Answer:3 units/s
Step-by-step explanation:
Given
[tex]y=\sqrt{x}[/tex]
Point P lie on this curve so any general point on curve can be written as [tex](x,\sqrt{x})[/tex]
and [tex]\frac{\mathrm{d} x}{\mathrm{d} t}=4 units/s[/tex]
Distance between Point P and (2,0)
[tex]P=\sqrt{(x-2)^2+(\sqrt{x}-0)^2}[/tex]
P at x=3 P=2
rate at which distance is changing is
[tex]\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{\mathrm{d} \sqrt{(x-2)^2+(\sqrt{x}-0)^2}}{\mathrm{d} t}[/tex]
[tex]\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}[/tex]
[tex]\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2\times 3-3}{2\times 2}\times 4=3 units/s[/tex]
