Question

find the equation of the line tangent to the curve at the point defined by the given value of t. x=sec(t),y=tan(t), at t=pi/6

Answer 1

first find dx/dt and dy/dt
can you ?

Answer 2

dx/dt = sec(t)tan(t); dy/dt=sec^2(t)

Answer 3

yes, now dy/dx is just the ratio of those,
\(\huge \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)
then just plug in t=pi/6

Answer 4

dy/dx = sec^2(t)/(sec(t)tan(t))
is that right?

Answer 5

dy/dx = sec(t)/tan(t)

Answer 6

yes, thats correct.
now you can either simplify that or just directly put t=pi/6

Answer 7

Note : dy/dx will give you slope of the required line.

Answer 8

ok I see! thank you!

Answer 9

so, could you solve this Question further on your own ?
if you need further help, you can just ask :)
welcome ^_^

Answer 10

yes, I think can solve the question now, thanks a lot!

Answer 11

:)

Answer 12

oh, and by the way
\(\huge \color{red}{\text{Welcome to Open Study}}\ddot\smile\)