Question

A particle moves along the curve \(y = {\bf{2sin}}\left( {\frac{{\pi x}}{{\bf{2}}}} \right)\). As the particle passes through the point \(\left( {\frac{{\bf{1}}}{{\bf{3}}},{\bf{1}}} \right)\), its x-coordinate increases at a rate of \(\sqrt {{\bf{10}}} \;\;{\bf{cm}}{\rm{/}}{\bf{s}}\). How fast is the distance from the particle to the origin changing at this instant?

Short answer

The distance between particle and origin is changing at a rate of \(1 + \frac{{3\sqrt 3 \pi }}{2}\;{\rm{cm/s}}\).

Step-by-step solution

Find the expression for the rate of change of distance

The distance of the particle from the origin can be calculated as,

\(z = \sqrt {{x^2} + {y^2}} \)

Substitute \(2\sin \left( {\frac{{\pi x}}{2}} \right)\) for y.

\({z^2} = {x^2} + {\left( {2\sin \left( {\frac{{\pi x}}{2}} \right)} \right)^2}\)

Differentiate the equation with respect to t.

\(\begin{aligned}2z\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= 2x\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + 4 \cdot 2\sin \left( {\frac{{\pi x}}{2}} \right)\cos \left( {\frac{{\pi x}}{2}} \right)\left( {\frac{\pi }{2}} \right)\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}t}}} \right)\\z\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= x\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + 2\pi \sin \left( {\frac{{\pi x}}{2}} \right)\cos \left( {\frac{{\pi x}}{2}} \right)\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}t}}} \right)\\z\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= x\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + \pi \sin \left( {\pi x} \right)\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}t}}} \right)\end{aligned}\)

Find the value of \(\frac{{{\bf{d}}z}}{{{\bf{d}}t}}\)

Find the value of z, for \(\left( {\frac{1}{3},1} \right)\).

\(\begin{aligned}z &= \sqrt {{{\left( {\frac{1}{3}} \right)}^2} + {1^2}} \\ &= \frac{{\sqrt {10} }}{3}\end{aligned}\)

Substitute \(\frac{{\sqrt {10} }}{3}\) for z and \(\sqrt {10} \) for \(\frac{{{\rm{d}}x}}{{{\rm{d}}t}}\) in the equation \(z\frac{{{\rm{d}}z}}{{{\rm{d}}t}} = x\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + \pi \sin \left( {\pi x} \right)\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}t}}} \right)\).

\(\begin{aligned}\left( {\frac{{\sqrt {10} }}{3}} \right)\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= \left( {\frac{1}{3}} \right)\left( {\sqrt {10} } \right) + \pi \sin \left( {\frac{\pi }{3}} \right)\left( {\sqrt {10} } \right)\\\left( {\frac{{\sqrt {10} }}{3}} \right)\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= \frac{{\sqrt {10} }}{3} + \pi \left( {\frac{{\sqrt 3 }}{2}} \right)\sqrt {10} \\\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= 1 + \pi \left( {\frac{{\sqrt 3 }}{2}} \right)\sqrt {10} \times \frac{3}{{\sqrt {10} }}\\ &= 1 + \frac{{3\sqrt 3 \pi }}{2}\;{\rm{cm/s}}\end{aligned}\)

Thus, the distance between particle and origin is changing at a rate of \(1 + \frac{{3\sqrt 3 \pi }}{2}\;{\rm{cm/s}}\).