Parametric questions help

Question

faiqraees · Answer

x = 1/cos^3 t   y = tan^3 t, dy/dx = sin t
show that the equation of the tangent to the curve at the point with parameter t is
y = x sin t − tan t.

faiqraees · Answer

I know that we have to use the equation y-y1=m(x-x1) I also know that the value of x will be  1/cos^3 t  and  y will be tan^3 t. The m will be the dy/dx. The problem is when I simplify the constant that emerges is sin^2t/cos^3t rather than -tan t

hartnn · Answer

when you plugged in 
x = 1/cos^3 t , y = tan^3 t

you got the equation like:
$\Large \tan^3 t = \dfrac{\sin t}{\cos^3x}+c$

right?

faiqraees · Answer

yes

hartnn · Answer

isolate 'c' 

write tan^3 = sin^3/ cos^3

faiqraees · Answer

although indenominator it should be be t not x

faiqraees · Answer

done

hartnn · Answer

oops yes

you'll get 
$\large c=\dfrac{\sin^3 t - \sin t}{\cos^3 t} $
ok?

faiqraees · Answer

done

hartnn · Answer

factor out sin t

hartnn · Answer

$\Large \sin^2 t - 1 = - (1-\sin^2 t) = -\cos^2 t$
this is the most critical step :)

faiqraees · Answer

oh okay thanks

faiqraees · Answer

is there any way i can get -tan x by going from sin^2 x / cos^3 x

hartnn · Answer

nopes,
`sin^2 x / cos^3 x` is not correct.

hartnn · Answer

how did you get that?

faiqraees · Answer

$$ \large c=\dfrac{\sin^3 t - \sin t}{\cos^3 t}$$
I simplified the numerator by subtacting to get sin^2t . Is that incorrect?

hartnn · Answer

$3a - a =2a$
but 
$a^3-a \ne a^2$

exponents don't work that way :)

faiqraees · Answer

Oh yeah rightttttt. Damn I feel so tupid now

faiqraees · Answer

stupid*

hartnn · Answer

lol happens to all of us :)