# Thread: Sorry to ask another question, but I need help with trig functions!

1. ## Sorry to ask another question, but I need help with trig functions!

The question:
When you use a sewing-machine, the needle moves in relation to the sewing plate (where the clothe is). To the time t, the needle's height y(t) over the sewing-plate is decided by:

$y(t)=\frac 32 \cdot sin(12\pi t)$

Where t is meazured in seconds and y(t) is meazured in centimeters.

The needle's velocity to the time t is y'(t). What is the velocity of the needle when it passes the sewing-plate?

I found the y'(t) previously with the help from you guys:

$y'(t)=\frac 32 \cdot 12 \pi \cdot \cos(12 \pi t)$

But what now? I'm a bit lost.

2. I'm kind of confused by the question. If it is asking what the velocity of the needle is when touching the plate you would find t by

$0=\frac 32 \cdot sin(12\pi t)$

Once you find t then plug it in your velocity function.

3. Well you know on a sewing machine, there is this plate where you put the clothes on, and then the needle goes down into the clothes and into this plate. I need to calculate the velocity of the needle when it goes through this plate.

I would set y(t) to zero as well, but it confuses me when it says that y'(t) is the velocity?

4. Ok then the way I told you is correct. The first function is your position function y(t). The position function tells you the height of the needle for any t, t being time. I set the position function to 0 becuase that would be when the needle is touching the plate.

Now y'(t) is your velocity function and tells you the velocity for any t.

Ok so find out what t makes the position equal to 0

$0=\frac 32 \cdot sin(12\pi t)$

So when you solve for t you will find at what time the needle is touching the plate

Now take your value for t and plug it in to your velocity function and it will tell you what the velocity is at the time the needle is touching the plate.

5. The way to solve this question is to calculate $y'(t_1)$, where $t_1$ is the time at which the needle hits the plate. That time is calculated by $0 = \frac{3}{2}\sin {(12\pi{t_1})}$. In other words, at what time is the sine equal to zero? A very simple answer is when t = 0. Therefore, to answer the question of velocity, you solve for $y'(t)$ by substituting $t = 0$ into the velocity equation.

6. Okay, thanks both of you. You are both saying the same, which is to find the value of t in the first equation when y(t) = 0, and then substitute t in the differentiated equation with the value I have found.

$0=\frac 32 \cdot sin(12\pi t)$

$\frac 0{\frac 32}=sin(12\pi t)$

$12\pi t=sin^{-1} (0)$

$t = \frac 0{12\pi}$

$t=0$

And so t is has the value 0. Then 0 substitutes t in the velocity equation:

$y'(t)=\frac 32 \cdot 12 \pi \cdot \cos(12 \pi \cdot 0)$

$y'(t)=\frac 32 \cdot 12 \pi \cdot \cos(0)$

$y'(t)=\frac 32 \cdot 12 \pi \cdot 1$

$y'(t)=56.55$

So that is basically the velocity, right?

One last question, is my answer then 56.55 cm/seconds?

7. Yep that's the right answer and yes $\frac{cm}{s}$

8. Thanks a whole bunch