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Sagot :
At [tex]$5 \mathrm{pm}$[/tex] the distance between the ships is changing at the speed of [tex]$\mathbf{3 0 . 2 1}$[/tex] knots
Firstly we need to an equation to represent both ships and the distance between each. A is moving [tex]$25 \mathrm{knots}$[/tex] west and [tex]$B$[/tex] is moving [tex]$17 \mathrm{knot}$[/tex] north. [tex]$D$[/tex]will be the distance between the two. Drawing it, you'll notice that it creates a right triangle.
So we use the Pythagorean Theorem:
[tex]$D^{\wedge} 2=A^{\wedge} 2+B^{\wedge} 2$[/tex]
Differentiate in relation to time:
[tex]$2 \mathrm{D}(\mathrm{dD} / \mathrm{dt})=2 \mathrm{~A}(\mathrm{dA} / \mathrm{dt})+2 \mathrm{~B}(\mathrm{~dB} / \mathrm{dt})$[/tex]
Now we must find all of our variables.
[tex]& A=\text { time(speed })+\text { original distance }=5(25)+10=135 \\[/tex]
[tex]& B=5(17)+0=85 \\[/tex]
[tex]& D=V\left(A^{\wedge} 2+B^{\wedge} 2\right)=159.53 \\[/tex]
[tex]& d A / d t=25 \\[/tex]
[tex]& d B / d t=17[/tex]
Plug in all your variables and solve for
[tex]$(\mathrm{dD} / \mathrm{dt})$ :[/tex]
[tex]& 2 \mathrm{D}(\mathrm{dD} / \mathrm{dt})=2 \mathrm{~A}(\mathrm{dA} / \mathrm{dt})+2 \mathrm{~B}(\mathrm{~dB} / \mathrm{dt}) \\[/tex]
[tex]& 2(159.53)(\mathrm{dD} / \mathrm{dt})=2(135)(25)+2(85)(17) \\[/tex]
[tex]& 319.061(\mathrm{dD} / \mathrm{dt})=9640 \\[/tex]
[tex]& \mathrm{dD} / \mathrm{dt}=9640 / 319.061 \\[/tex]
[tex]& \mathrm{dD} / \mathrm{dt}=30.21 \text { knots }\end{aligned}[/tex]
At [tex]$5 \mathrm{pm}$[/tex] the distance between the ships is changing at the speed of [tex]$\mathbf{3 0 . 2 1}$[/tex] knots
To know more about Pythagoras Theorem
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