Find the points of horizontal tangency to the polar curve. r = a sin ose<, a...
rose 3 sin (40) - Find all points 0 <0 < 27 where the curve r = 2 - 4 cos 0 has vertical or horizontal unes.
1. Find the points of horizontal and vertical tangency (if any) to the polar curve r = 2 – 2 cos(0)
6. Let a curve be parameterized by x = t3 – 9t, y=t+3 for 1 st < 2. Find the xy coordinates of the points of horizontal tangency and vertical tangency.
3 TT Find the slope of the tangent line to polar curve r = 7 – 6 sin 0 at the point ( 7 – 6- 2 2 3 TT TT Find the points (x, y) at which the polar curve r = 1 + sin(e), 0 < has a vertical 4 4. and horizontal tangent line. Vertical Tangent Line: Horizontal Tangent Line:
Find two sets of polar Coordinates for the point for os @ <211. - r = smaller Value large ualere
Let C be the closed curve defined by r(t) = costi + sin tj + sin 2tk for 0 <t< 27. (a) (5 pts] Show that this curve C lies on the surface S defined by z = 2xy. (b) (20 pts] By using Stokes' Theorem, evaluate the line integral F. dr C where F(x, y, z) = (y2 + cos x)i + (sin y +22)j + xk
(2 points) Find the exact length of the curve y = In(sin(x)) for #/6 <</2. Arc Length Hint: You will need to use the fact that ſesc(x) dx = In|csc() - cot(3) + C.
-/5 points Trig2 5.3.061. in radians. Let ose < 2.) Write 21 and 22 in polar form. (Express z = √3+1, 32= 1 + √3i 1 = Find the product z122 and the quotients and (Express your answers in polar form with expressed in radians.) Z1 Z2 = Need Help? Read it Watch it
1. Given the polar curve: r = 2coso, OSO < 21 : a) Name the shape and sketch the graph: m2 21/3 TT/3 511/B 1/6 TT 0 711/6 11T/6 411/3 5/3 31/2 I c) Determine where the curve has horizontal tangents.
3. Find the length of the curve y = for 0 < I<2.