Sinusoidal Graphs word problem help

[perpetual-burn]

Hi, I've done most of my homework, but can't figure this one out. If anyone could explain what to do, I would really appreciate it.

*i got 78 for an answer, but its wrong*

It is exactly noon. The hour hand of a clock is 8cm long. If the top of the clock is 50 cm from the ceiling and the radius of the clock is 15 cm, how far is the tip of the hour hand from the ceiling at 12:35?

I really need help on this, I am really not sure what to do.

 

[perpetual-burn]

Please help... :(

 

[Ouabache]

At first you said it was 12Noon. At that time the hour hand is vertical and the distance from tip to the ceiling would be (50+15)-8 = 57cm

Then you ask, what is the distance from hour hand tip to ceiling at 12:35. The hour hand moves every minute? How much does it move in 1 hour? (in degrees or radians, just be consistent). What fraction of an hour do you have at 35min. Well the 1hr is 60min, so the fraction would be 35/60. Multiply that times the angle (you calculated) the small hand traverses in 1 hour.
Now you have the angle of the hour hand from the vertical at 12:35.

Using basic trigonometry, you can find the height of the tip relative to the center of the clock face.. Try that... if you get stuck just come back and we can give you some more help..

 

[perpetual-burn]

Its supposed to be solved with a cosine graph. Not how you said. The radius has to be incorporated as well. Anyone else givea shot? Thanks anyways for the effort.

 

[jtbell]

Its supposed to be solved with a cosine graph.

What exactly does this mean? Are you supposed to draw a graph of distance (between tip of hour hand and the ceiling) versus time, and read off the distance at 12:35 from that?

Using the information given, you can plot some of the points of the graph with little or no calculation. For example, at t = 0 hours, d = 50 cm. At t = 6 hours, d = 80 cm. Similarly 3 hr, 9 hr, 12 hr, etc. are easy. But what about the intermediate points? Have you been given any guidelines about this? Is it OK simply to draw a freehand sketch of a graph between the points you can find easily, and read the approximate answer off of that? Or do you need something precise? Do you have to come up with a formula for the graph, and calculate the answer from that?

 

[HallsofIvy]

Its supposed to be solved with a cosine graph. Not how you said. The radius has to be incorporated as well. Anyone else givea shot? Thanks anyways for the effort.

Outabache suggested "What fraction of an hour do you have at 35min. Well the 1hr is 60min, so the fraction would be 35/60. Multiply that times the angle (you calculated) the small hand traverses in 1 hour.
Now you have the angle of the hour hand from the vertical at 12:35. "

Ok, if you MUST "use a cosine graph" look up that angle on your cosine graph!

 

[Ouabache]

As HallsofIvy suggested (and I implied by suggesting use of trigonometry $ht_{hourhand}=8cos\theta$), you can use a cosine graph to look up the amplitude at the angle you found for the hour hand at 12:35. But not just any cosine curve. You will need to scale your graph properly.
HINT: Find the graph for cosine in your text. Scale the amplitude of that curve by the length of the hour hand (max amplit = 8cm, min amplit = -8cm).

Once you find the angle of the hour hand, (as I described previously), you can read off the height (of the hour hand) from the graph. (The accuracy of this value will impove by drawing a larger graph). But that is not your final solution. You still need to find the height from the tip of the hour hand, to the ceiling. HINT: you will be doing a similar calculation as I did when the time was exactly 12Noon. Do you think it will be greater or less than the value I calcuated for 12Noon?

Extra: You are lucky the time is not between 3pm and 9pm. You would still be subtracting the height of the hour hand but remember, the relative height is also negative (below the center of clock). Example, at 6pm the distance from 'hour-hand tip' to ceiling is 50+15-(-8) = 73 (but you should have guessed that intuitively).