Scalar Diagram (involving distances only):

math expression

We can see that:

`sin\ theta=5/x`, so

`theta=sin^-1(5/x)`

Vector Diagram (involving velocities):

Related rates winch problem

[See the section on Vector concepts for more on vectors and scalars.]

We are also given that:

`(dx)/(dt)=-2\ "ms"^-1`

Since we need `(d theta)/(dt)`, we use:

`(d theta)/(dt)=(d theta)/(dx)(dx)/(dt)`

Because

`theta=sin^-1(5/x)`

we have:

`(d theta)/(dx)=1/(sqrt(1-(5/x)^2))(d(5/x))/(dx)`

`=(-5)/(x^2sqrt(1-(5/x)^2))`

So

`(d theta)/(dt)=(d theta)/(dx)(dx)/(dt)`

`=(-5)/(x^2sqrt(1-(5/x)^2))(-2)`

`=10/(x^2sqrt(1-(5/x)^2))`

We want to know the rate of change of θ when x = 10 m, so we substitute, as follows:

`(d theta)/(dt)=10/(10^2sqrt(1-(5/10)^2))`

`=0.1155\ "rad"//"s"`

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