piston and flywheel trig

[bluekitten]

Hi! I'm having problems with a trig problem, and I hope that you can help.

The problem I'm working on shows a setup of a piston and flywheel



First we are asked to express point Q in terms of sin and cos. (45 cos t, 45 sin t)

Then we are asked to express point p as a function of t (they suggest using the distance formula)

doing this, I can get to
2025(sin^2 t + cos^2 t) - 90p cos t +p^2 = 22500

Using a pythagorean id, I can reduce this to 2025 - 90p cos t +p^2 = 22500

and then to - 90p cos t +p^2 = 20475

but I cannot for the life of me figure out how to get p by it's self. can anyone help??

 

[galactus]

You are right, since it is a right triangle, you can just use Pythagoras to find the length of p.

But you can also use the trig identities.

For instance, \(\displaystyle \L\\p=\frac{150}{sin(t)}\)

Of course, we know \(\displaystyle \L\\p=15\sqrt{109}\)

Therefore, \(\displaystyle \L\\t\approx{73.3^{\circ}}\)

Is this what you're getting at?.

 

[bluekitten]

You are right, since it is a right triangle, you can just use Pythagoras to find the length of p.

But you can also use the trig identities.

For instance, \(\displaystyle \L\\p=\frac{150}{sin(t)}\)

Of course, we know \(\displaystyle \L\\p=15\sqrt{109}\)

Therefore, \(\displaystyle \L\\t\approx{73.3^{\circ}}\)

Is this what you're getting at?.

not quite. Here is the full page


I am on # 3. I need to write a function that expresses p in terms of t. the \(\displaystyle \L\\p=\frac{150}{sin(t)}\) might do it, but how did you get there?

 

[galactus]

That's just a basic trig function. Nothing fancy. I assume you know how sine, cosine, tangent work?.

You could also use \(\displaystyle \L\\p=\frac{45}{cos(t)}\)


BUT, I see what you mean now. Atfer looking at the page.

It would appear you have it. You should see the quadratic.

The distance formula would be:

\(\displaystyle \L\\22500=(45cos(t)-p)^{2}+(45sin(t))^{2}\)

\(\displaystyle \L\\22500=2025-90pcos(t)+p^{2}\)

\(\displaystyle \L\\20475=p^{2}-90pcos(t)\)

But, we found that \(\displaystyle \L\\t=73.3\) degrees.

Therefore, we have a quadratic:

\(\displaystyle \L\\p^{2}-90cos(73.3)p-20475=0\)

Solve the quadratic.

One of the solutions will be the one we found the easy way with

\(\displaystyle \L\\\frac{150}{sin(73.3)}\)

The other will be a negative, hence, extraneous.

 

[bluekitten]

That's just a basic trig function. Nothing fancy. I assume you know how sine, cosine, tangent work?.

You could also use \(\displaystyle \L\\p=\frac{45}{cos(t)}\)


BUT, I see what you mean now. Atfer looking at the page.

It would appear you have it. You should see the quadratic.

The distance formula would be:

\(\displaystyle \L\\22500=(45cos(t)-p)^{2}+(45sin(t))^{2}\)

\(\displaystyle \L\\22500=2025-90pcos(t)+p^{2}\)

\(\displaystyle \L\\20475=p^{2}-90pcos(t)\)

But, we found that \(\displaystyle \L\\t=73.3\) degrees.

Therefore, we have a quadratic:

\(\displaystyle \L\\p^{2}-90cos(73.3)p-20475=0\)

Solve the quadratic.

One of the solutions will be the one we found the easy way with

\(\displaystyle \L\\\frac{150}{sin(73.3)}\)

The other will be a negative, hence, extraneous.

That's it! I wasn't seeing the quadratic there. I feel dumb LOL talk about not seeing the forest for the trees...

 

[wjm11]

express point p as a function of t

Perhaps I’m missing something in interpreting the problem, but here’s what I came up with:

Take the large triangle OQP and split it into two smaller triangles by dropping a vertical line from Q down to the x-axis. Let this intersection point be called C. The distance from the origin to C is 45cost. The distance from C to P (by the Pythagorean formula) is (150^2 – (45^2)(sin^2t))^(1/2). Therefore the distance from O to P is the sum of these two distances, and the value of p is:

p = 45cost + (150^2 – (45^2)(sin^2t))^(1/2)

 

[bluekitten]

Perhaps I’m missing something in interpreting the problem, but here’s what I came up with:

Take the large triangle OQP and split it into two smaller triangles by dropping a vertical line from Q down to the x-axis. Let this intersection point be called C. The distance from the origin to C is 45cost. The distance from C to P (by the Pythagorean formula) is (150^2 – (45^2)(sin^2t))^(1/2). Therefore the distance from O to P is the sum of these two distances, and the value of p is:

p = 45cost + (150^2 – (45^2)(sin^2t))^(1/2)

Well, using point-in-the-plane, we found the coordinates of point Q to be (45cos(t), 45 sin(t))

We know that the distance from point Q to point p is 150 centimeters. Thus, we can use the distance formula of

$\displaystyle Length = \sqrt {(x_1 - x_2 )^2 + (y_1 - y_2 )^2 }$

and we get

$\displaystyle 150 = \sqrt {(45\cos (t) - p)^2 + (45\sin (t) - 0)^2 }$

which becomes
$\displaystyle 150^2 = \left( {\sqrt {(45\cos (t) - p)^2 + (45\sin (t) - 0)^2 } } \right)^2$

$\displaystyle 22500 = (45\cos (t) - p)^2 + (45\sin (t) - 0)^2$

$\displaystyle 22500 = 2025\cos ^2 (t) - 90\cos (t)p + p^2 + 2025\sin ^2 (t)$

$\displaystyle 22500 = 2025(\sin ^2 (t) + \cos ^2 (t)) - 90\cos (t)p + p^2$

Using the pythagorian identity, we know that

$\displaystyle \sin ^2 (t) + \cos ^2 (t)$

is equal to 1. so, therefore, this can be re-written as

$\displaystyle 0 = p^2 - 90\cos (t)p - 20475$

and then plugged into the quadratic equation, I get my function.

$\displaystyle p(t) = {{90\cos (t) + \sqrt {8100\cos ^2 (t) + 81900} } \over 2}$

Does that make sense?

A bit off topic, but I've finished my project, anyone want to take a look at it within the next couple of hours and tell me what they think? I can e-mail it - it's in word format.