A circle is defined by the parametric equations
x = 2cos(2t) and y = 2sin(2t) for t is all real numbers.
(a) Find the coordinates of the point P on the circle when t = (4*pi)/3.
(b) Find the equation of the tangent to the circle at P.
Can you please show your working to help me understand how you reach your answer? Thanks - Jichu
(a) surely you can just plug in a value for t!
x = 2cos(8π/3) = 2(-1/2) = -1
y = 2sin(8π/3) = 2(√3/2) = √3
(b) Now you have a point and a slope, so the tangent line is
y-√3 = 1/√3 (x+1)
Certainly, Jichu! Let's solve the problem step by step.
(a) To find the coordinates of the point P on the circle when t = (4π)/3, we can substitute t into the parametric equations x = 2cos(2t) and y = 2sin(2t).
Given: t = (4π)/3
For x, substitute t into the equation x = 2cos(2t):
x = 2cos(2(4π)/3)
x = 2cos(8π/3)
Now let's simplify the expression inside the cosine function:
x = 2cos(2π + (2π/3))
We know that cos(2π + theta) = cos(theta), so we can simplify further:
x = 2cos(2π/3)
It is known that cos(2π/3) = -1/2, so substituting the value:
x = 2 * (-1/2)
x = -1
For y, substitute t into the equation y = 2sin(2t):
y = 2sin(2(4π)/3)
y = 2sin(8π/3)
Similar to x, let's simplify the expression inside the sine function:
y = 2sin(2π + (2π/3))
sin(2π + theta) = -sin(theta), so we can simplify further:
y = 2 * (-sin(2π/3))
It is known that sin(2π/3) = √3/2, so substituting the value:
y = 2 * (-√3/2)
y = -√3
Therefore, the coordinates of point P on the circle when t = (4π)/3 are P(-1, -√3).
(b) To find the equation of the tangent to the circle at point P(-1, -√3), we can start by finding the slope of the tangent line using the derivative of the parametric equations.
The derivatives of x and y with respect to t are:
dx/dt = -4sin(2t)
dy/dt = 4cos(2t)
To find the slope of the tangent line at P, substitute t = (4π)/3 into the derivatives:
m = dy/dx = (dy/dt)/(dx/dt)
= (4cos(2(4π)/3))/(-4sin(2(4π)/3))
= cos(8π/3)/-sin(8π/3)
Now, let's simplify the expression by using the trigonometric identities:
cos(8π/3) = cos(2π + (2π/3)) = cos(2π/3) = -1/2
sin(8π/3) = sin(2π + (2π/3)) = -sin(2π/3) = -√3/2
Substituting these values into the slope equation:
m = (-1/2) / (-√3/2)
m = 1/√3
So the slope of the tangent line is 1/√3.
Now, we have a point (-1, -√3) on the line and its slope is 1/√3. We can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values:
y - (-√3) = (1/√3)(x - (-1))
y + √3 = (1/√3)(x + 1)
To simplify further, let's rationalize the denominator √3:
y + √3 = (x + 1) / √3
√3(y + √3) = x + 1
Expanding the left side:
√3y + 3 = x + 1
Rearranging the equation to the standard form:
x - √3y = -2
Therefore, the equation of the tangent to the circle at point P is x - √3y = -2.