To find the parametric equations for the tangent line to the curve at a specified point, we need to find the derivative of the parametric equations and evaluate it at the given point.
1. First, let's find the derivative of each equation.
Take the derivative of x(t), y(t), and z(t) with respect to t:
dx/dt = -4sin(t)
dy/dt = 4cos(t)
dz/dt = -20sin(2t)
2. Now, we need to find the value of t that corresponds to the given point (2√2, 2√2, 0).
Plug in x = 2√2, y = 2√2, and z = 0 into the original equations:
2√2 = 4cos(t)
2√2 = 4sin(t)
0 = 10cos(2t)
3. Solve the equations simultaneously to find the value of t.
Divide the first equation by 4: √2 = cos(t)
Divide the second equation by 4: √2 = sin(t)
From the unit circle, you can see that t = π/4 or t = 5π/4.
4. Evaluate the derivatives at t = π/4 or t = 5π/4.
At t = π/4:
dx/dt = -4sin(π/4) = -2√2
dy/dt = 4cos(π/4) = 2√2
dz/dt = -20sin(2(Ï€/4)) = -20sin(Ï€/2) = -20
5. Now we have the slope of the tangent line, which is the derivative dz/dt divided by dx/dt and dy/dt.
Slope = dz/dt / dx/dt = -20 / (-2√2) = 10/√2 = 5√2.
6. Finally, we can write the parametric equations for the tangent line.
The point-slope form of the line is given by:
x - x1 = t(dx/dt)
y - y1 = t(dy/dt)
z - z1 = t(dz/dt)
where (x1, y1, z1) is the given point.
Substituting the values:
x - 2√2 = t(-2√2)
y - 2√2 = t(2√2)
z - 0 = t(-20)
Simplifying, we get:
x = 2√2 - 2√2t
y = 2√2 + 2√2t
z = -20t
These are the parametric equations for the tangent line to the curve at the point (2√2, 2√2, 0).