Find the equation for the tangent line to the curve at the point defined by the given value of t.also find d²y/dx² at this point x=SECt, y=TANt,where t=π/4.
x(?/4) = ?2
y(?/4) = 1
dy/dx = (dy/dt)/(dx/dt) = sec^2(t)/(sect tant) = sect/tant = csc t
so at t=?/4, y' = ?2 and the tangent line is
y-1 = ?2 (x-?2)
y = ?2 x - 1
Note that x^2 = 1+y^2. So, to check the graphs, see
http://www.wolframalpha.com/input/?i=plot+x%5E2-y%5E2%3D1,+y%3D%E2%88%9A2+x+-+1
for the 2nd derivative,
d²y/dx² = d/dx (dy/dx)
= d/dt(dy/dx) / dx/dt
= -cot^2(t)/(sect tant)
= -cos^4(t)/sin(t)
To find the equation for the tangent line to the curve at the point (x, y) with a given value of t, you need to find the derivative of y with respect to x, and then evaluate it at that specific point.
Given x = sec(t) and y = tan(t), where t = π/4:
1. Start by finding dy/dx, which represents the derivative of y with respect to x.
To do this, you can use the chain rule as follows:
dy/dt = dy/dx * dx/dt
Differentiating x = sec(t) with respect to t gives:
dx/dt = sec(t) * tan(t)
Differentiating y = tan(t) with respect to t gives:
dy/dt = sec^2(t)
2. Now, solve for dy/dx. Divide both sides of the equation dx/dt = sec(t) * tan(t) by dy/dt = sec^2(t):
dy/dx = dy/dt / dx/dt
= sec^2(t) / (sec(t) * tan(t))
= 1/tan(t)
= cos(t)/sin(t)
= cos(t) * csc(t)
3. Evaluate dy/dx at the point x = sec(t), y = tan(t), and t = π/4.
Since sec(t) = 1/cos(t) and tan(t) = sin(t)/cos(t), we have:
x = 1/cos(t) = √2
y = sin(t)/cos(t) = 1
Substituting these values into dy/dx = cos(t) * csc(t):
dy/dx = cos(π/4) * csc(π/4)
= (√2/2) * (2/√2)
= 1
So, at the point (x, y) = (√2, 1) with t = π/4, the value of dy/dx is equal to 1.
4. Finally, find d²y/dx², the second derivative of y with respect to x.
Differentiating dy/dx = cos(t) * csc(t) with respect to x gives:
d²y/dx² = d/dx [cos(t) * csc(t)]
= -cos(t)^2 * sec(t) * csc(t)
= -cos(t)^2 / sin(t)
= -cos(t) * cot(t)
Now, evaluate d²y/dx² at the point (x, y) = (√2, 1) with t = π/4:
d²y/dx² = -cos(π/4) * cot(π/4)
= -(√2/2) * (1/1)
= -√2/2
Therefore, at the point (x, y) = (√2, 1) with t = π/4, the value of d²y/dx² is equal to -√2/2.
In summary, the equation for the tangent line to the curve at the point (x, y) = (√2, 1) with t = π/4 is y = mx + b, where m represents the value of dy/dx = 1 and b represents the y-intercept. To find the value of b, substitute the point (x, y) = (√2, 1) into the equation:
1 = (1)(√2) + b
b = 1 - √2
Therefore, the equation of the tangent line is y = x + (1 - √2).