Question-162561

Question Number 162561 by Mathematification last updated on 30/Dec/21

Answered by MJS_new last updated on 30/Dec/21

we have to find the minimum value of ∣ A B ∣

with A = (\begin{matrix} p \\ p^{2} \end{matrix}) and B = (\begin{matrix} q \\ \ln q \end{matrix}) . this can only

be approximated . I get

p \approx . 538167

q \approx . 929078

∣ A B ∣= d \approx . 533588

from the parallel tangents we get q = \frac{1}{2 p} \Rightarrow

now {it}^{'} s easier to approximate

p \approx . 538167720574

q \approx . 929078391150

d \approx . 533587574933

Answered by mr W last updated on 30/Dec/21

p o i n t o n y = x^{2} : P (p, p^{2})

p o i n t o n y = \ln x : Q (q, \ln q)

w e s h o u l d f i n d P, Q s u c h t h a t P Q

i s m i n i m u m .

S = {P Q}^{2} = {(p - q)}^{2} + {(p^{2} - \ln q)}^{2}

\frac{\partial S}{\partial p} = 2 (p - q) + 4 (p^{2} - \ln q) p = 0

\Rightarrow (p - q) + 2 (p^{2} - \ln q) p = 0 \dots (i)

\frac{\partial S}{\partial q} = - 2 (p - q) - 2 (p^{2} - \ln q) \frac{1}{q} = 0

\Rightarrow (p - q) q + (p^{2} - \ln q) = 0 \dots (i i)

(i) - (i i) \times 2 p :

(p - q) (1 - 2 p q) = 0

\Rightarrow 1 - 2 p q = 0 \Rightarrow q = \frac{1}{2 p}

(p - \frac{1}{2 p}) + 2 p (p^{2} - \ln \frac{1}{2 p}) = 0

p^{2} (2 p^{2} + 1 + 2 \ln 2 p) = \frac{1}{2}

\Rightarrow p \approx 0.5382

\Rightarrow {P Q}_{m i n} = d_{m a x} \approx 0.5336

Commented by mr W last updated on 30/Dec/21

Commented by mr W last updated on 30/Dec/21

Commented by Tawa11 last updated on 30/Dec/21

Great sir

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