Few stupid questions about polar curves and stuff

[formula107]

Few stupid questions about polar curves and stuff...

Okay, here's the first dumb question. I have to find the tangent line where r=2-3sin(T) at the polar point (2,pi). To find the slope, you just take dy/dx, and I come up with 2/3. I know that part is right.

T=theta (I know it's spelled wrong)

But how do I turn this into an equation for the tangent line? Do I just use y=mx+b? That seems too simple for this, and I would think she would want it in polar coordinates, not cartesian.

Secondly, what (cos2(T))^2. is it 2cos(2T) or 2 cos(4T) or what? I know we learned this like 2 years ago, but I just can't remember anymore.

Thanks guys!

 

[dextercioby]

Okay, here's the first dumb question. I have to find the tangent line where r=2-3sin(T) at the polar point (2,pi). To find the slope, you just take dy/dx, and I come up with 2/3. I know that part is right.

T=theta (I know it's spelled wrong)

But how do I turn this into an equation for the tangent line? Do I just use y=mx+b? That seems too simple for this, and I would think she would want it in polar coordinates, not cartesian.

Secondly, what (cos2(T))^2. is it 2cos(2T) or 2 cos(4T) or what? I know we learned this like 2 years ago, but I just can't remember anymore.

Thanks guys!

For the first part,i'm really lazy to check whether your calculations are correct,but the answer is yes,once u got the slope,u simply plug it in an straight line's equation.

\cos^{2}2\theta=\frac{1+\cos 4\theta}{2}.

 

[formula107]

For the first part,i'm really lazy to check whether your calculations are correct,but the answer is yes,once u got the slope,u simply plug it in an straight line's equation.

\cos^{2}2\theta=\frac{1+\cos 4\theta}{2}.

Sorry man, but I'm still confused on the first part. So if I have a slope of 2/3, would I just make the equation r=2-3sin(2/3)?

The original was r=2-3sin(T)

Thanks again

 

[HallsofIvy]

To find the slope, you just take dy/dx, and I come up with 2/3. I know that part is right.

There's your first problem: I get -1/3 for the slope of the tangent line at (2, pi). How did you get 2/3?

Once you have the equation of the tangent line in y= mx+ b form, it should be easy to convert to polar coordinates by using x= r cos(θ), y= r sin(θ).

 

[formula107]

There's your first problem: I get -1/3 for the slope of the tangent line at (2, pi). How did you get 2/3?

Once you have the equation of the tangent line in y= mx+ b form, it should be easy to convert to polar coordinates by using x= r cos(θ), y= r sin(θ).

Even the book has the answer as 2/3.

I had: 2cos(T)-6sin(t)cos(T) / -2sin(T) -3(cos^2(T)-sin^2(T)), with T = pi I get -2/-3 = 2/3.

 

[formula107]

There's your first problem: I get -1/3 for the slope of the tangent line at (2, pi). How did you get 2/3?

Once you have the equation of the tangent line in y= mx+ b form, it should be easy to convert to polar coordinates by using x= r cos(θ), y= r sin(θ).

Okay, here's my next problem. Assuming the slope is 2/3, which I sure hope it is, being that that is at -2, the y-intercept would be 4/3, right?

So then I would have y = (2/3)x + 4/3?

How do I convert this to a polar equation? Would I put -2 in for x, and 0 in for y, since that is the cartesian point @ the polar point (2, pi)?

Ah crap, I have have no idea, now I'm lost.

 

[HallsofIvy]

Oops, I dropped the sign from x= -2! Yes, the slope of the tangent line at (2,pi) is, in fact, 2/3. Since we know the line goes through (-2, 0) ((2,pi) converted to rectangular coords), the equation is y= (2/3)(x- 2)= (2/3)x- 4/3 just as you say.

Now, as I said before: " by using x= r cos(θ), y= r sin(θ)"

r sin(θ)= (2/3)r cos(θ)- 4/3.

You can simplify that by solving for r if you like.