Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

x=4cos(t)
y=4sin(t)
z= 10cos(2t)

(2sqrt2,2sqrt2,0)

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(π/4)) = -20sin(π/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).

To find the parametric equations for the tangent line at the specified point, we need to find the derivative of each parameter with respect to the parameter variable (t), substitute the value of t into the derivatives, and then use the point and the derivatives to form the equation of the tangent line.

Given:
x = 4cos(t)
y = 4sin(t)
z = 10cos(2t)

To find the derivative of each parameter, differentiate x, y, and z with respect to t:

dx/dt = -4sin(t)
dy/dt = 4cos(t)
dz/dt = -20sin(2t)

Next, substitute t = π/4 (corresponding to the point (2√2, 2√2, 0)) into the derivatives:

dx/dt = -4sin(π/4) = -2√2
dy/dt = 4cos(π/4) = 2√2
dz/dt = -20sin(2(π/4)) = 0

The point (2√2, 2√2, 0) corresponds to t = π/4, and we have the derivatives at this point: dx/dt = -2√2, dy/dt = 2√2, and dz/dt = 0.

The parametric equations for the tangent line are as follows:

x = (2√2) + (-2√2)t
y = (2√2) + (2√2)t
z = 0 + 0t

Simplifying the equations, we get:

x = 2√2 - 2√2t
y = 2√2 + 2√2t
z = 0

Therefore, the parametric equations for the tangent line to the curve with the given parametric equations at the point (2√2, 2√2, 0) are:
x = 2√2 - 2√2t
y = 2√2 + 2√2t
z = 0