angle of elevation: rate of change in angle

[abby_07]

A fishing line is reeled in at a rate of 1 foot per second from a bridge 15 feet above the water. At what rate is the angle between the line and the water changing when 25 feet of line is out?

i have no idea how to solve this problem

 

[galactus]

There are various ways to tackle this problem. Let's do it this way:


\(\displaystyle \L\\sin({\theta})=\frac{15}{y}\)

\(\displaystyle \L\\cos({\theta})\frac{d{\theta}}{dt}=\frac{-15}{y^{2}}\frac{dy}{dt}\)

By Pythagoras, x=20 when y=25.

Therefore, \(\displaystyle \L\\cos({\theta})=\frac{20}{25}\)

\(\displaystyle \L\\(\frac{20}{25})(\frac{d{\theta}}{dt})=\frac{-15}{625}(-1)=\frac{3}{100}\;\ \frac{rad}{sec}\)


You could also use some other trig functions.

\(\displaystyle \L\\15csc({\theta})=y\)

\(\displaystyle \L\\-15csc({\theta})cot({\theta})\frac{d{\theta}}{dt}=\frac{dy}{dt}\)

When y=25, \(\displaystyle \L\\csc({\theta})=\frac{25}{15}=\frac{5}{3}\)

And \(\displaystyle \L\\cot({\theta})=\frac{20}{15}=\frac{4}{3}\)

\(\displaystyle \L\\-15(\frac{5}{3})(\frac{4}{3})\frac{d{\theta}}{dt}=-1\)

\(\displaystyle \L\\\frac{d{\theta}}{dt}=\frac{3}{100}\;\ rad/sec\)