How do you find parametric equations for the tangent line to the curve with the given parametric equations at the specified point #x = 1+10 * sqrt(t)#, #y = t^5 - t#, and #z=t^5 + t# ; (11 , 0 , 2)?

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Answer 1 · 2016-11-14T18:38:18

Please see the explanation.

Find the value of t that creates the point #(11, 0, 2)#

Substitute 11 for x:

#11 = 1 + 10 sqrt(t)#

#10sqrt(t) = 10#

#t = 1#

check that #t = 1# works for the y and z values:

#1^5 - 1 = 0#
#1^5 + 1 = 2#

These check, #t = 1#

Find the tangent vector:

#dx/dt = 5/sqrt(t)#
#dy/dt = 5t^4 - 1#
#dz/dt = 5t^4 + 1#

The tangent vector for all points, #barv(t)#, is:

#barv(t) = {5/sqrt(t)}hati + {5t^4 - 1}hatj + {5t^4 + 1}hatk#

We are interested in the tangent vector at the given point:

#barv(1) = 5hati + 4hatj + 6hatk#

The vector equation of the tangent line is

#(x,y,z) = (11, 0, 2) + s(5hati + 4hatj + 6hatk)#

The parametric equations for this line are:

#x = 11 + 5s#
#y = 4s#
#z = 2 + 6s#