C4 2017 7 Edexcel

(a)

Since the height of the water is decreasing as a rate of 1.1cm per minute when h = 130 we have:

\displaystyle -1.1 = \frac {\mathrm dh} {\mathrm dt} = k\sqrt{130 - 9} = 11k

So:

\displaystyle k = -\frac {1.1} {11} = -\frac 1 {10}

(b)

Note that:

\displaystyle \frac 1 {\sqrt {h - 9}} \frac {\mathrm dh} {\mathrm dt} = -\frac 1 {10}

Integrating with respect to t:

\displaystyle \int \frac 1 {\sqrt {h -9}} \mathrm dh = \int \left(-\frac 1 {10}\right) \mathrm dt

So:

\displaystyle 2 \sqrt {h - 9} = -\frac 1 {10} t + C

At t = 0 we have h = 200, so:

\displaystyle C = 2 \sqrt {191}

So:

\displaystyle 2 \sqrt {h - 9} =-\frac 1 {10} t + 2 \sqrt {191}

When h = 50 we have:

\displaystyle 2 \sqrt {41} = -\frac 1 {10} t + 2 \sqrt {191}

So:

\displaystyle \frac 1 {10} t = 2(\sqrt {191} - \sqrt {41})

Giving:

t = 20(\sqrt{191} - \sqrt{41})

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