UPSC 2019 Maths Optional Paper 1 Q5e — Step-by-Step Solution

10 marks · Section B

Gradient: definition, geometric meaning, computation · Vector Analysis · asked 6× in 13 yrs · Read the full method →

Question

Find the directional derivative of the function $xy^2+yz^2+zx^2$ along the tangent to the curve $x=t,\ y=t^2,\ z=t^3$ at the point $(1,1,1)$ .

Directional derivative $=\nabla\phi\cdot\hat T$ , with $\hat T$ the unit tangent $\vec r\,'(t)/|\vec r\,'(t)|$ at the given parameter value.

Let $\phi(x,y,z)=xy^2+yz^2+zx^2$ .

\nabla\phi=\big(\phi_x,\phi_y,\phi_z\big)=\big(y^2+2zx,\ 2xy+z^2,\ 2yz+x^2\big).

At $(1,1,1)$ :

\nabla\phi\big|_{(1,1,1)}=(1+2,\ 2+1,\ 2+1)=(3,\ 3,\ 3).

$\vec r(t)=(t,\,t^2,\,t^3)$ , so $\vec r\,'(t)=(1,\,2t,\,3t^2)$ . The point $(1,1,1)$ corresponds to $t=1$ :

\vec r\,'(1)=(1,\ 2,\ 3),\qquad |\vec r\,'(1)|=\sqrt{1+4+9}=\sqrt{14}.

Unit tangent:

\hat T=\frac{1}{\sqrt{14}}(1,2,3).

\frac{d\phi}{ds}=\nabla\phi\cdot\hat T=\frac{(3,3,3)\cdot(1,2,3)}{\sqrt{14}}=\frac{3+6+9}{\sqrt{14}}=\frac{18}{\sqrt{14}}.

Rationalising:

\boxed{\;\frac{d\phi}{ds}=\frac{18}{\sqrt{14}}=\frac{9\sqrt{14}}{7}\approx 4.811.\;}