Give the equation of the tangent line to the curve at given point x(t)=2 cos(t) y=2 sin(t) at t=pi/6?
x = 2cos t
y = 2sin t
dx/dt = -2sin t
dy/dt = 2cos t
slope = dy/dx
= (dy/dt) / )dx/dt)
= 2cost/-2sint
= -cost/sint
when t = pi/6
sint = 1/2
cost= √3/2
slope = (-√3/2)/(1/2) = -√3
point: (2cos pi/6, 2sin pi/6)
or (√3 , 1)
So now you have m = -√3, and the point (√3,1)
equation of tangent:
y - 1 = √3(x - √3)
y - 1 = √3x - 3
or
y = √3x - 2
To find the equation of the tangent line to the curve at the given point, we need to find the slope of the tangent line first.
The slope of the tangent line at any point on the curve can be found using the derivative of the curve.
Given:
x(t) = 2cos(t)
y(t) = 2sin(t)
To find the slope of the tangent line, we differentiate each equation with respect to t:
dx/dt = -2sin(t)
dy/dt = 2cos(t)
To find the slope of the tangent line at t=π/6, substitute t=π/6 into the derivatives:
dx/dt = -2sin(π/6) = -2(1/2) = -1
dy/dt = 2cos(π/6) = 2(√3/2) = √3
Therefore, the slope of the tangent line at t=π/6 is -1/√3.
Now, we have the slope of the tangent line and the point at which it intersects (x=2cos(π/6), y=2sin(π/6)).
Using the point-slope form of the equation of a line:
y - y₁ = m(x - x₁),
where m is the slope and (x₁, y₁) is a point on the line, substitute the values:
y - 2sin(π/6) = (-1/√3)(x - 2cos(π/6)).
Simplifying the equation:
y - √3/2 = -x/√3 + 1/√3.
Rearranging the equation to get it in slope-intercept form (y = mx + b):
y = -x/√3 + √3/2 + √3/2.
Combining like terms:
y = -x/√3 + √3.
Therefore, the equation of the tangent line to the curve at (x(t), y(t)) = (2cos(t), 2sin(t)) at t=π/6 is y = -x/√3 + √3.
To find the equation of the tangent line to a curve at a given point, we need to find the slope of the tangent line and the coordinates of the given point.
First, let's find the coordinates of the given point when t = π/6:
x(t) = 2cos(t)
x(π/6) = 2cos(π/6)
x(π/6) = 2 * √3 / 2
x(π/6) = √3
y(t) = 2sin(t)
y(π/6) = 2sin(π/6)
y(π/6) = 2 * 1 / 2
y(π/6) = 1
So, the coordinates of the given point are (x, y) = (√3, 1).
Now, let's find the slope of the tangent line at this point. The slope of a curve at a given point can be found by taking the derivative of the curve and evaluating it at that point.
The derivatives of x(t) and y(t) with respect to t are:
dx/dt = -2sin(t)
dy/dt = 2cos(t)
Now, let's evaluate the derivatives at t = π/6:
dx/dt (π/6) = -2sin(π/6)
dx/dt (π/6) = -2 * 1 / 2
dx/dt (π/6) = -1
dy/dt (π/6) = 2cos(π/6)
dy/dt (π/6) = 2 * √3 / 2
dy/dt (π/6) = √3
So, the slope of the tangent line at t = π/6 is dy/dt (π/6) / dx/dt (π/6) = √3 / -1 = -√3.
Now that we have the slope (-√3) and the coordinates of the given point (√3, 1), we can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values, we have:
y - 1 = -√3(x - √3)
Simplifying:
y - 1 = -√3x + 3
The equation of the tangent line to the curve at the given point is y = -√3x + 4.