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Question Number 134511 by mohammad17 last updated on 04/Mar/21

Commented by mohammad17 last updated on 04/Mar/21

how can solve this

$${how}\:{can}\:{solve}\:{this} \\ $$

Commented by mohammad17 last updated on 04/Mar/21

??????

$$?????? \\ $$

Commented by mohammad17 last updated on 04/Mar/21

i want question(2) A please

$${i}\:{want}\:{question}\left(\mathrm{2}\right)\:{A}\:{please} \\ $$

Commented by mr W last updated on 04/Mar/21

distance from the point (u,v) to line ac+by+c=0 is d=((∣au+bv+c∣)/( (√(a^2 +b^2 ))))

$${distance}\:{from}\:{the}\:{point}\:\left({u},{v}\right)\:{to}\:{line} \\ $$$${ac}+{by}+{c}=\mathrm{0}\:{is} \\ $$$${d}=\frac{\mid{au}+{bv}+{c}\mid}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$

Commented by Dwaipayan Shikari last updated on 06/Mar/21

d=((∣−8−2−5∣)/( (√(4+1))))=3(√5)

$${d}=\frac{\mid−\mathrm{8}−\mathrm{2}−\mathrm{5}\mid}{\:\sqrt{\mathrm{4}+\mathrm{1}}}=\mathrm{3}\sqrt{\mathrm{5}} \\ $$

Answered by mathmax by abdo last updated on 06/Mar/21

a) d=((∣2x_0 −y_0 −5∣)/( (√(2^2 +(−1)^2 ))))=((∣−8−2−5∣)/( (√5)))=((15)/( (√5)))

$$\left.\mathrm{a}\right)\:\mathrm{d}=\frac{\mid\mathrm{2x}_{\mathrm{0}} −\mathrm{y}_{\mathrm{0}} −\mathrm{5}\mid}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }}=\frac{\mid−\mathrm{8}−\mathrm{2}−\mathrm{5}\mid}{\:\sqrt{\mathrm{5}}}=\frac{\mathrm{15}}{\:\sqrt{\mathrm{5}}} \\ $$

Answered by mathmax by abdo last updated on 06/Mar/21

b)x=sin(3z) and y =((sin^2 (3z))/(cos^2 (3z)))=(x^2 /(1−x^2 )) ⇒(dy/dx)=((2x(1−x^2 )−x^2 (−2x))/((1−x^2 )^2 )) =((2x−2x^3 +2x^3 )/((1−x^2 )^2 ))=((2x)/((1−x^2 )^2 )) and (d^2 y/dx^2 )=((2(1−x^2 )^2 −2x.2(−2x)(1−x^2 ))/((1−x^2 )^4 ))=((2(1−x^2 )+8x^2 )/((1−x^2 )^3 )) =((2−2x^2 +8x^2 )/((1−x^2 )^3 ))=((6x^2 +2)/((1−x^2 )^3 ))

$$\left.\mathrm{b}\right)\mathrm{x}=\mathrm{sin}\left(\mathrm{3z}\right)\:\mathrm{and}\:\mathrm{y}\:=\frac{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{3z}\right)}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{3z}\right)}=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{2x}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)−\mathrm{x}^{\mathrm{2}} \left(−\mathrm{2x}\right)}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2x}−\mathrm{2x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{3}} }{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{2x}}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\mathrm{and} \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2x}.\mathrm{2}\left(−\mathrm{2x}\right)\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{4}} }=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)+\mathrm{8x}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{2}−\mathrm{2x}^{\mathrm{2}} \:+\mathrm{8x}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }=\frac{\mathrm{6x}^{\mathrm{2}} \:+\mathrm{2}}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$

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