A point is moving along the curve $y = \sqrt{x}$ $y=sqrt(x)$ 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?

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A point is moving along the curve $y = \sqrt{x}$ $y=sqrt(x)$ 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?

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Answer 1 · 2017-06-03T02:19:49

$x = 1 \Rightarrow \frac{d}{d t} (\frac{d y}{d x}) = - \frac{1}{2} . {min}^{- 1}$ $x=1 => d/dt(dy/dx) = -1/2 color(white)".""min"^-1$

$x = 4 \Rightarrow \frac{d}{d t} (\frac{d y}{d x}) = - \frac{1}{16} . {min}^{- 1}$ $x=4 => d/dt(dy/dx) = -1/16 color(white)".""min"^-1$

Explanation:

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

$\frac{d x}{d t} = 2 . {min}^{- 1}$ $dx/dt = 2 color(white)".""min"^-1$

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

$\frac{d}{d t} (\frac{d y}{d x})$ $d/dt(dy/dx)$

First, let's evaluate $\frac{d y}{d x}$ $dy/dx$ :

$\frac{d y}{d x} = \frac{d}{d x} (\sqrt{x}) = \frac{1}{2 \sqrt{x}}$ $dy/dx = d/dx(sqrt(x)) = 1/(2sqrtx)$

Now, we need to evaluate $\frac{d}{d t} (\frac{d y}{d x})$ $d/dt(dy/dx)$ :

$\frac{d}{d t} (\frac{1}{2 \sqrt{x}}) = \frac{1}{2} (\frac{d}{d t} \frac{1}{\sqrt{x}})$ $d/dt(1/(2sqrt(x))) = 1/2(d/dt1/sqrtx)$

$= \frac{1}{2} (\frac{d}{d x} \frac{1}{\sqrt{x}}) (\frac{d x}{d t})$ $= 1/2(d/dx1/sqrtx)(dx/dt)$

$= \frac{1}{2} (- \frac{1}{2 x^{\frac{3}{2}}}) (2 . {min}^{- 1})$ $= 1/2(-1/(2x^(3/2)))(2 color(white)"." "min"^-1)$

$= - \frac{1}{2 x^{\frac{3}{2}}} . {min}^{- 1}$ $= -1/(2x^(3/2)) color(white)".""min"^-1$

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

$x = 1 \Rightarrow - \frac{1}{2 x^{\frac{3}{2}}} . {min}^{- 1} = - \frac{1}{2} . {min}^{- 1}$ $x=1 =>-1/(2x^(3/2)) color(white)".""min"^-1= -1/(2)color(white)".""min"^-1$

$x = 4 \Rightarrow - \frac{1}{2 x^{\frac{3}{2}}} . {min}^{- 1} = - \frac{1}{2 \cdot 8} . {min}^{- 1} = - \frac{1}{16} . {min}^{- 1}$ $x=4 => -1/(2x^(3/2)) color(white)".""min"^-1 = -1/(2*8) color(white)".""min"^-1 = -1/16 color(white)".""min"^-1$

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A point is moving along the curve $y = \sqrt{x}$ $y=sqrt(x)$ 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?

1 Answer

Explanation:

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A point is moving along the curve y=√xy=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?

1 Answer

Explanation:

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A point is moving along the curve $y = \sqrt{x}$ $y=sqrt(x)$ 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?