3 years ago

HELP IVE FALLEN AND I CANT GET UP

Mathematics

2 answers:

zavuch27 [327]3 years ago

7 0

Answer:

JUST GET UP

Step-by-step explanation:

Pani-rosa [81]3 years ago

3 0

Answer:

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Step-by-step explanation:kk

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A point P is moving along the curve whose equation is y = \sqrt x . Suppose that x is increasing at the rate of 4 units/s when x

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Given

$y=\sqrt{x}$

Point P lie on this curve so any general point on curve can be written as $(x,\sqrt{x})$

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Distance between Point P and (2,0)

$P=\sqrt{(x-2)^2+(\sqrt{x}-0)^2}$

P at x=3 P=2

rate at which distance is changing is

$\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{\mathrm{d} \sqrt{(x-2)^2+(\sqrt{x}-0)^2}}{\mathrm{d} t}$

$\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2x-3}{\sqrt{(x-2)^2+(\sqrt{x}-0)^2}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2\times 3-3}{2\times 2}\times 4=3 units/s$

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