2 = a^2 + b^2 - 2ab*cos(c), where c is the angle between the minute and hour hands and a and b are the lengths of the minute and hour hands, respectively.
At 10:00, the angle between the minute and hour hands is 300 degrees (or 5π/6 radians). However, we need to find the current rate of change of the distance, which means we need to differentiate the equation with respect to time. Let's denote the distance between the tip of the hands as d, and the rate of change of d with respect to time as dd/dt.
Differentiating both sides of the equation c^2 = 80 - 64*cos(c) with respect to time gives:
2c(dc/dt) = 0 + 64*sin(c)(dc/dt)
(Recall that differentiating cos(c) gives -sin(c)(dc/dt))
Now, we need to find the value of sin(c). Since sin(c) = opposite/hypotenuse, we can use the small angle approximation sin(c) ≈ c (in radians) when c is small. In this case, c is 5π/6 radians, which is not small, so we can't use the approximation. Instead, we need to find an expression for sin(c) using the lengths of the hour and minute hands.
To do this, we can geometrically relate the lengths of the sides with the angle c. Consider the right triangle formed by the watch face, with the minute hand, hour hand, and the line connecting their tips as the sides. The side lengths of this triangle are a (minute hand), b (hour hand), and c (distance we are interested in). We can use the Pythagorean theorem to relate the lengths of these sides:
a^2 + b^2 = c^2
Solving for c, we have:
c^2 = a^2 + b^2
Substituting the given values a = 8 and b = 4, we find:
c^2 = 8^2 + 4^2
c^2 = 80
Taking the square root of both sides, we get:
c = √80
Now that we have the value of c, we can calculate sin(c) using the inverse trigonometric function sin^-1:
sin(c) = sin(sin^-1(√80))
Note that sin^-1(√80) gives us the angle in radians whose sine is √80. Since this angle is not a nice, round number, you may need to use a calculator or a mathematical software to find its value.
Once you have the value of sin(c), you can substitute it back into the equation 2c(dc/dt) = 64*sin(c)(dc/dt) that we obtained from differentiating the original equation. Solve for dc/dt to find the rate at which the distance between the two tips is changing at 10:00.