- Thread starter
- #1

- Jul 18, 2013

- 715

I have a few trivial questions regarding finding equations of the tangent line:

1) Find an equation of the tangent line to the parametric curve:

x = 2 sin 2t y= 2 sin t

at the point \(\displaystyle (\sqrt{3}, 1)\)

The textbook says that point "corresponds to the parameter value \(\displaystyle t= \frac{pi}{6}\)"

How do they know that? After testing for a few minutes, I noticed that tan-1\(\displaystyle (\sqrt{3}/1) \) = \(\displaystyle t= \frac{pi}{6}\) Is that how they determined it?

2) Similar question but

x = t sin t

y = t cos t

point (0, -pi)

Using \(\displaystyle \frac{dy}{dx}\) = \(\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), I got

\(\displaystyle \frac{(cos t -sint)}{(sin t + cos t)}\) = slope of tangent line

I then plugged the "0" from the point (0, -pi) into the equation...

(cos 0 - sin 0) / (sin 0 + cos 0) = 1

Is that right/am I on the right track?

1) Find an equation of the tangent line to the parametric curve:

x = 2 sin 2t y= 2 sin t

at the point \(\displaystyle (\sqrt{3}, 1)\)

The textbook says that point "corresponds to the parameter value \(\displaystyle t= \frac{pi}{6}\)"

How do they know that? After testing for a few minutes, I noticed that tan-1\(\displaystyle (\sqrt{3}/1) \) = \(\displaystyle t= \frac{pi}{6}\) Is that how they determined it?

2) Similar question but

x = t sin t

y = t cos t

point (0, -pi)

Using \(\displaystyle \frac{dy}{dx}\) = \(\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), I got

\(\displaystyle \frac{(cos t -sint)}{(sin t + cos t)}\) = slope of tangent line

I then plugged the "0" from the point (0, -pi) into the equation...

(cos 0 - sin 0) / (sin 0 + cos 0) = 1

Is that right/am I on the right track?

Last edited: