Question

Find the gradient of the normal to the curve, at the specified value of t.

\[ (b)~~~ x=\cos^2t~~~~~~~ y=\sin^2t~~~~~~when~~ t = {1\over3}\pi\]

Answer 1

x' = -2costsint?

Answer 2

i think its asking for the same thing? gradient is dy/dx ?

Answer 3

Yes, but I have trig problems :/

Answer 4

x' is right

Answer 5

Ok

y'= 2sintcost?

Answer 6

yep

Answer 7

Ok, now how do I place in the value of x so that

-2costsint
--------- = something?
2sintcost

Answer 8

Would it be -1?

Answer 9

yeah looks like it all cancels
dy/dx = -1

Answer 10

So, it would be

m(-1)=-1

Answer 11

also note sin^2 +cos^2 = 1

--> x+y = 1
y = 1-x this has slope of -1

Answer 12

Ok, true.
Well, since it's at the normal, the answer would be 1?

Answer 13

what is m?

Answer 14

m=slope

Answer 15

yes the normal is perpendicular

Answer 16

Great, thank you!

Answer 17

I might place another one up for reassurance.. and I show you what I do, and see if it's right?

Answer 18

ok

Answer 19

I placed it up