Respuesta :
check the picture below.
now, keep in mind that ship B is going at 20kph, thus from noon to 4pm, is 4 hours, so it has travelled by then 20 * 4 or 80 kilometers, thus b = 80.
whilst the ship B is moving north, the distance "a" is not really changing, and thus is a constant, that matters because the derivative of a constant is 0.
[tex]\bf c^2=a^2+b^2\implies \stackrel{chain~rule}{2c\cfrac{dc}{dt}}=0+2b\cfrac{db}{dt}\implies \cfrac{dc}{dt}=\cfrac{b\frac{db}{dt}}{c} \\\\\\ \begin{cases} \frac{db}{dt}=20\\ c=10\sqrt{353}\\ b=80 \end{cases}\implies \cfrac{dc}{dt}=\cfrac{80\cdot 20}{10\sqrt{353}}\implies \cfrac{dc}{dt}=\cfrac{160}{\sqrt{353}} \\\\\\ \textit{and rationalizing the denominator}\implies \cfrac{160\sqrt{353}}{353}[/tex]
now, keep in mind that ship B is going at 20kph, thus from noon to 4pm, is 4 hours, so it has travelled by then 20 * 4 or 80 kilometers, thus b = 80.
whilst the ship B is moving north, the distance "a" is not really changing, and thus is a constant, that matters because the derivative of a constant is 0.
[tex]\bf c^2=a^2+b^2\implies \stackrel{chain~rule}{2c\cfrac{dc}{dt}}=0+2b\cfrac{db}{dt}\implies \cfrac{dc}{dt}=\cfrac{b\frac{db}{dt}}{c} \\\\\\ \begin{cases} \frac{db}{dt}=20\\ c=10\sqrt{353}\\ b=80 \end{cases}\implies \cfrac{dc}{dt}=\cfrac{80\cdot 20}{10\sqrt{353}}\implies \cfrac{dc}{dt}=\cfrac{160}{\sqrt{353}} \\\\\\ \textit{and rationalizing the denominator}\implies \cfrac{160\sqrt{353}}{353}[/tex]
