<p>Does anybody know how to find the horizontal and vertical tangent lines when
r=2(1-cos (theta))? </p>
<p>Thanks ;)</p>
<p>anybody? please?</p>
<p>Haha, I was thinking about the problem as a was signing in, and I typed my username as “nontangent.”</p>
<p>Anyway, this is a polar curve. A polar curve is a kind of parametric equation, where y = rsin(theta) and x = rcos(theta). So, to find the slopes of the tangent lines, just find dy/d(theta) over dx/d(theta).
Horizontal tangents occur when dy/d(theta) equals zero, because then the whole derivative equals zero.
Vertical tangents occur when dx/d(theta) equals zero, because then the derivative is undefined.</p>
<p>y = rsin(theta)
y = (2(1-cos(theta))(sin(theta))
I’m going to start using T instead of the word “theta” to make this easier to see and type.
dy/d(T) = 2 (sin(T)sin(T) + (1-cos(T))cos(T) )
Simplifying,
dy/d(T) = 2 [ (sin(T)^2) + cos (T) - (cos(T))^2 ]
dy/d(T) = 2 [ 1 - cos(T)^2 + cos (T) - (cos(T))^2 ]
dy/d(T) = 2 [ -2cos(T)^2 + cos(T) + 1 ]
Factoring,
dy/d(T) = 2 (cos(T) - 1) (2cos(T)+ 1)
dy/d(T) equals zero when cost(T) = 1 and cos(T) = -1/2.
So, horizontal tangents may occur at T = 0, 2pi/3, and 4pi/3, as long as dx/d(T) does not equal zero at that point, too–because then you would have a vertical tangent, instead. Looking at the graph, that might be the case with T = 0.</p>
<p>Follow the same process with dx/d(T) to find the vertical tangents.</p>
<p>lols.</p>
<p>omgosh thanks so much! </p>
<p>i was actually having trouble with the vertical tangents…b/c i know ihave to use l’hoptal’s rule somehow, but if you figure out the denominator of dy/dx, i think it’s -2sin(theta)?</p>
<p>and in that case, theta is 0 or pi…but i don’t understand what to do next…</p>
<p>Okay, let’s see…</p>
<p>x = rcos(T)
x = 2(1-cos(T))cos(T) = 2 [ (cos)T - cos(T)^2 ]
Now take the derivative of that.
dx/d(T) = 2 ( -sin(T) - 2cos(T)(-sin(T)) )
dx/d(T) = 2 (2cos(T)sin(T) - sin(T))
Factoring out a sin(T)…
dx/d(T) = 2 (sinT) (2cosT -1)
So, now we solve for T assuming that dx/d(T) = 0.
sin(T) = 0 and cosT = 1/2.
Therefore, T = 0, pi, pi/3, and 5pi/3.
And those are your vertical tangents.</p>
<p>Hi nonexistent, did u get the message by any chance?</p>