Question

HELP!!! Find the curvature of r=2+2sin(t) at t=5pi/4

Answer 1

http://en.wikipedia.org/wiki/Curvature#Curvature_of_a_graph

use the formula for polar form

Answer 2

the derivatives aren't bad.... derivative of sin is cos , derivative of cos is -sin

Answer 3

so I can use r as y equation?

Answer 4

no its different for polar equation....look below it gives the polar formula

\kappa(\theta) = \frac{|r^2 + 2r'^2 - r r''|}{\left(r^2+r'^2 \right)^{3/2}}

Answer 5

\[\kappa(\theta) = \frac{|r^2 + 2r'^2 - r r''|}{\left(r^2+r'^2 \right)^{3/2}}\]

Answer 6

Do I plug in 5pi/4 to get r, r', and r''?

Answer 7

yeah

Answer 8

Ok but first I have to do the derivatives before I plug it in right?

Answer 9

correct

Answer 10

Ok let me see if I can figure it, do you know the answer if I ask?

Answer 11

yes

Answer 12

Ok give me a sec

Answer 13

Um.... is it 0.9799?

Answer 14

i get .135

Answer 15

Ok so for r'=2cos(t) and r''=-2sin(t) right?

Answer 16

r = 2 - sqrt(2)

r' = -sqrt(2)

r'' = sqrt(2)

Answer 17

yes you are right

Answer 18

Ok let me redo this

Answer 19

hold on, i apologize I made a mistake
answer is .9799

Answer 20

i did by hand getting an exact answer first and i went too fast lol

Answer 21

When I redid it I got .199, so my first answer was right?

Answer 22

hmm what did you change?

Answer 23

I just plugged in sqrt2, 2-sqrt2 and -sqrt2. Not the actual decimal.

Answer 24

well the answers should be very close, .199 is way off from .97
check how you entered them into calculator

yes i believe .9799 is correct

Answer 25

Ok I'll redo it again I hate getting different answers lol

Answer 26

i should have done this first
here is confirmation
http://www.wolframalpha.com/input/?i=abs%28a%5E2+%2B2b%5E2+-ac%29%2F%28a%5E2+%2Bb%5E2%29%5E1.5+%2C+a+%3D+2-sqrt%282%29%2C+b+%3D+-sqrt%282%29%2C+c+%3D+sqrt%282%29

Answer 27

Ok I got it thanks for the help :)

Answer 28

:)

Answer 29

Can you help me with one more?

Answer 30

ok

Answer 31

Find the parametric equations of the tangent lines to C at t = pi/6, curve C is r(t) = (4sin(3t))I+(2t)j+(4cos(3t))k.

Answer 32

ok so we need 3 lines, x(t) y(t) z(t)

do each component the same way, the slope of tangent line is derivative at point pi/6
equation of line
\[x(t) - r_x(\pi/6) = r_x'(\pi/6) (t - \pi/6)\]
similar for y,z as well

Answer 33

So would I set it up like this?
\[4\sin(t)-r(\pi/6)=r'(\pi/6)(t-\pi/6)\]

Answer 34

Oops I mean 4sin(3t)

Answer 35

almost, 4sin(3t) = r(t) for x component

Answer 36

x(t) is the dependent variable, leave it alone

Answer 37

Oh so I would plug pi/6 into the 4sin(3t)?

Answer 38

yes

Answer 39

So to find y(t) and z(t) I would just replace the x(t) right?

Answer 40

yes the general form stays the same

also change the r(t) and r'(t) to match the component

Answer 41

Ok which is r(t)=4sin(3t) and r'(t)=12cos(3t)?

Answer 42

yes, plug in pi/6

Answer 43

Is it x(t)=4?

Answer 44

correct

Answer 45

Ah so now I just do y(t) and z(t) the same way?

Answer 46

yep, so find derivative of 2t to get r'(t) for y component

Answer 47

If r'(t)=2 then how do I plug in pi/6?

Answer 48

you cant, no need, the slope is just 2

Answer 49

Oh so y(t)=2t?

Answer 50

correct

Answer 51

Ok let me do the z(t) and see if its right, one min

Answer 52

Is it z(t)=-12t+2pi?

Answer 53

you got it

Answer 54

YAY thank you so much!! I really appreciate it!

Answer 55

yw