Exercise 11.3 Class 12 Math Solutions - Unit 11 Vector Valued Functions, Punjab Board
Step-by-step solutions to Exercise 11.3, Unit 11: Vector Valued Functions and Their Derivations, for Class 12 Punjab Board students. Covers calculating velocity vectors v(t) = r'(t), speed ||v(t)||, acceleration vectors a(t) = r''(t), and direction of motion unit vectors, along with real-life 3D kinematics applications including an autonomous drone inspecting a cylindrical water tower along a helical flight path, worked out question by question.
Exercise Overview
Exercise 11.3 of Class 12 Mathematics connects vector calculus with classical 3D kinematics under the Punjab Board PCTB / PECTAA curriculum. When a particle or object moves through two- or three-dimensional space along a curved path defined by the position vector (\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k}), its physical motion is fully characterized by vector derivatives with respect to time (t).
Kinematic Vector Formulations:
- Velocity Vector: The first time derivative (\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = x'(t)\mathbf{i} + y'(t)\mathbf{j} + z'(t)\mathbf{k}), representing instantaneous rate and direction of position change.
- Scalar Speed: The scalar magnitude of the velocity vector, given by (v(t) = \|\mathbf{v}(t)\| = \sqrt{[x'(t)]^2 + [y'(t)]^2 + [z'(t)]^2}).
- Direction of Motion: The dimensionless unit tangent vector along the trajectory: (\mathbf{u}_v(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|}).
- Acceleration Vector: The second time derivative (\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d^2\mathbf{r}}{dt^2} = x''(t)\mathbf{i} + y''(t)\mathbf{j} + z''(t)\mathbf{k}).
Real-Life Modeling - Drone Inspection of a Cylindrical Water Tower: A featured real-world application in Exercise 11.3 is the mathematical trajectory of an inspection drone flying around a cylindrical structure of radius (R). The drone maintains a constant distance (R) from the vertical axis while climbing at a constant vertical rate (c) with angular velocity (\omega), tracing a circular helix: \[ \mathbf{r}(t) = (R\cos\omega t)\mathbf{i} + (R\sin\omega t)\mathbf{j} + ct\,\mathbf{k} \] Differentiating yields the velocity vector (\mathbf{v}(t) = (-R\omega\sin\omega t)\mathbf{i} + (R\omega\cos\omega t)\mathbf{j} + c\,\mathbf{k}) with constant speed (\|\mathbf{v}(t)\| = \sqrt{R^2\omega^2 + c^2}), and acceleration (\mathbf{a}(t) = -R\omega^2(\cos\omega t\,\mathbf{i} + \sin\omega t\,\mathbf{j})), pointing directly towards the central cylinder axis with constant magnitude (\|\mathbf{a}(t)\| = R\omega^2).
For Punjab Board HSSC Part 2 examinations across all nine educational divisions (Lahore, Rawalpindi, Gujranwala, Faisalabad, Multan, Bahawalpur, Sargodha, Sahiwal, and D.G. Khan), Exercise 11.3 is a high-yield source of conceptual Section B and applied Section C exam questions. These step-by-step solutions provide rigorous component differentiation, speed magnitudes, and complete physical trajectory proofs for every textbook question.