Question

A point is moving along the curve `y=sqrt(x)` in such a way that its x coordinate id increasing at the rate of 2 units per minute. At what rate is its slope changing (a) when x=1? (b) when x=4?

Answer 1

`x=1 => d/dt(dy/dx) = -1/2 color(white)".""min"^-1`

`x=4 => d/dt(dy/dx) = -1/16 color(white)".""min"^-1`

Explanation:

Let's call our time variable `t`, in minutes. We know that:

`dx/dt = 2 color(white)".""min"^-1`

And we're trying to find the rate at which the slope is changing with respect to time:

`d/dt(dy/dx)`

First, let's evaluate `dy/dx`:

`dy/dx = d/dx(sqrt(x)) = 1/(2sqrtx)`

Now, we need to evaluate `d/dt(dy/dx)`:

`d/dt(1/(2sqrt(x))) = 1/2(d/dt1/sqrtx)`

`= 1/2(d/dx1/sqrtx)(dx/dt)`

`= 1/2(-1/(2x^(3/2)))(2 color(white)"." "min"^-1)`

`= -1/(2x^(3/2)) color(white)".""min"^-1`

Finally, all we have to do is plug in `x=1` and `x=4` to answer the two parts to the problem.

`x=1 =>-1/(2x^(3/2)) color(white)".""min"^-1= -1/(2)color(white)".""min"^-1`

`x=4 => -1/(2x^(3/2)) color(white)".""min"^-1 = -1/(2*8) color(white)".""min"^-1 = -1/16 color(white)".""min"^-1`

Final Answer