Question

part c) Give the equation for the tangent line in slope-intercept(ax+by=c)form.
part a gave me a function r(t)= from 0<=t<=2pi. part b gave me the parametric equation for the tangent line to the curve at t=pi/4. the parametric equation was e^(pi/4)<(sqrt(2)/2),(sqrt(2)/2)+(t)sqrt(2)> i believe that the answer should be x=e^(pi/4)sqrt(2)/2 but i have no idea how to get to it

Answer 1

did you derive the answer to b yourself? because it looks off to me

Answer 2

well my professor went over it in class
I may have written it down wrong but i dont think so

Answer 3

i can type out the steps if you would like

Answer 4

r'(t)<e^t(cost-sint),e^t(sin t+cos t>
right?

Answer 5

correct

Answer 6

then r'(pi/4)=e^(pi/4)<0,sqrt2>

Answer 7

right so then the tangent line should be =e^(pi/4)<sqrt2,sqrt2>+te^(pi/4)<0,sqrt2>

Answer 8

which is e^(pi/4)<(sqrt2)/2,(sqrt2/2)+te^(pi/4)sqrt2>

Answer 9

ok now I agree, I didn't see which step you were on
so write out the formulas for x and y explicitly

Answer 10

hey you dropped the t in the x component

Answer 11

i dont think there is one is there?because its te^(pi/4)(0)

Answer 12

first off r(pi/4)=e^(pi/4)<sqrt2/2,sqrt2/2>
you forgot the /2 in one line but you seem to have fixed it in the next

Answer 13

now the point plus t times the tangent vector is what we need for the equation

Answer 14

oh shnap you are right, so that's
x=sqrt2/2
y=sqrt2/2+e^(pi/4)(sqrt2)t

Answer 15

x=e^(pi/4)sqrt2/2
y=e^(pi/4)sqrt2/2+e^(pi/4)(sqrt2)t

Answer 16

ya so i know how to do it when i have 2 t's but now when i have just one

Answer 17

I don't get it, this should just be a vertical line
t does not change x, only y
so it should be the line x=e^(pi/4)sqrt2/2 by my reasoning

Answer 18

ya ok thats what i was told the answer should be but i dont understand how to justify it

Answer 19

like how do you take the paramertic equations we had and get to that reasoning?

Answer 20

didn't I?
at time t=0 y=x
as t changes y does and x does not
that must be a vertical line

Answer 21

you want some more algebra, eh?

Answer 22

haha idk about that but that does make sense

Answer 23

thank you

Answer 24

another way to see it is just by looking at r'(pi/4) it is a vertical direction vector (x-component 0), so again we know it is a vertical line
like I said, plugin in t=0 into the parametric equation for the tangent line shows you where y=0, and hence tells you which vertical line it is
that is the best I can reason it :P

Answer 25

welcome!

Answer 26

ok I thought of a more rigorous way to show it, just find a defined point by plugging in t=0 and you get x=y=e^(pi/4)sqrt2/2
then find the slope by finding\[m={y(b)-y(a)\over x(b)-x(a)}\]for two points t=a and t=b
you will of course see that the slope is undefined since x does not depend on t, which means a vertical line passing through the point we found above