Question

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)

Answer 1

Answer:

$x=4\sqrt{3}-4t, y=3+4\sqrt{3}t, z=4-6\sqrt{3} t$

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:

$r(t) = 8 cos (t) i + 8 sin (t) j + 6 cos(2t) k$

We can find the derivate like this:

$r'(t)= -8 sin(t) i + 8 cos (t) j -12 sin(2t) k$

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:

$4\sqrt{3}= 8 cos (t)$

$\frac{\sqrt{3}}{2} = cos(t)$

$t = \pi/6$

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

$r'(\pi/6)=-8 sin(\pi/6) i + 8 cos (\pi/6) j -12 sin(2 \pi/6) k$

$r'(\pi/6)=-4 i + 4\sqrt{3} j -6\sqrt{3} k$

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

$r(t) = a + tb$

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

$r(t) = (4 \sqrt{3},3,4) + t(-4, 4\sqrt{3},-6\sqrt{3})$

$r(t)=(4\sqrt{3}-4t, 3+4\sqrt{3}t, 4-6\sqrt{3} t)$

And then the parametric equations are given by

$x=4\sqrt{3}-4t, y=3+4\sqrt{3}t, z=4-6\sqrt{3} t$