Answer:
The equation of the tangent line to the curve at (-1, 4) is y - 4 =
= (-2e^4/9)(x + 1).
Explanation:
First, we need to find dy/dt and dx/dt using the given parametric equations:
dx/dt = -2e^(-2t)
dy/dt = 8e^(8t) - 6
Then, we can find dy/dx by dividing dy/dt by dx/dt:
dy/dx = (dy/dt)/(dx/dt)
dy/dx = [8e^(8t) - 6] / [-2e^(-2t)]
dy/dx = -3e^(10t) + 4e^(2t)
To find d2y/dx2, we need to take the derivative of dy/dx with respect to t and divide by dx/dt:
d/dt(dy/dx) = d/dt[-3e^(10t) + 4e^(2t)]
d/dt(dy/dx) = -30e^(10t) + 8e^(2t)
d2y/dx2 = [d/dt(dy/dx)] / dx/dt
d2y/dx2 = (-30e^(10t) + 8e^(2t)) / (-2e^(-2t))
d2y/dx2 = 15e^(12t) - 4e^(4t)
Next, we can find the point on the curve corresponding to t=2:
x = (2^2 - 1)6 = 30
y = 2cos(pi*2) = -2
To find the equation of the tangent line, we need the slope of the tangent, which is given by dy/dt at t=2:
dy/dt = -2pi sin(pi*2) = 0
Therefore, the slope of the tangent is 0, which means the tangent line is a horizontal line passing through the point (30,-2). The equation of the tangent line is y = -2.
Finally, to find the equation of the tangent(s) to the curve at (-1, 4), we need to find the value(s) of t that correspond to this point. We can set x = -1 and solve for t:
t^3 - 4t - 1 = -1
t^3 - 4t = 0
t(t^2 - 4) = 0
t = 0 or t = ±2
Since the point (-1,4) corresponds to t=2, we can find dy/dt and dx/dt at t=2:
dx/dt = 18
dy/dt = -4e^4
The slope of the tangent is dy/dx, which is given by dy/dt divided by dx/dt:
dy/dx = (dy/dt)/(dx/dt)
dy/dx = (-4e^4)/18
dy/dx = -2e^4/9
So the equation of the tangent line is:
y - 4 = (-2e^4/9)(x + 1)
Therefore, the equation of the tangent line to the curve at (-1, 4) is y - 4 = (-2e^4/9)(x + 1).