find the tangent line equation at x=cost+sin2t, y=sint+cos2t, t=0
y = sin t + cos^2 t - sin^2 t
dy/dt = cos t -2cost sint -2 sint cost
at t = 0 dy/dt = 1
x = cos t +2 sin t cos t
dx/dt = -sint -2sin^2t+2 cos^2t
at t = 0 dx/dt = 2
dy/dt / dx/dt = 1/2 = slope
x = 1 and y = 1 at t = 0
y = 1 x + b
b = 0
y = x
x = 1 and y = 1 at t = 0
y = .5 x + b
1 = .5 + b
b = .5
y = .5 x + .5
the 2s aren't exponents, is the process the same?
To find the equation of the tangent line at a specific point on a curve, you need to determine the slope of the curve at that point.
Given the parametric equations x = cos(t) + sin(2t) and y = sin(t) + cos(2t), we can find the tangent line equation at t = 0 by following these steps:
1. Calculate the derivatives dx/dt and dy/dt.
The derivative of x with respect to t, dx/dt, represents the rate of change of x with respect to t. Similarly, the derivative of y with respect to t, dy/dt, represents the rate of change of y with respect to t.
dx/dt = -sin(t) + 2cos(2t)
dy/dt = cos(t) - 2sin(2t)
2. Evaluate the derivatives at t = 0.
Substituting t = 0 into the derivatives, we get:
dx/dt = -sin(0) + 2cos(0) = -1 + 2 = 1
dy/dt = cos(0) - 2sin(0) = 1 - 0 = 1
So, at t = 0, dx/dt = 1 and dy/dt = 1.
3. Use the derivatives to calculate the slope of the tangent line.
The slope of the tangent line is given by the derivative of y with respect to x, which is dy/dx. We can calculate dy/dx using the formula:
dy/dx = (dy/dt) / (dx/dt)
Substituting the values we found earlier:
dy/dx = 1 / 1 = 1
Therefore, the slope of the tangent line at t = 0 is 1.
4. Find the coordinates of the point on the curve at t = 0.
Substituting t = 0 into the given parametric equations, we get:
x = cos(0) + sin(2(0)) = 1 + sin(0) = 1 + 0 = 1
y = sin(0) + cos(2(0)) = sin(0) + cos(0) = 0 + 1 = 1
So, the point on the curve at t = 0 is (1, 1).
5. Write the equation of the tangent line.
We now have the slope of the tangent line (m = 1) and the point on the curve (1, 1). Using the point-slope form of a linear equation:
y - y1 = m(x - x1)
Substituting the values:
y - 1 = 1(x - 1)
Simplifying:
y - 1 = x - 1
y = x
Hence, the equation of the tangent line at x = cos(t) + sin(2t), y = sin(t) + cos(2t), t = 0 is y = x.