Let the side of the square be $s$.
The perimeter of the square is $P = 4s$.
We are given that the rate of decrease of the side is $\frac{ds}{dt} = -1.5~cm/s$.
We need to find the rate of decrease of its perimeter, which is $\frac{dP}{dt}$.
Differentiate the perimeter equation with respect to time $t$:
$\frac{dP}{dt} = \frac{d}{dt}(4s)$
$\frac{dP}{dt} = 4 \frac{ds}{dt}$
Substitute the given value of $\frac{ds}{dt}$:
$\frac{dP}{dt} = 4 (-1.5~cm/s)$
$\frac{dP}{dt} = -6~cm/s$
The rate of decrease of its perimeter is $6~cm/s$.
The final answer is $\boxed{6~cm/s}$.
AI Tutor Explanation
Powered by Gemini
Step-by-Step Solution
Let \(s\) be the side of the square and \(P\) be its perimeter.
We are given that \(\frac{ds}{dt} = -1.5\) cm/s (decreasing).
The perimeter of a square is given by \(P = 4s\).
We want to find \(\frac{dP}{dt}\).
Differentiating \(P = 4s\) with respect to time \(t\), we get:
\(\frac{dP}{dt} = 4 \frac{ds}{dt}\)
Substituting the given value of \(\frac{ds}{dt}\):
\(\frac{dP}{dt} = 4 \times (-1.5)\)
\(\frac{dP}{dt} = -6\) cm/s
The rate of decrease of the perimeter is 6 cm/s.
Correct Answer: 6 cm/s
AI Suggestion: Option B
AI generated content. Review strictly for academic accuracy.
Pedagogical Audit
Bloom's Analysis:
This is an APPLY question because it requires students to apply their understanding of related rates and differentiation to solve a practical problem. They need to use the given rate of change of the side of a square to find the rate of change of its perimeter.
Knowledge Dimension:PROCEDURAL
Justification:The question requires the student to apply a specific procedure (differentiation and related rates) to arrive at the solution. It involves knowing how to relate the rate of change of the side to the rate of change of the perimeter.
Syllabus Audit:
In the context of CBSE Class 12, this is classified as COMPETENCY. It assesses the student's ability to apply calculus concepts to solve real-world problems, which aligns with competency-based education.