UPSC Maths 2013 Paper 1 Q8b — Solution

10 marks · Section B

Question

A curve in space is defined by the vector equation $\vec{r}=t^{2}\hat{i}+2t\hat{j}-t^{3}\hat{k}$ . Determine the angle between the tangents to this curve at the points $t=+1$ and $t=-1$ .

Technique

Standard cosine-formula for angle between vectors.

Solution

Strategy. The tangent vector is $\vec r\,'(t)$ . Compute at $t=\pm 1$ ; use $\cos\theta=\dfrac{\vec u\cdot\vec v}{|\vec u||\vec v|}$ .

Step 1 — Tangent vector

\vec r\,'(t)=2t\,\hat i+2\,\hat j-3t^{2}\,\hat k.

At $t=+1$ : $\vec r\,'(1)=(2,\,2,\,-3)$ . At $t=-1$ : $\vec r\,'(-1)=(-2,\,2,\,-3)$ .

Step 2 — Dot product and magnitudes

\vec r\,'(1)\cdot\vec r\,'(-1)=(2)(-2)+(2)(2)+(-3)(-3)=-4+4+9=9.

|\vec r\,'(1)|^{2}=4+4+9=17,\qquad|\vec r\,'(-1)|^{2}=4+4+9=17.

Step 3 — Angle

\cos\theta=\frac{9}{\sqrt{17}\cdot\sqrt{17}}=\frac{9}{17}.

Answer

\boxed{\;\theta=\cos^{-1}\!\left(\tfrac{9}{17}\right)\approx 58.03°.\;}