Question

a particle is moving along the curve ( y = sqrt{x} ). as the particle passes through the point ( (4,2) ), its ( x )-coordinate increases at a rate of ( 3 mathrm{~cm} / mathrm{s} ). how fast is the distance from the particle to the origin changing at this instant?

Explanation:

Step1: Find the distance formula

The distance \(d\) from a point \((x,y)\) to the origin \((0,0)\) is given by \(d=\sqrt{x^{2}+y^{2}}\). Since \(y = \sqrt{x}\), we substitute \(y\) into the distance formula: \(d=\sqrt{x^{2}+x}\).

Step2: Differentiate with respect to time \(t\)

Using the chain rule \(\frac{dd}{dt}=\frac{1}{2\sqrt{x^{2}+x}}\cdot(2x + 1)\cdot\frac{dx}{dt}\).

Step3: Substitute \(x = 4\) and \(\frac{dx}{dt}=3\)

When \(x = 4\), \(\frac{dd}{dt}=\frac{1}{2\sqrt{4^{2}+4}}\cdot(2\times4 + 1)\cdot3\). First, calculate the denominator \(\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}\). Then the numerator: \((8 + 1)\times3=27\). So \(\frac{dd}{dt}=\frac{27}{4\sqrt{5}}=\frac{27\sqrt{5}}{20}\approx3.02\).

Answer:

\(\frac{27\sqrt{5}}{20}\text{ cm/s}\)