Question

As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes through the point (ƒ(t0), g(t0), h(t0)) parallel to v(t0), the curve’s velocity vector at t0. In Exercises 23–26, find parametric equations for the line that is tangent to the given curve at the given parameter value t = t0.

Answer 1

Answer:

$x = t$

$y = \frac{1}{3}t$

$z =t$

Explanation:

Given

$r(t) = f(t)i + g(t)j + h(t)k$ at $t = 0$

Point: $(f(t0), g(t0), h(t0))$

$r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk$, $t0 = 1$ -- Missing Information

Required

Determine the parametric equations

$r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk$

Differentiate with respect to t

$r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k$

Let t = 1 (i.e $t0 = 1$)

$r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k$

$r'(1) = i +\frac{3}{3^2}j + (0 + 1)k$

$r'(1) = i +\frac{3}{9}j + (1)k$

$r'(1) = i +\frac{1}{3}j + (1)k$

$r'(1) = i +\frac{1}{3}j + k$

To solve for x, y and z, we make use of:

$r(t) = f(t)i + g(t)j + h(t)k$

This implies that:

$r'(1)t = xi + yj + zk$

So, we have:

$xi + yj + zk = (i +\frac{1}{3}j + k)t$

$xi + yj + zk = it +\frac{1}{3}jt + kt$

By comparison:

$xi = it$

Divide by i

$x = t$

$yj = \frac{1}{3}jt$

Divide by j

$y = \frac{1}{3}t$

$zk = kt$

Divide by k

$z = t$

Hence, the parametric equations are:

$x = t$

$y = \frac{1}{3}t$

$z =t$