Question

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)

Answer 1

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

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

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

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