BC Calculus Help!

[byalexandr]

Hopefully someone out there can help, I have no idea where to start.

 

Say I have this polar equation, where r=θ+sin(2θ)

 

I want to find what angle (θ) corresponds to the point on the curve where x=-2.

 

 

[kingkickolas]

As far as I can tell, there are infinitely many solutions for x=-2. They have to specify a y-value if they want just one angle.

In any case, I would expand sin(2t) to 2 sin(t) cos(t), and multiply both sides by r², so you can replace r cos(t) and r sin(t) with x an y, respectively. Then it's just solve for t. But again, they need to specify a y, or there's a family of solutions. 

[byalexandr]

1 minute ago, kingkickolas said:

As far as I can tell, there are infinitely many solutions for x=-2. They have to specify a y-value if they want just one angle.

In any case, I would expand sin(2t) to 2 sin(t) cos(t), and multiply both sides by r², so you can replace r cos(t) and r sin(t) with x an y, respectively. Then it's just solve for t. But again, they need to specify a y, or there's a family of solutions. 

Well it's implied that the y valur is the point that the two lines intersect, so the point would be (-2,y).

[kingkickolas]

4 minutes ago, byalexandr said:

Well it's implied that the y valur is the point that the two lines intersect, so the point would be (-2,y).

Yeah but there's infinitely many such y values (see the link in my post). So you need to know which one they want, or the best you can do is leave the answer in terms of y.

[byalexandr]

11 hours ago, kingkickolas said:

Yeah but there's infinitely many such y values (see the link in my post). So you need to know which one they want, or the best you can do is leave the answer in terms of y.

So, I figured it out.

 

To find the angle, you use the formula x=r*cos(theta).

 

x is given as -2 and r is given in the original problem. Simply plug in the two and solve for theta.

 

And I see your point in your first post, I forgot to mention that theta max is 2pi

[kingkickolas]

1 hour ago, byalexandr said:

So, I figured it out.

 

To find the angle, you use the formula x=r*cos(theta).

 

x is given as -2 and r is given in the original problem. Simply plug in the two and solve for theta.

 

And I see your point in your first post, I forgot to mention that theta max is 2pi

Yeahhhh you're gonna get a transcendental equation for theta if you try that. Unless they further specify a y or an r or something, then it's not going to be solvable algebraically. That's a pretty weird question for a Calc BC class. Can you post a picture of the problem or something?

 

Side note: if you want more fun with transcendental equations, try finding all 3 solutions to the equation x² = 2^x. Two of the solutions are more or less obvious, but the third is tricky! 

[byalexandr]

51 minutes ago, kingkickolas said:

Yeahhhh you're gonna get a transcendental equation for theta if you try that. Unless they further specify a y or an r or something, then it's not going to be solvable algebraically. That's a pretty weird question for a Calc BC class. Can you post a picture of the problem or something?

 

Side note: if you want more fun with transcendental equations, try finding all 3 solutions to the equation x² = 2^x. Two of the solutions are more or less obvious, but the third is tricky! 

I feel like I'm missing something then from my original explanation. My classmate got 2.786 as the solution for θ. But they used the same formula I did, I'm not sure if I missed a step or something.

[kingkickolas]

9 minutes ago, byalexandr said:

I feel like I'm missing something then from my original explanation. My classmate got 2.786 as the solution for θ. But they used the same formula I did, I'm not sure if I missed a step or something.

That's the correct answer, but there's just no closed form for that solution (you know, something like "sqrt(arccos(1/3))"). You'll have to get the answer numerically. If that's what the question is asking, then you're golden (as long as you know some numerical method to get the answer). Otherwise, you're shit outta luck. That is, unless I'm missing something too. @Dash Lambda, care to comment?

[Dash Lambda]

1 hour ago, kingkickolas said:

That's the correct answer, but there's just no closed form for that solution (you know, something like "sqrt(arccos(1/3))"). You'll have to get the answer numerically. If that's what the question is asking, then you're golden (as long as you know some numerical method to get the answer). Otherwise, you're shit outta luck. That is, unless I'm missing something too. @Dash Lambda, care to comment?

You're not missing anything.

The polar function contains the component r=θ, which cannot be solved explicitly for θ in terms of x or y. Best way to approach it is either numerically or graphically, I'd personally just do it graphically. (Specifically by plotting the function for x you guys already got, x=(θ+sin(2θ))cos(θ).)