Question
Download Solution PDFयदि \(x=a\left(t+\frac{1}{t}\right)\) और \(y=a\left(t-\frac{1}{t}\right)\), तो \(\frac{d x}{d y}\) है:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
\(x=a\left(t+\frac{1}{t}\right)\) और \(y=a\left(t-\frac{1}{t}\right)\)
संकल्पना:
सूत्र \(\rm \frac{d}{dt}t^n=nt^{n-1}\) का प्रयोग करने पर
और \(\rm \frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dx}{dt}}\)
गणना:
\(x=a\left(t+\frac{1}{t}\right)\) और \(y=a\left(t-\frac{1}{t}\right)\)
दोनों समीकरणों का t के सापेक्ष अवकलन करने पर
\(\rm \frac{dx}{dt}=\frac{d}{dt}[a({t+\frac{1}{t}})]\) और \(\rm \frac{dy}{dt}=\frac{d}{dt}[a({t-\frac{1}{t}})]\)
\(\rm \frac{dx}{dt}=a({1-\frac{1}{t^2}})\) और \(\rm \frac{dy}{dt}=a({1+\frac{1}{t^2}})\)
अब, हम यह जानते हैं
\(\rm \frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}\)
तब
\(\rm \frac{dx}{dy}=\frac{a({1-\frac{1}{t^2}})}{a({1+\frac{1}{t^2}})}\)
अब दाहिनी ओर अंश और हर दोनों में t से गुणा करने पर
\(\rm \frac{dx}{dy}=\frac{a({t-\frac{1}{t}})}{a({t+\frac{1}{t}})}\)
\(\rm \frac{dx}{dy}=\frac{y}{x}\)
अतः विकल्प (2) सही है।
Last updated on May 26, 2025
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