Before You Begin

Question #41294

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{u} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{v} = 12\hat{i} + 2\hat{j}$ . A point $P$ moves along the curve $C$ given by the position vector $\vec{r}(t) = t^2\hat{i} + t\hat{j} + \hat{k}$ , where $t \in \mathbb{R}$ . Let $S$ be the region in the first octant bounded by the sphere $x^2 + y^2 + z^2 = 651$ . If $P$ is restricted to lie within or on the boundary of $S$ , what is the maximum volume of the parallelepiped formed by the vectors $\vec{u}$ , $\vec{v}$ , and $\vec{r}(t)$ ?

Options

A.
8
B.
10
C.
50
D.
120

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Determinant as Geometric Area/Volume|Maximum/Minimum of Quadratic Expressions|Boundary Value Theorem

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to optimize a geometric volume represented by a determinant, specifically targeting the 'Boundary Value Theorem' trap.

The determinant yields an absolute quadratic function $V(t) = |-2t^2 + 12t - 10|$ .

Students frequently rely solely on calculus, setting the derivative of the inner quadratic to zero to find the vertex at $t=3$ , which gives an interior local maximum volume of 8.

However, they fail to evaluate the function at the boundaries of the domain $t \in [0, 5]$ (derived from the first octant and sphere constraints).

The true global maximum occurs at the boundary $t=0$ , yielding a volume of 10.

An additional trap is ignoring the first octant constraint, which would lead to an incorrect global maximum of 120 at $t=-5$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Determinant as Geometric Area/Volume
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free