Respuesta :
Answer:
[tex]y = 0.75x-0.0026[/tex]
Step-by-step explanation:
Given that a function is parametrized as
[tex]x=sect \\y = cost[/tex]
we have to find the equation of the line tangent at
[tex]t=\frac{\pi}{6}[/tex]
First let us find the point of contact by substituting values of t
[tex]x=sec \frac{\pi}{6} = 1.157\\y = cos \frac{\pi}{6}=0.8660[/tex]}
Now let us find derivatives[tex]dx=sect tant dt\\dy = -sint dt\\\frac{dy}{dx} =\frac{-sint}{sect tant}\\=-0.75[/tex]
Slope = 0.75 and one point = (1.157, 0.8660)
Using point slope formula we find equation of tangentline is
[tex]y-0.8660 = 0.75(x-1.157)\\y = 0.75x-0.8676+0.8660\\y = 0.75x-0.0026[/tex]
Answer:
a) 3x + 4y = 4√3
b) [tex]\frac{d^{2}y }{dx^{2} } = \frac{3\sqrt{3} }{4}[/tex]
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
[tex]dy/dx = \frac{dy/dt }{dx/dt}[/tex]
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:
[tex]y - y_{1} = m(x - x_{1} )[/tex]
At t = π/6
[tex]y - \frac{\sqrt{3} }{2} = \frac{-3}{4} (x - \frac{2}{\sqrt{3} } )[/tex]
By simplifying the equation above, the equation for the tangent line becomes:
3x + 4y = 4√3
b) Find the value of [tex]\frac{d^{2}y }{dx^{2} }[/tex] at t = π/6
[tex]y = cost = 1/sect[/tex]
since [tex]x = sect[/tex]
y = 1/x
[tex]dy/dx = \frac{-1}{x^{2} }[/tex]
[tex]\frac{d^{2}y }{dx^{2} } = \frac{2}{x^{3} } \\\frac{d^{2}y }{dx^{2} } = \frac{2}{sec^{3}t }[/tex]
At t = π,6, sect = 2/√3
[tex]\frac{d^{2}y }{dx^{2} } = \frac{2}{(2/\sqrt{2}) ^{3}}\\\frac{d^{2}y }{dx^{2} } = \frac{3\sqrt{3} }{4}[/tex]
