Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = 8 cos(t), y = 8 sin(t), z = 6 cos(2t); (4 3 , 4, 3)

Step-by-step explanation:

We need to find first th tangent vector. And we know that th tangent vector is the derivate of the vectorial function r(t). If we define r(t) like this:

We can find the derivate like this:

We can see that at the point given (4 sqrt (3),3,4) the value for t= pi/6. Because from the equation for x we have this:

Now since we have the value for t we can find the derivate at the point t=pi/6

And from the definition of an equation of a line passing thorugh a point with position vector a and parallel to th bector b:

Then the equation of th tangent line at (4 sqrt (3),3,4) is:

And then the parametric equations are given by

step-by-step explanation:

the answer is 128

1 and 0

step-by-step explanation: