UPSC 2018 Maths Optional Paper 1 Q5b — Step-by-Step Solution

10 marks · Section B

Curves in space: tangent, normal, binormal · Vector Analysis · asked 2× in 13 yrs · Read the full method →

Question

Find the angle between the tangent at a general point of the curve whose equations are $x=3t,\ y=3t^2,\ z=3t^3$ and the line $y=z-x=0$ .

Tangent $=\vec r\,'(t)$ ; line direction from intersecting planes; $\cos\theta=\dfrac{\vec T\cdot\vec L}{|\vec T||\vec L|}$ .

$\vec r(t)=(3t,\,3t^2,\,3t^3)$ , so the tangent vector at a general point is

\vec T=\vec r\,'(t)=(3,\;6t,\;9t^2)\ \parallel\ (1,\;2t,\;3t^2).

The line is given by the two planes $y=0$ and $z-x=0$ . A point on it satisfies $y=0,\ z=x$ , so the line is $\{(s,0,s)\}$ with direction

\vec L=(1,\;0,\;1).

\cos\theta=\frac{\vec T\cdot\vec L}{|\vec T|\,|\vec L|} =\frac{(1)(1)+(2t)(0)+(3t^2)(1)}{\sqrt{1+4t^2+9t^4}\,\sqrt{2}} =\frac{1+3t^2}{\sqrt2\,\sqrt{1+4t^2+9t^4}}.

Equivalently, using $\vec T\times\vec L=(2t,\,3t^2-1,\,-2t)$ with $|\vec T\times\vec L|=\sqrt{9t^4+2t^2+1}$ ,

\tan\theta=\frac{|\vec T\times\vec L|}{\vec T\cdot\vec L}=\frac{\sqrt{9t^4+2t^2+1}}{1+3t^2}.

The angle therefore depends on the parameter $t$ (the general point):

\boxed{\;\theta=\cos^{-1}\!\frac{1+3t^2}{\sqrt{2\,(1+4t^2+9t^4)}} =\tan^{-1}\frac{\sqrt{9t^4+2t^2+1}}{1+3t^2}.\;}