# Trigonometric functions - angular measures and tangent curves

1. Dec 17, 2008

### fp252

I encountered a few problems for a few questions while doing my homework.

1. Angular measure problem:
A Ferris wheel with a radius of 25.3m makes 2 rotations every minute.
a) Find the average angular speed of the Ferris wheel in radians per second.
b) How far does a rider travel if the ride lasts 5 min?

For a), I did this:
(2π radians/hr) x (2 rotations/min) x (1min/60s)

That answer was correct one; however, I'm not sure how it works, because I did the question by looking at an example.

for b),
5 x (π/15) x (1 rotation/2π radians) x (2π(25.3)/1 rotation)
= 126.5π/15
= 26.4 m

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The second set of problems deal with tangent curves and their periods. They ask to determine the period and phase shift with respect to y = tan x for each function. The phase shifts I can get right. I know that you can find the period by using (pi/abs(b)), but I the answer in the textbook doesn't match mine, and I don't know why.

a) y = tan (x-(π/4)) ; my answer: π ; book's answer: π/2
b) y = tan (2x-π) ; my answer: (π/2) ; book's answer: π
c) y = tan ((x/2)+(π/2)) ; my answer: (π/3) ; book's answer: π/2

I thought that if something's in a format like tan (2x+1), where there's a number in front of x, then you have to change it to tan 2(x+0.5).

Can anyone explain why my answers are wrong?

2. Dec 18, 2008

### Staff: Mentor

Your answer is numerically correct, but your units are off. The first factor 2π radians/rotation. From the dimensional analysis perspective, you should have radians/rotation * rotations/min * min/sec. The rotations cancel and the minutes cancel, and you are left with radians/sec. Hopefully, understanding how the units work out will help your understanding of this problem.
For b, you'll need to use your answer from part a, and the fact that arc length = radius * angle in radians.

So the total arc length (the distance the rider travels) = radius * radians/sec * time in seconds. The units will be meters * radians/sec * seconds. The seconds units will cancel, and you can ignore the radians units, since they are essentially dimensionless units. You'll end up in units of meters for the distance travelled.
If you know the graph of y = f(x), the graph of y = f(kx), where k > 0, will be compressed toward the y-axis if k > 1, and expanded away from the y-axis if k < 1.
Also, if you know the graph of y = f(x), the graph of y = f(x - a), where a > 0, is translated to the right by a units.
If you have both kinds of transformations going on, do the expansions/stretches first, and then the translations.
a) period: π; phase shift: π/4 (This graph is shifted by π/4 units to the right compared to the basic tangent function.)
b) period: π/2; phase shift: π/2 to the right
c) period: π; phase shift: π to the left.

3. Dec 18, 2008

### fp252

So was the book wrong for the answers of the tangent periods?

Also, I still keep getting 2π as an answer for c)... (through [π/abs(B)])

4. Dec 18, 2008

### Tedjn

I believe your tangent period are right.

5. Dec 18, 2008

### Staff: Mentor

The book's answer for periods appear to be incorrect. Also, my answer for the period in c should have been 2π, not π as I wrote.

6. Dec 18, 2008

### fp252

I have asked the teacher about the answers of the periods of the tangent curves, and he has confirmed that the book's answers were incorrect.

Oh, and thanks again for the explanation for angular speed... now if only I could remember how to do those types of questions on my own. I understand the dimensional analysis part, but I wouldn't know what order I need them to be in (eg. [2π radians/1 rotation] x ???)

7. Dec 18, 2008

### Staff: Mentor

You work with what you have toward where you want to end up. For example, in your first problem, you were trying to find the angular speed in radians/sec. You were given that it makes two rotations per minute.

So that's $$2 * \frac{2 \pi radians}{min}$$, so you need to multiply by units (things that don't change what you multiply) so as to end up with radians/sec.

$$2 * \frac{2 \pi radians}{min} * \frac{1 min}{60 sec}$$

The fraction 1 min/ 60 sec comes from the equation 1 min = 60 sec <==> 1 min / 60 sec = 1. In the previous equation, the min dimensions cancel, and you're left with dimensions of radians/seconds.

In part b, I started with dimensions of radians/sec and I wanted to end up with dimensions of meters, so the dimensions were meters * radians/sec * sec, where the last factor is the number of seconds in 5 minutes. As I said earlier, you can ignore the radians dimension, since it is really a dimensionless quantity.