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tampuk1
@tampuk1
August 2022
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В какой точке кривой [tex] y^{2}=2x^{3} [/tex] касательная перпендикулярна к прямой [tex]4x-3y+2=0[/tex]
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mikael2
3y=4x+2 y=4/3x+2/3 k1=4/3
y=√2x³ x>0 y=√2*x^3/2 y'=3/√2*√x
y=-√2x³ x<0 y'=-3/√2*√x
k=-1/k1
3√x/√2=-3/4 не подходит
-3√x/√2=-3/4 √x/√2=1/4 x/2=1/16 x=1/8
6 votes
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Answers & Comments
y=√2x³ x>0 y=√2*x^3/2 y'=3/√2*√x
y=-√2x³ x<0 y'=-3/√2*√x
k=-1/k1
3√x/√2=-3/4 не подходит
-3√x/√2=-3/4 √x/√2=1/4 x/2=1/16 x=1/8