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Ad libitum [116K]
3 years ago
10

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

= 3. How fast is the distance between P and the point (2, 0) changing at this instant? (enter numerical answer as an integer or decimal to three decimal places, do not enter units)
Mathematics
1 answer:
Mice21 [21]3 years ago
8 0

Answer:3 units/s

Step-by-step explanation:

Given

y=\sqrt{x}

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

and \frac{\mathrm{d} x}{\mathrm{d} t}=4 units/s

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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