Find an equation for the line tangent to the curve at the point defined by the given value of t. also, find the value of startfraction d squared y over dx squared endfraction d2y dx2 at this point. x equals=sec t , y equals=cos t; tequals= pi over 6

Step-by-step explanation:

Given that a function is parametrized as

we have to find the equation of the line tangent at

First let us find the point of contact by substituting values of t

}

Now let us find derivatives

Slope = 0.75 and one point = (1.157, 0.8660)

Using point slope formula we find equation of tangentline is

a) 3x + 4y = 4√3

b)

Step-by-step explanation:

a) x = sec t

y = cos t

t = π/6

dy/dt = -sin t

dx/dt = sect tant = (1/cost) * (sint /cost)

dx/dt = sint/cos²t

dy/dx = -sint * cos²t/sint

dy/dx = - cos²t

At t = π/6

dy/dx = - cos²(π/6) = -3/4

x = sec(π/6) = 2/√3

y = cos(π/6) = √3/2

The slope, m = dy/de

Equation of the tangent line is:

At t = π/6

By simplifying the equation above, the equation for the tangent line becomes:

3x + 4y = 4√3

b) Find the value of at t = π/6

since

y = 1/x

At t = π,6, sect = 2/√3

i would say b) is your answer

sorry if im wrong

step-by-step explanation:

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