How do you find parametric equations for the tangent line to the curve with the given parametric equations #x=7t^2-4# and #y=7t^2+4# and #z=6t+5# and (3,11,11)?

1 Answer
Feb 22, 2015

The answer is:

#x=3+14t#
#y=11+14t#
#z=11+6t#

The point #(3,11,11)# is for #t=1#, as you can see substituting it in the three equations of the curve.

Now let's search the generic vector tangent to the curve:

#x'=14t#
#y'=14t#
#z'=6#

So, for #t=1# it is: #vecv(14,14,6)#.

So, remembering that given a point #P(x_P,y_P,z_P)# and a direction #vecv(a,b,c)# the line that passes from that point with that direction is:

#x=x_P+at#
#y=y_P+bt#
#z=z_P+ct#

so the tangent is:

#x=3+14t#
#y=11+14t#
#z=11+6t#

N.B. The direction #(14,14,6)# is the same that #(7,7,3)#, so the line could be written also:

#x=3+7t#
#y=11+7t#
#z=11+3t#

that's simplier!