Q1251: If $x~cos(p+y)+cos~p~sin(p+y)=0$ prove that $cos~p\frac{dy}{...

Class CBSE Class 12 Mathematics Derivatives Q #1251

KNOWLEDGE BASED

UNDERSTAND

3 Marks 2024 AISSCE(Board Exam) SA

If $x~cos(p+y)+cos~p~sin(p+y)=0$ prove that $cos~p\frac{dy}{dx}=-cos^{2}(p+y),$ where p is a constant.

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AI Tutor Explanation

Detailed Solution

Step 1: Rewrite the given equation

Given the equation: $x \cos(p+y) + \cos p \sin(p+y) = 0$. We want to find $\frac{dy}{dx}$.

Step 2: Differentiate both sides with respect to x

Differentiating both sides of the equation with respect to $x$, we get: $$ \frac{d}{dx} [x \cos(p+y) + \cos p \sin(p+y)] = \frac{d}{dx} [0] $$ $$ \frac{d}{dx} [x \cos(p+y)] + \frac{d}{dx} [\cos p \sin(p+y)] = 0 $$

Step 3: Apply the product rule and chain rule

Using the product rule for the first term and the chain rule for both terms: $$ \left[ (1) \cos(p+y) + x (-\sin(p+y)) \frac{dy}{dx} \right] + \cos p \left[ \cos(p+y) \frac{dy}{dx} \right] = 0 $$ $$ \cos(p+y) - x \sin(p+y) \frac{dy}{dx} + \cos p \cos(p+y) \frac{dy}{dx} = 0 $$

Step 4: Isolate dy/dx

Now, we isolate $\frac{dy}{dx}$: $$ \frac{dy}{dx} [ -x \sin(p+y) + \cos p \cos(p+y) ] = - \cos(p+y) $$ $$ \frac{dy}{dx} = \frac{- \cos(p+y)}{-x \sin(p+y) + \cos p \cos(p+y)} $$

Step 5: Substitute x from the original equation

From the original equation, we have $x \cos(p+y) = - \cos p \sin(p+y)$, so $x = - \frac{\cos p \sin(p+y)}{\cos(p+y)}$. Substituting this into the expression for $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{- \cos(p+y)}{- \left( - \frac{\cos p \sin(p+y)}{\cos(p+y)} \right) \sin(p+y) + \cos p \cos(p+y)} $$ $$ \frac{dy}{dx} = \frac{- \cos(p+y)}{\frac{\cos p \sin^2(p+y)}{\cos(p+y)} + \cos p \cos(p+y)} $$ $$ \frac{dy}{dx} = \frac{- \cos(p+y)}{\frac{\cos p \sin^2(p+y) + \cos p \cos^2(p+y)}{\cos(p+y)}} $$ $$ \frac{dy}{dx} = \frac{- \cos^2(p+y)}{\cos p (\sin^2(p+y) + \cos^2(p+y))} $$ $$ \frac{dy}{dx} = \frac{- \cos^2(p+y)}{\cos p (1)} $$ $$ \frac{dy}{dx} = \frac{- \cos^2(p+y)}{\cos p} $$

Step 6: Rearrange to get the desired result

Multiplying both sides by $\cos p$, we get: $$ \cos p \frac{dy}{dx} = - \cos^2(p+y) $$

Final Answer: $\cos p \frac{dy}{dx} = - \cos^2(p+y)$

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to comprehend the concepts of differentiation, product rule, and chain rule to solve the problem. They must also understand how to manipulate trigonometric identities and algebraic expressions to arrive at the final answer.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the relationships between different mathematical concepts such as differentiation rules (product and chain rule) and trigonometric identities. It's not just about recalling facts or following a specific algorithm, but about understanding the underlying principles.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of differentiation techniques, which are core concepts covered in the syllabus. The question is a standard application of these concepts, rather than a complex, real-world application.