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Home » r^2 + z^2 = c r \quad \Rightarrow \quad r^2 - c r + z^2 = 0 - Get link 4share

r^2 + z^2 = c r \quad \Rightarrow \quad r^2 - c r + z^2 = 0 - Get link 4share

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 a = \frac{x + 2}{x - 2}, \quad \frac{x - 2}{x + 2} = \frac{1}{a}.

📌 The equation becomes:

📌 a + \frac{1}{a} = 2.

📌 Multiply both sides by \( a \):

📌 a^2 + 1 = 2a \quad \Rightarrow \quad a^2 - 2a + 1 = 0 \quad \Rightarrow \quad (a - 1)^2 = 0.

📌 So, \( a = 1 \). Thus:

📌 \frac{x + 2}{x - 2} = 1 \quad \Rightarrow \quad x + 2 = x - 2,

📌 which simplifies to \( 2 = -2 \), a contradiction. This implies no solution unless we consider domain restrictions.

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\[ C(t) = \frac{kt}{t^2 + 1} \]
If \(u = kt\) and \(v = t^2 + 1\), then
\[ u' = k, \quad v' = 2t \]
\[ C'(t) = \frac{(k)(t^2 + 1) - (kt)(2t)}{(t^2 + 1)^2} \]
\[ C'(t) = \frac{k(t^2 + 1) - 2kt^2}{(t^2 + 1)^2} \]
\[ C'(t) = \frac{kt^2 + k - 2kt^2}{(t^2 + 1)^2} \]
\[ C'(t) = \frac{-kt^2 + k}{(t^2 + 1)^2} \]
\[ C'(t) = \frac{k(1 - t^2)}{(t^2 + 1)^2} \]
Set \(C'(t) = 0\):
\[ \frac{k(1 - t^2)}{(t^2 + 1)^2} = 0 \]

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