Question

y

Answer 1

In your question, do you mean, "with respect to y"?

Answer 2

@jycmbonu ?

Answer 3

Oh, okay, thanks!

Answer 4

So we have to start by making x a function of theta. Then we can take the derivative of x, with respect to theta.
Do you have a good starting point?

Answer 5

Well, let's look at JUST the triangle in the middle of the bigger triangle.

Answer 6

You know your trigonometric functions sine and cosine, right?

Answer 7

Well it looks like you have a right triangle (we should assume so, I believe) and an angle, and the hypotenuse. So what is the length of the side that is x?

Answer 8

Hint: x is the non-hypotenuse side that is ADJACENT to the angle.

Answer 9

One moment please! I was talking as I went, but I saw how to do it. I'll calculate now.
I didn't get that.. But maybe I'm off.
Wolfram alpha and I got 15cos(theta). Wolfram alpha is a computational search engine, at wolframalpha.com. However, it's step by step solution is more complicated than necessary if you use a trig identity.
What work did you do so far? We can go from there!

Answer 10

Wolfram alpha uses "product rule". I simplified the equation before derivating.
For and equation, did you get:
\[x=y* cos(\theta)\]and since y=15*sin(theta)\[x=(15*sin(\theta))*cos(\theta)\]and, with less parenthesis,\[x=15*sin\theta*cos\theta\]

Answer 11

What I did was say \[sin\theta*cos\theta=\frac{1}{2}*sin(2\theta)\]But I can't always remember these trig functions, so product rule is helpful too.

Answer 12

\[[15*sin\theta*cos\theta]\frac{d}{dx}\]

Answer 13

@jycmbonu , I have to leave soon, but I wish you luck. I will be able to help for the next 5 minutes.

Answer 14

Is there anything you can do to make that expression above simpler?

Answer 15

Oh! I've always meant \[\frac{d}{d\theta}\], sorry!

You can factor any constants out of what you're taking the derivative of. Then you can compute\[15[sin\theta*cos\theta]\frac{d}{d\theta}\]

Answer 16

Take care!