# Trig function word problem #2

• Apr 25th 2009, 05:22 PM
skeske1234
Trig function word problem #2
A boat tied up at a dock bobs up and down with passing waves. THe vertical distance between its high and low point is 1.8 m and the cycle is repeated every 4 s.

a) Determine a sinusoidal equation to model the vertical position in metres of the boat versus the time in seconds.

b) Use your model to determine when, during each cycle the boat is 0.5 m above its mean position. Round your answers to the nearest hundredth of a second.

Ok, for part a) my equation is:

V(t)=0.9sin(pi/2)(t)

for part b) I subbed in 0.5 for V(t)
and my answer was 0.37 sec which is correct. However, there are two answers in the back of the book. 0.37 sec and 1.63 sec. I am not sure how to get the second answer: 1.63 sec.

I can't figure out how to get this second time, because since sin is positive in 1st and 2nd quadrant. I got first quadrant, but if you take the 2nd quadrant, pi-0.55555 gives you the angle, which is too big. Sine cannot be over 1, in this case it is 2.58.

How would I figure out the second time?

Thank you
• Apr 25th 2009, 06:20 PM
Shyam
Quote:

Originally Posted by skeske1234
A boat tied up at a dock bobs up and down with passing waves. THe vertical distance between its high and low point is 1.8 m and the cycle is repeated every 4 s.

a) Determine a sinusoidal equation to model the vertical position in metres of the boat versus the time in seconds.

b) Use your model to determine when, during each cycle the boat is 0.5 m above its mean position. Round your answers to the nearest hundredth of a second.

Ok, for part a) my equation is:

V(t)=0.9sin(pi/2)(t)

for part b) I subbed in 0.5 for V(t)
and my answer was 0.37 sec which is correct. However, there are two answers in the back of the book. 0.37 sec and 1.63 sec. I am not sure how to get the second answer: 1.63 sec.

I can't figure out how to get this second time, because since sin is positive in 1st and 2nd quadrant. I got first quadrant, but if you take the 2nd quadrant, pi-0.55555 gives you the angle, which is too big. Sine cannot be over 1, in this case it is 2.58.

How would I figure out the second time?

Thank you

$0.5=0.9 \sin \left(\frac{\pi t}{2}\right)$

$\sin \left(\frac{\pi t}{2}\right)=\frac{5}{9}=0.55556$

$\frac{\pi t}{2}=0.589 \;\;and \;\;\pi-0.589$

$\frac{\pi t}{2}=0.589 \;and \;2.552$

t=0.37, 1.63

Also if you see the graph, (attached), the line at y = 0.5 cuts the positive cycle at two points, first at 0.37 sec, and, second at 2 - 0.37 = 1.63 sec