Question

Can you please check my answer for this problem?

Answer 1

Find all points of horizontal and vertical tangency to the curve
\[x=4\cos ^{2}\theta \] \[y=2\sin \theta \]

Answer 2

I got (4,0), (4, pi), (4, 2pi), (0,0), (0, pi) and (0, 2pi)

Answer 3

@AlexandervonHumboldt2 i think your answer is correct i'm not sure though

Answer 4

find dy/dx right?

Answer 5

yep

Answer 6

y' = 2 cos(t)
x' = -8 cos(t) sin(t)

y'=0 gives us horizontals, and x'=0 gives us verticals

Answer 7

Right, and there aren't any horizontals

Answer 8

the issue of course is that when y'=0, x'=0 since they have a common factor

Answer 9

then sin(t) = 0 should define the verticals

Answer 10

Right

Answer 11

I guess I'm mainly not sure if I plugged in the theta values in the right equations... did I?

Answer 12

I plugged 0, pi and 2pi into the given equations.

Answer 13

consider substitution

cos^2 = 1 + sin^2

x = 4(1+sin^2(x)), and y^2/4 = sin^2

x = 4(1+y^2/4) = 4+y^2

Answer 14

its a sideways parabola ...

Answer 15

Okay! Does that affect the solution?

Answer 16

it might help understand the solution

x' = 2y = 0 when y=0 is the vertical tangent

Answer 17

y = 2sin(t) = 0 when t is an integer multiple of pi

Answer 18

right

Answer 19

so the solution set is:

(x,y) such that x=4, y=0

Answer 20

ohh okay

Answer 21

4 cos^2(t) = 4 when is an integer multiple of pi
2 sin(t) = 0 when t is an integer multiple of pi

now the odd ball is i think this thing is only part of a parabola and that it may be bouncing back and forth along the curve .... but i cant scale the graphing program i have to determine that

Answer 22

Oh okay.

Answer 23

I think I got it, actually

Answer 24

cos^2 ranges from 0 to 1, so x would be between 0 and 4

sin ranges from -1 to 1, so y would bounce between -2 and 2