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Given-a-function-R-R-with-2-x-2-Find-the-value-of-d-2-dx-2-when-8-

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Question Number 117704 by bemath last updated on 13/Oct/20
Given a function ψ:R→R with ψ(θ) = θ^2 −x^2 . Find the value of ((d^2 ψ(θ))/dx^2 ) when θ=8
$$\mathrm{Given}\:\mathrm{a}\:\mathrm{function}\:\psi:\mathbb{R}\rightarrow\mathbb{R} \\ $$$$\mathrm{with}\:\psi\left(\theta\right)\:=\:\theta^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} .\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\frac{\mathrm{d}^{\mathrm{2}} \psi\left(\theta\right)}{\mathrm{dx}^{\mathrm{2}} }\:\:\mathrm{when}\:\theta=\mathrm{8} \\ $$
Commented by prakash jain last updated on 13/Oct/20
θ independent variable?
$$\theta\:\mathrm{independent}\:\mathrm{variable}? \\ $$
Commented by bemath last updated on 13/Oct/20
no sir
$$\mathrm{no}\:\mathrm{sir}\: \\ $$
Commented by prakash jain last updated on 13/Oct/20
(d/dx)ψ=2θ(d/dx)θ−2x (d^2 /dx^2 )ψ=2((dθ/dx))^2 +2θ(d^2 θ/dx^2 )−2 i think if θ is not independent of yiu will need to know θ(x)
$$\frac{{d}}{{dx}}\psi=\mathrm{2}\theta\frac{{d}}{{dx}}\theta−\mathrm{2}{x} \\ $$$$\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\psi=\mathrm{2}\left(\frac{{d}\theta}{{dx}}\right)^{\mathrm{2}} +\mathrm{2}\theta\frac{{d}^{\mathrm{2}} \theta}{{dx}^{\mathrm{2}} }−\mathrm{2} \\ $$$${i}\:{think}\:{if}\:\theta\:{is}\:{not}\:{independent}\:{of} \\ $$$${yiu}\:{will}\:{need}\:{to}\:{know}\:\theta\left({x}\right) \\ $$
Commented by bemath last updated on 13/Oct/20
sir in line 2 it should be (d^2 ψ/dx^2 ) = 2((dθ/dx))^2 +2θ (d^2 θ/dx^2 ) −2 ?
$$\mathrm{sir}\:\mathrm{in}\:\mathrm{line}\:\mathrm{2}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\: \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \psi}{\mathrm{dx}^{\mathrm{2}} }\:=\:\mathrm{2}\left(\frac{\mathrm{d}\theta}{\mathrm{dx}}\right)^{\mathrm{2}} +\mathrm{2}\theta\:\frac{\mathrm{d}^{\mathrm{2}} \theta}{\mathrm{dx}^{\mathrm{2}} }\:−\mathrm{2}\:? \\ $$
Commented by prakash jain last updated on 13/Oct/20
yes. corrected in red
$$\mathrm{yes}.\:\mathrm{corrected}\:\mathrm{in}\:\mathrm{red} \\ $$
Commented by Dwaipayan Shikari last updated on 13/Oct/20
But in the question θ is independent of x ψ(θ)=θ^2 −x^2 (And ψ(θ)=θ^2 −x^2 Where −x^2 is constant) ψ′(θ)=2θ(dθ/dx)−2x ψ′′(θ)=2(dθ/dx)+2θ(d^2 θ/dx^2 )−2 (dθ/dx)=0 ψ′′(θ)=−2
$${But}\:{in}\:{the}\:{question}\:\theta\:{is}\:{independent}\:{of}\:{x} \\ $$$$\psi\left(\theta\right)=\theta^{\mathrm{2}} −{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\left({And}\:\psi\left(\theta\right)=\theta^{\mathrm{2}} −{x}^{\mathrm{2}} \:\:\:\:\:\:{Where}\:−{x}^{\mathrm{2}} \:{is}\:\:{constant}\right) \\ $$$$\psi'\left(\theta\right)=\mathrm{2}\theta\frac{{d}\theta}{{dx}}−\mathrm{2}{x} \\ $$$$\psi''\left(\theta\right)=\mathrm{2}\frac{{d}\theta}{{dx}}+\mathrm{2}\theta\frac{{d}^{\mathrm{2}} \theta}{{dx}^{\mathrm{2}} }−\mathrm{2}\:\:\:\:\:\:\:\frac{{d}\theta}{{dx}}=\mathrm{0} \\ $$$$\psi''\left(\theta\right)=−\mathrm{2} \\ $$
Commented by prakash jain last updated on 13/Oct/20
It was already replied that θ is not independent variable.
$$\mathrm{It}\:\mathrm{was}\:\mathrm{already}\:\mathrm{replied}\:\mathrm{that}\:\theta\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{independent}\:\mathrm{variable}. \\ $$

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