Class CBSE Class 12 Mathematics Applications of Derivatives Q #1405
COMPETENCY BASED
UNDERSTAND
2 Marks 2025 AISSCE(Board Exam) VSA
For the curve $y=5x-2x^{3}$ if x increases at the rate of $2\text{ units/s}$, then how fast is the slope of the curve changing when $x=2$?

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Detailed Solution

Step 1: Find the slope of the curve

The slope of the curve is given by the derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$. $$y = 5x - 2x^3$$ $$\frac{dy}{dx} = 5 - 6x^2$$ Let $m$ be the slope, so $m = 5 - 6x^2$.

Step 2: Find the rate of change of the slope

We need to find how fast the slope is changing, which is $\frac{dm}{dt}$. We are given that $\frac{dx}{dt} = 2$. We can use the chain rule to find $\frac{dm}{dt}$: $$\frac{dm}{dt} = \frac{dm}{dx} \cdot \frac{dx}{dt}$$ First, find $\frac{dm}{dx}$: $$m = 5 - 6x^2$$ $$\frac{dm}{dx} = -12x$$

Step 3: Calculate the rate of change of the slope when x=2

Now, we can find $\frac{dm}{dt}$ when $x=2$ and $\frac{dx}{dt} = 2$: $$\frac{dm}{dt} = \frac{dm}{dx} \cdot \frac{dx}{dt} = (-12x) \cdot (2) = -24x$$ When $x=2$: $$\frac{dm}{dt} = -24(2) = -48$$

Step 4: State the final answer

The slope of the curve is changing at a rate of $-48$ units/s when $x=2$.

Final Answer: -48 units/s

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the concept of derivatives and related rates to solve the problem. They need to interpret the given information and apply the chain rule.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific procedure (differentiation and chain rule) to find the rate of change of the slope.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply the concepts of derivatives to solve a related rates problem, which is a practical application of calculus.