Question

t=2 ആണെങ്കിൽ x=t²+ 2, y=4t-5, z=2t²-6t² എന്നീ കരുതലുള്ള കർവിൽ ഉള്ള യൂണിറ്റ് ടാൻജന്റ വെക്ടർ ആണ്

• A. $$\frac{11}{\sqrt 3}(i+j+k)$$
• B. $$\frac{2}{\sqrt 5}(2i+j+k)$$
• C. $$\frac{1}{3}(2i+2j+k)$$
• D. $$\frac{11}{3}(i+j+k)$$

Answer 1

Answer:

$$\frac{1}{3}(2i+2j+k)$$

Given curve:
$x = t^2 + 2,\quad y = 4t - 5,\quad z = 2t^2 - 6t$

Position vector

$\vec{r}(t) = (t^2 + 2),i + (4t - 5),j + (2t^2 - 6t),k$

Differentiate to get tangent vector

$\frac{d\vec{r}}{dt} = (2t),i + (4),j + (4t - 6),k$

At (t = 2)

$= (2×2)i + 4j + (4×2 - 6)k$
$= 4i + 4j + (8 - 6)k$
$= 4i + 4j + 2k$

Magnitude

$|\vec{v}| = \sqrt{4^2 + 4^2 + 2^2}$
$= \sqrt{16 + 16 + 4}$
$= \sqrt{36} = 6$

Unit tangent vector

$\hat{T} = \frac{1}{6}(4i + 4j + 2k)$

Simplify:
$= \frac{1}{3}(2i + 2j + k)$
Final Answer:
$\boxed{\frac{1}{3}(2i + 2j + k)}$