Question

What is the unit tangent vector at t = 2 on the curve $x = {t^2} + 2$, $y = 4t - 5$, $z = 2{t^2} - 6t$?
(a). $\dfrac{1}{{\sqrt 3 }}(\hat i + \hat j + \hat k)$
(b). $\dfrac{1}{3}(2\hat i + 2\hat j + \hat k)$
(c). $\dfrac{1}{{\sqrt 6 }}(2\hat i + \hat j + \hat k)$

Answer 1

Hint: Find the position vector of any point on the given curve. Then, tangent to the curve at this point is the derivative of the position vector with respect to t. Then, normalize the position vector to get a unit vector by dividing its magnitude.

Complete step-by-step answer:
The tangent to a curve is the line that touches the curve at a point but does not intersect the curve at any other point even when extended infinitely.
The derivative of a vector-valued function gives a new vector that is tangent to the defined curve.
The curve is defined as $x = {t^2} + 2$, $y = 4t - 5$, $z = 2{t^2} - 6t$. The position vector of the point on the curve is given as:
$\overrightarrow r = x\hat i + y\hat j + z\hat k$
Hence, we get the following position vector in terms of t.
$\overrightarrow r = ({t^2} + 2)\hat i + (4t - 5)\hat j + (2{t^2} - 6t)\hat k..........(1)$
The tangent vector to the curve at a point (x, y, z) is obtained by differentiating equation (1) with respect to t.
$\dfrac{{d\overrightarrow r }}{{dt}} = \dfrac{d}{{dt}}[({t^2} + 2)\hat i + (4t - 5)\hat j + (2{t^2} - 6t)\hat k]$
The differentiation of a vector is obtained by differentiating the coefficients of the vectors alone.
$\dfrac{{d\overrightarrow r }}{{dt}} = \dfrac{d}{{dt}}({t^2} + 2)\hat i + \dfrac{d}{{dt}}(4t - 5)\hat j + \dfrac{d}{{dt}}(2{t^2} - 6t)\hat k$
Simplifying, we get:
$\dfrac{{d\overrightarrow r }}{{dt}} = 2t\hat i + 4\hat j + (4t - 6)\hat k$
We need to find the tangent vector at t = 2, hence, we substitute t = 2 in the above equation to get as follows:
$\dfrac{{d\overrightarrow r }}{{dt}} = 2(2)\hat i + 4\hat j + (4(2) - 6)\hat k$
Simplifying, we get:
$\dfrac{{d\overrightarrow r }}{{dt}} = 4\hat i + 4\hat j + (8 - 6)\hat k$
$\dfrac{{d\overrightarrow r }}{{dt}} = 4\hat i + 4\hat j + 2\hat k............(2)$
We now need to normalize equation (2) to get the unit tangent vector. The unit vector is obtained by dividing the vector by its magnitude.
$\overrightarrow T = \dfrac{{\dfrac{{d\overrightarrow r }}{{dt}}}}{{\left| {\dfrac{{d\overrightarrow r }}{{dt}}} \right|}}$
$\overrightarrow T = \dfrac{{4\hat i + 4\hat j + 2\hat k}}{{\left| {4\hat i + 4\hat j + 2\hat k} \right|}}$
$\overrightarrow T = \dfrac{{4\hat i + 4\hat j + 2\hat k}}{{\sqrt {{4^2} + {4^2} + {2^2}} }}$
$\overrightarrow T = \dfrac{{4\hat i + 4\hat j + 2\hat k}}{{\sqrt {16 + 16 + 4} }}$
$\overrightarrow T = \dfrac{{4\hat i + 4\hat j + 2\hat k}}{{\sqrt {36} }}$
We know that $\sqrt {36}$ is 6.
$\overrightarrow T = \dfrac{{4\hat i + 4\hat j + 2\hat k}}{6}$
$\overrightarrow T = \dfrac{1}{3}(2\hat i + 2\hat j + \hat k)$
Hence, option (b) is the correct answer.

Note: You might not find any difficulty doing this question but if you do not normalize the tangent vector to find the unit tangent vector, then your answer will be wrong.Students should remember that the tangent vector to the curve at a point (x, y, z) is obtained by differentiating equation with respect to t and unit tangent vector is obtained by normalizing the vector i.e. dividing the position vector by its magnitude.