# Related Rates QuestionA particle moves from right to left along the parabolic curve y=sqrt-x in such a way that its x-coordinate (in meters) decreases at the rate of 8 m/sec. How fast is the...

Related Rates Question

A particle moves from right to left along the parabolic curve y=sqrt-x in such a way that its x-coordinate (in meters) decreases at the rate of 8 m/sec. How fast is the angle of inclination theta of the line joining the particle to the origin changing when x=-4?

### 1 Answer | Add Yours

You need to remember that the angle that the tangent line makes to x axis may be found using the tangent function definition such that:

`tantheta = y/x`

The problem provides the information that `y=sqrt(-x), ` hence you need to substitute `sqrt(-x)` for `y ` in the formula `tantheta = y/x` such that:

`tantheta = sqrt(-x) /x =gttheta = tan^(-1) (sqrt(-x) /x)`

You need to differentiate theta with respect to time such that:

`(d theta )/(dt) = (1/(1 + (-x)/x^2))*(sqrt(-x) /x)'*(dx)/(dt)`

The problem provides the information that x coordinate decreases at a rate of -8 m/sec, hence you need to substitute -8 m/sec for `(dx)/(dt)` in formula above such that:

`(d theta )/(dt) = (1/(1 + (-x)/x^2))*((-x/(2sqrt(-x)) - sqrt(-x))/x^2)*(-8)`

`(d theta )/(dt) = ((x^2)/(x^2 - x))*(-8x/((2sqrt(-x))*x^2))`

`(d theta )/(dt) = -4x/((x^2 - x)(sqrt(-x)))`

You need to find how fast the angle theta is changing at `x=-4` , hence you need to substitute -4 for x in equation `(d theta)/(dt) = 4x/((x^2 - x)(sqrt(-x)))` such that:

`(d theta)/(dt) = -4(-4)/((16+4)(sqrt4)) =gt (d theta)/(dt) = 16/40`

`(d theta)/(dt) = 2/5 =gt(d theta)/(dt) = 0.4` angle/sec

**Hence, evaluating how fast the angle of inclination theta is changing at `x=-4` yields `(d theta)/(dt) = 0.4` angle/sec.**