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Question

Given x=1/2Tan t, y= 1/2 Sec t, t=pi/3

a. find an equation for the line tangent to the curve at the given value of t
b. find the value of d^2y / dx^2 at this point

Answer 1

x = (1/2)tan(t)
y = (1/2)sec(t)

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

(a)
When t = pi/3:
dy/dx = 2 / sqrt(3) = 2sqrt(3) / 3
x = sqrt(3) / 2
y = 1
The equation of the tangent is:
y = ( 2sqrt(3) / 3 ) x + k .............(1)
Putting x = sqrt(3) / 2:
1 = ( 2sqrt(3) / 3) (sqrt(3) / 2 ) + k
k = 0
Hence (1) becomes:
y = ( 2sqrt(3) / 3 ) x.

(b)
d^2y / dx^2
= (d^2y / dt^2) / (dx / dt)
= ( tan^2(t) sec(t) - sec^3(t) ) / tan^3(t)sec(t)
= sec(t) ( tan^2(t) - sec^2(t) ) / tan^3(t)sec(t)
= -1/ tan^3(t)
When t = pi /3, this evaluates to -sqrt(3) / 9.

Answer 2

Question a)
dx/dt = (1/2).sec² t and dy/dt = (1/2).sec t.tan t
dy / dx = (dy / dt) / (dx / dt) = tan t / sec t
dy / dx = (√3 / 2) / 2 = √3 / 4 is gradient , m
At t = π / 3 :-
x = √3 / 2 and y = 1 / (2 cos t) = 1
Tangent passes thro` (√3/2 , 1)
Gradient , m = √3 / 4
y - 1 = (√3 / 4).(x - √3 / 2)
4y - 4 = √3.x - 3 / 2
8y - 8 = 2.√3.x - 3
2.√3.x - 8y + 5 = 0
Question b)
dy / dt = tan t / sec t
d²y/dt² = (sec t.sec t - sec t.tan t) / sec² t
d²y / dt² = 1 - tan t / sec t
d²y / dt² = 1 - sin t = 1 - √3 / 2 = (1/2).(2 - √3)

Answer 3

A= 1/2tansq + 3sec +C

B= d2y= Sec3 squared
d2x= Cos8x