The hands of a clock in some tower are 4.5m and 2m in length. How fast is the distance between the tips of the hands changing at 9:00? (Hint: Use the law of cosines)
The distance between the tips of the hands is changing at a rate of _______ m/hr at 9:00?
I tried several times, mathlab is stating the correct answer as a two digit integer with the following decimals round to the tenths place.
After following several different paths of approach I am getting a three digit number.
Any help would be great,
Thanks
I followed this example, where am i missing something or going about it wrong?
If we let y be the angle between the two hands and x be the distance between the two tips, then, by the law of cosines, we have:
x^2 = 5^2 + 1.5^2 - 2*5*1.5cos(y)
x^2 = 27.5 - 15cos(y)
Take the derivative of both sides with respect to t, time.
2x*dx/dt = 15sin(y)*dy/dt
Since it is 9:00, the angle between the two hands must be y = π/2. And since there is a right triangle, x = √(5^2 + 1.5^2) = √27.5. In order to find dy/dt, consider the fact that the hours hand goes around 2π in one hour and the minutes hand goes around 2π in 1/60 hour, therefore we have dy/dt = 2π - 2π/(1/60) = -118π. Plug that all in:
2(√27.5)*dx/dt = 15sin(π/2)*(-118π)
2(√27.5)*dx/dt = 15(1)*(-118π)
2(√27.5)*dx/dt = -1770π
(√27.5)*dx/dt = -885π
dx/dt = -885π/√27.5 ≈ -530.184
let Ø be the angle between them
the angular velocity of the minute hand = 2π/60 rad/min = π/30 rad/min
the angular velicity of the hour hand = 2π/(12(60)) or π/720 rad/min
then, so dØ/dt = (π/30 - π/720) rad/min
dØ/dt = 23π/720 rad/min
let the distance between the tips of the hands be d m
d^2 = 4.5^2 + 2^2 - 2(2)(4.5)cosØ
d^2 = 24.25 - 18cosØ
differentiate with respect to t
2d dd/dt = 0 + 18sinØ dØ/dt
Now at 9:00, the angle Ø = 90, and
the angle Ø between the minute and the hour hand is increasing at 23π/720 rad/min
d^2 = 24.25 - 18cos9°0 = 24.25 - 0
d = √24.25
dd/dt = 18 sin90° (23π/720) / 2√24.25
= .1834 m/min
or
11.005 m/hr
check my arithmetic.
Thank you for the response,
i tried the method mentioned above three times and was incorrect each time, i double checked all my work to match the method above.
The correct answer is always just .3 under the answer
All of your arithmetic is right as well, so it is not that.
Could it be do/dt?
o being the angle between the clock hands?
Please any help is greatly appreciated
ive been stuck on this one for days
Well, well, well. It seems like you've got yourself into quite the mathematical pickle, my friend. Don't worry, Clown Bot is here to help you out and hopefully bring a smile back to your face!
Now, let's tackle this problem together. We'll be using the trusty law of cosines to determine the rate at which the distance between the tips of the clock hands is changing at 9:00. But before we dive in, let me remind you that math can be a little sneaky sometimes. So, keep your eyes peeled for any hidden clown tricks along the way!
Alright, let's start solving this equation. The law of cosines states that c^2 = a^2 + b^2 - 2ab * cos(C), where c is the side opposite angle C in a triangle. In this case, we can consider the tips of the hands as two sides of a triangle.
We know that the length of one hand is 4.5m and the length of the other hand is 2m. The angle between the hands at 9:00 is 90 degrees (because it's a right angle).
Plugging these values into the equation, we get:
c^2 = 4.5^2 + 2^2 - 2 * 4.5 * 2 * cos(90)
Now, let me ask you a little clown question: What's the cosine of 90 degrees? It's zero, my friend! So now our equation becomes:
c^2 = 4.5^2 + 2^2 - 2 * 4.5 * 2 * 0
You see, when the cosine is zero, the whole term disappears like magic! So we're left with:
c^2 = 4.5^2 + 2^2
Let's calculate that:
c^2 = 20.25 + 4
c^2 = 24.25
Now, we just need to find the square root of both sides to find the value of c:
c = √24.25
c ≈ 4.93 (rounded to the nearest hundredth)
Great! We have the length of the distance between the tips of the hands. But remember, we need to find how fast it's changing at 9:00. For that, we need to consider the rates of change of each hand.
Now, I hate to break it to you, but without additional information about how fast each hand is moving, we cannot determine the rate at which the distance is changing at 9:00. It's like trying to tell a joke without a punchline, my friend. It just doesn't work!
So, unfortunately, we cannot calculate the rate of change without more information. But hey, at least we had a chance to do some math and poke a little fun along the way, right? Keep that clown spirit alive, my friend, and don't let math get you down!
If you have any other questions or need more math humor, I'm here to help. Just give Clown Bot a shout!
To find the rate at which the distance between the tips of the clock hands is changing, we can use the law of cosines. Let's break down the problem and go step by step.
Let's assume that the distance between the tips of the hands is represented by the letter "D". We want to find the rate at which "D" is changing at 9:00.
First, we need to visualize the situation. The longer hand of the clock (hour hand) has a length of 4.5m, and the shorter hand (minute hand) has a length of 2m. At 9:00, the hour hand points directly at the 9, while the minute hand points to the 12.
Now, let's draw a triangle using the clock hands as two sides. The third side of this triangle represents the distance "D" between the tips of the hands. At 9:00, this triangle is a right triangle.
Next, let's introduce some variables. Let "x" represent the angle between the hour hand and the vertical line, and "y" represent the angle between the minute hand and the vertical line.
Since the hour hand points at the 9, it is perpendicular to the vertical line, so "x" is 90 degrees.
To find the values of "y" and "D", we need to relate them to "x" using trigonometry. We know that the sine of an angle is equal to the length of the opposite side divided by the length of the hypotenuse. In this case, the hypotenuse is the length of the hour hand.
sin(y) = D / 4.5
To find the value of "y", we need to take the inverse sine (arcsine) of both sides:
y = arcsin(D / 4.5)
Now, let's use the law of cosines to find a relation between "D", "x", and "y". The law of cosines states that:
c^2 = a^2 + b^2 - 2ab * cos(C)
In our case, "a" represents the length of the hour hand (4.5m), "b" represents the length of the minute hand (2m), and "c" represents the distance "D" between the tips of the hands.
D^2 = 4.5^2 + 2^2 - 2 * 4.5 * 2 * cos(90 - y)
Simplifying the equation:
D^2 = 20.25 + 4 - 18 * cos(90 - y)
D^2 = 24.25 - 18 * cos(90 - y)
At 9:00, "y" is equal to the angle between the minute hand and the vertical line, which is the complement of "x". In other words, "y" is equal to 90 - x. Since "x" is 90 degrees at 9:00, "y" is 0 degrees.
Plugging in "y" into the equation:
D^2 = 24.25 - 18 * cos(90 - 0)
D^2 = 24.25 - 18 * cos(90)
D^2 = 24.25 - 0
D^2 = 24.25
Taking the square root of both sides:
D = √24.25
D = 4.925 m
Now that we know the distance "D" at 9:00, we need to find the rate at which "D" is changing. To do this, we need to differentiate the equation with respect to time.
Differentiating implicitly with respect to time "t":
2D * dD/dt = 0
2 * 4.925 * dD/dt = 0
9.85 * dD/dt = 0
dD/dt = 0 / 9.85
dD/dt = 0 m/hr
Therefore, the rate at which the distance between the tips of the hands is changing at 9:00 is 0 m/hr.
I hope this explanation helps! Let me know if you have any further questions.