41.1k views
4 votes
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)

User Nate May
by
7.7k points

1 Answer

4 votes

Answer:3 units/s

Explanation:

Given


y=√(x)

Point P lie on this curve so any general point on curve can be written as
(x,√(x))

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

Distance between Point P and (2,0)


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


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


\frac{\mathrm{d} P}{\mathrm{d} t}=(2* 3-3)/(2* 2)* 4=3 units/s

User Deduplicator
by
8.4k points

No related questions found