Exercise 11.2 Class 12 Math Solutions - Unit 11 Vector Valued Functions, Punjab Board

Step-by-step solutions to Exercise 11.2, Unit 11: Vector Valued Functions and Their Derivations, for Class 12 Punjab Board students. Covers first-principles definition of vector derivatives, component-wise differentiation of vector functions r'(t) = f'(t)i + g'(t)j + h'(t)k, finding tangent vectors, computing unit tangent vectors T(t), evaluating velocity, speed, and acceleration vectors, and applying product rules for dot and cross products of vector functions, worked out question by question.

This page provides complete step-by-step solved solutions for Exercise 11.2 of Unit 11: Vector Valued Functions and Their Derivations, Class 12 Mathematics, for Punjab Board (PCTB / PECTAA). Every question is solved with complete analytical rigor, covering the limit definition of vector derivatives \(\mathbf{r}'(t) = \lim_{\Delta t \to 0} \frac{\mathbf{r}(t+\Delta t) - \mathbf{r}(t)}{\Delta t}\), component-wise differentiation \(\mathbf{r}'(t) = f'(t)\mathbf{i} + g'(t)\mathbf{j} + h'(t)\mathbf{k}\), determining tangent vectors and unit tangent vectors \(\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}\), finding velocity \(\mathbf{v}(t)\), speed \(v(t) = \|\mathbf{v}(t)\|\), and acceleration \(\mathbf{a}(t) = \mathbf{r}''(t)\), and applying differentiation rules to scalar multiples, dot products \(\frac{d}{dt}[\mathbf{u} \cdot \mathbf{v}]\), and cross products \(\frac{d}{dt}[\mathbf{u} \times \mathbf{v}]\).
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Exercise Overview

Exercise 11.2 is the central analytical exercise of Unit 11 in Class 12 Mathematics under the Punjab Board PCTB / PECTAA curriculum, extending differential calculus into vector spaces. If (\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}) is a position vector function describing a curve in 3D space, its derivative from first principles is defined as: \[ \mathbf{r}'(t) = \frac{d\mathbf{r}}{dt} = \lim_{\Delta t \to 0} \frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{\Delta t} \] Provided the component derivatives (f'(t)), (g'(t)), and (h'(t)) exist, differentiation is carried out component-by-component: \[ \mathbf{r}'(t) = f'(t)\mathbf{i} + g'(t)\mathbf{j} + h'(t)\mathbf{k} = \langle f'(t), g'(t), h'(t) \rangle \]

Geometric and Physical Significance: The vector (\mathbf{r}'(t)) is tangent to the space curve at the point corresponding to (t), pointing in the direction of increasing (t). Normalizing this tangent vector yields the Unit Tangent Vector: \[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \quad (\text{provided } \mathbf{r}'(t) \neq \mathbf{0}) \] In physical kinematics where (t) represents time, (\mathbf{r}(t)) is position, (\mathbf{v}(t) = \mathbf{r}'(t)) is the instantaneous velocity vector, the magnitude (\|\mathbf{v}(t)\|) is the scalar speed, and the second derivative (\mathbf{a}(t) = \mathbf{r}''(t) = \mathbf{v}'(t)) is the acceleration vector.

Differentiation Rules for Vector Functions: Exercise 11.2 details the calculus operational rules for vector operations: \[ \frac{d}{dt}[c(t)\mathbf{u}(t)] = c'(t)\mathbf{u}(t) + c(t)\mathbf{u}'(t) \] \[ \frac{d}{dt}[\mathbf{u}(t) \cdot \mathbf{v}(t)] = \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t) \] \[ \frac{d}{dt}[\mathbf{u}(t) \times \mathbf{v}(t)] = \mathbf{u}'(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}'(t) \quad (\text{preserving cross product order}) \] An important fundamental theorem verified in this exercise is that if (\|\mathbf{r}(t)\| = c) (constant magnitude), then (\mathbf{r}(t) \cdot \mathbf{r}'(t) = 0), proving that the position vector and its velocity tangent vector are strictly orthogonal.

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 11.2 is a prime source of calculus long questions, short derivations, and entry test (ECAT) problems. These step-by-step solutions provide clean differentiation steps, exact unit tangent evaluations, and complete kinematic vector calculations for every textbook question.

FAQs

Are these Exercise 11.2 solutions according to the Punjab Board syllabus?
Yes. These solutions strictly follow the Punjab Curriculum and Textbook Board (PCTB / PECTAA) syllabus for Class 12 Mathematics, matching all vector differentiation rules and exercises in Unit 11, Exercise 11.2 for the 2026-2027 session.
Do these solutions cover every question in Exercise 11.2?
Yes, every question from Exercise 11.2 is solved step by step, including component-wise derivatives, tangent vectors, unit tangent vectors, velocity and acceleration vectors, and differentiation of dot and cross products.
How is the derivative of a vector-valued function computed?
The derivative of a vector-valued function r(t) = f(t)i + g(t)j + h(t)k is found by differentiating each scalar component function with respect to t: r'(t) = f'(t)i + g'(t)j + h'(t)k, provided each component function is differentiable.
What is the geometric meaning of the derivative vector r'(t)?
Geometrically, r'(t) represents the tangent vector to the space curve traced by r(t) at parameter t, pointing in the direction of increasing parameter values. In kinematics, r'(t) represents the instantaneous velocity vector v(t).
Can students read these solutions online?
Yes, the full Exercise 11.2 solution set is available to read online directly on this page on MaryamNotes.pk with zero software download or registration required.
Are these solutions free for students?
Yes, Exercise 11.2 solutions on this page are completely free to read for all Intermediate Part 2 students across Lahore, Rawalpindi, Gujranwala, Faisalabad, Multan, and all other Punjab boards.