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)?

Question

Christopher Allers · Accepted Answer

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!

Samantha Adkison · Answer

undefined To find the parametric equations for the tangent line to the curve at the point (3,11,11), follow these steps:

1. Find the values of t at the point of interest by substituting x=3, y=11, and z=11 into the parametric equations.
2. Differentiate the parametric equations for x, y, and z with respect to t to find the derivatives dx/dt, dy/dt, and dz/dt.
3. Evaluate the derivatives at the t-value found in step 1 to get the slopes of the tangent line.
4. Use the point-slope form of a line to write the parametric equations of the tangent line.

Let's go through the steps:

1. Substitute x=3, y=11, and z=11 into the parametric equations:
   $ 3 = 7t^2 - 4 $,
   $ 11 = 7t^2 + 4 $,
   $ 11 = 6t + 5 $.

Solve the third equation to find $ t = 1 $.

2. Differentiate the parametric equations:
   $ \frac{dx}{dt} = 14t $,
   $ \frac{dy}{dt} = 14t $,
   $ \frac{dz}{dt} = 6 $.

3. Evaluate the derivatives at $ t = 1 $:
   $ \frac{dx}{dt} = 14 $,
   $ \frac{dy}{dt} = 14 $,
   $ \frac{dz}{dt} = 6 $.

4. Using the point-slope form of a line $ \frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} $, where $ (x_0, y_0, z_0) $ is the given point and $ a, b, c $ are the directional cosines of the tangent vector, we get:
   $ \frac{x-3}{14} = \frac{y-11}{14} = \frac{z-11}{6} $.

Thus, the parametric equations for the tangent line to the curve at the point (3,11,11) are:
   $ x = 3 + 14t $,
   $ y = 11 + 14t $,
   $ z = 11 + 6t $.

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)?