Q1273: If $x=e^{cos~3t}$ and $y=e^{sin~3t}$ , prove that $\frac{dy}...

Class CBSE Class 12 Mathematics Derivatives Q #1273

KNOWLEDGE BASED

REMEMBER

3 Marks 2024 AISSCE(Board Exam) SA

If $x=e^{cos~3t}$ and $y=e^{sin~3t}$ , prove that $\frac{dy}{dx}=-\frac{y~log~x}{x~log~y}$

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Detailed Solution

Step 1: Find $\frac{dx}{dt}$

Given $x = e^{\cos 3t}$. Differentiating with respect to $t$, we get: $$ \frac{dx}{dt} = e^{\cos 3t} \cdot \frac{d}{dt}(\cos 3t) = e^{\cos 3t} \cdot (-\sin 3t) \cdot 3 = -3e^{\cos 3t} \sin 3t $$ Since $x = e^{\cos 3t}$, we can write: $$ \frac{dx}{dt} = -3x \sin 3t $$

Step 2: Find $\frac{dy}{dt}$

Given $y = e^{\sin 3t}$. Differentiating with respect to $t$, we get: $$ \frac{dy}{dt} = e^{\sin 3t} \cdot \frac{d}{dt}(\sin 3t) = e^{\sin 3t} \cdot (\cos 3t) \cdot 3 = 3e^{\sin 3t} \cos 3t $$ Since $y = e^{\sin 3t}$, we can write: $$ \frac{dy}{dt} = 3y \cos 3t $$

Step 3: Find $\frac{dy}{dx}$

Using the chain rule, we have: $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3y \cos 3t}{-3x \sin 3t} = -\frac{y \cos 3t}{x \sin 3t} $$

Step 4: Express $\cos 3t$ and $\sin 3t$ in terms of $x$ and $y$

We have $x = e^{\cos 3t}$, so $\ln x = \cos 3t$. Similarly, $y = e^{\sin 3t}$, so $\ln y = \sin 3t$. Substituting these into the expression for $\frac{dy}{dx}$, we get: $$ \frac{dy}{dx} = -\frac{y \ln x}{x \ln y} $$

Final Answer: $\frac{dy}{dx} = -\frac{y \ln x}{x \ln y}$

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Pedagogical Audit

Bloom's Analysis: This is an REMEMBER question because it requires recalling the chain rule and differentiation formulas to solve the problem.

Knowledge Dimension: PROCEDURAL

Justification: The question requires applying differentiation rules and the chain rule, which are specific procedures in calculus.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests the student's ability to apply differentiation techniques learned in the textbook.