Suppose r(t)= cos(t)i+sin(t)j+(2t)k represents the position of a particle on a helix, where z is the height of the particle above the ground. when it is 20 units above the ground, the particle leaves the helix and moves along the tangent. find parametric equations for this tangent line (pick t so that it is continuous through the time when the particle leaves the helix).

The parametric equations for the tangent line are :

x = Cos(10) - t×Sin(10)

y = Sin(10) + t×Cos(10)

z = 20 + 2t

Step-by-step explanation:

When Z=20:

Z=2t=20 ⇒ t=10

The point of tangency is:

r(10)= Cos(10) i + Sin(10) j + 20 k

We have to find the derivative of r(t) to get the tangent line:

r'(t)= -Sin(t) i + Cos(t) j + 2 k

The direction vector at t=10 is:

r'(10)= -Sin(10) i + Cos(10) j + 2 k

So, the equation of the tangent line is given by:

x = cos 10 -t×Sin(10)

y = sin 10 + t×Cos(10)

z = 20 + 2t

alright well the answers to your question is

2xy/9 hope this have a nice day : )

step-by-step explanation:

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.