Implicit differentiation; reproducing textbook derivation

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The discussion focuses on implicit differentiation related to the equations \(\frac{1}{r}=\frac{a}{b^2}(1+ecosθ)\) and \(b^2=a^2(1-e^2)\). The goal is to find \(\frac{dθ}{dψ}\) through implicit differentiation, leading to the book's result of \(\frac{dθ}{dψ}=\frac{b}{a(1-ecosψ)}\). An initial attempt at differentiation yielded an equation involving both sine functions, but the author struggled to match the textbook result. Ultimately, the issue was resolved by incorporating a relationship between \(\sinψ\) and \(\sinθ\). This highlights the importance of recognizing and applying relevant trigonometric identities in implicit differentiation problems.
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Homework Statement



(The fourth equation is the central one)

first, we have \frac{1}{r}=\frac{a}{b^2}(1+ecosθ) and b^2=a^2(1-e^2)

now using these two, we transform

acosψ=ae+rcosθ into (1-ecosψ)(1+ecosθ)=\frac{b^2}{a^2}

we want to find dθ/dψ, and the author performs an inplicit differentiation, and the result in the book is (treating θ as a function of ψ)
dθ/dψ=\frac{b}{a(1-ecosψ)}

Homework Equations


stated above

The Attempt at a Solution



I performed the implicit differentiation and got:
esinψ(1+ecosθ)-esinθ\frac{dθ}{dψ}(1-ecosψ)=0

Is my implicit differentiation wrong or are some transformations needed? I tried to match the result in the book using the first two equations together with my result without success..
 
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Never mind, I found it. I just needed to use a relation between sinψ and sinθ (which I have not posted here).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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