UPSC Maths 2013 Paper 1 Q5e — Solution

10 marks · Section B

Question

Show that the curve

\vec{x}(t)=t\hat{i}+\left(\frac{1+t}{t}\right)\hat{j}+\left(\frac{1-t^{2}}{t}\right)\hat{k}

lies in a plane.

Technique

Spot a linear combination of $x,y,z$ that simplifies to a constant by examining the component formulas; verify identically in $t$ .

Solution

Strategy. Find a fixed plane $\alpha x+\beta y+\gamma z+\delta=0$ that the curve satisfies for all $t$ . (Equivalent to showing the torsion vanishes, but spotting the plane is faster.)

Step 1 — Write components

x(t)=t,\qquad y(t)=\frac{1+t}{t}=\frac{1}{t}+1,\qquad z(t)=\frac{1-t^{2}}{t}=\frac{1}{t}-t.

Step 2 — Eliminate $t$

From the $y$ formula: $y-1=\dfrac{1}{t}$ .

From the $z$ formula: $z+t=\dfrac{1}{t}$ , i.e. $z+x=\dfrac{1}{t}$ (using $x=t$ ).

Both equal $1/t$ , so

y-1=z+x\;\Longrightarrow\;x-y+z+1=0.

Step 3 — Verify the plane equation holds identically

For every $t\ne 0$ :

x-y+z+1=t-\!\left(\frac{1}{t}+1\right)+\!\left(\frac{1}{t}-t\right)+1=t-\frac{1}{t}-1+\frac{1}{t}-t+1=0\;\checkmark.

So the curve lies in the plane $\Pi:\;x-y+z+1=0$ .

Answer

\boxed{\;\text{The curve lies in the plane }x-y+z+1=0.\;}

UPSC Maths 2013 Paper 1 Q5e — Solution

Question

Technique

Solution

Step 1 — Write components

Step 2 — Eliminate ttt

Step 3 — Verify the plane equation holds identically

Answer

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Step 2 — Eliminate $t$