Rates of Change... (1 Viewer)

JAult · Jun 17, 2011

I did question 7 b) of 7H in cambridge 3U and I though it was correct. But, it isn't. Can someone please show the steps of this question...

Much appreciation

A spherical balloon is to be filled with water so that its surface area increases at a constant rate of 1 cm2 /s.
(a) Find, when the radius is 3cm: (i) the required rate of increase of the radius, (ii) the rate at which the water must be flowing in at that time.
(b) Find the volume when the volume is increasing at 10 cm3 /s.

Drongoski · Jun 17, 2011

JAult said:
I did question 7 b) of 7H in cambridge 3U and I though it was correct. But, it isn't. Can someone please show the steps of this question...

Much appreciation

A spherical balloon is to be filled with water so that its surface area increases at a constant rate of 1 cm2 /s.
(a) Find, when the radius is 3cm: (i) the required rate of increase of the radius, (ii) the rate at which the water must be flowing in at that time.
(b) Find the volume when the volume is increasing at 10 cm3 /s.

$V(r) = \frac {4}{3}\pi r^3 \implies \frac {dV}{dt} = 4 \pi r^2$

$A(r) = 4 \pi r^2 \implies \frac {dA}{dt} = 8 \pi r$

a)

$\frac {dA}{dt} = \frac {dA}{dr} \times \frac {dr}{dt} = 1 \implies \frac {dr}{dt} = \left ( { \frac {dA}{dr} \right ) ^{-1} = \frac {1}{8 \pi r} = \frac {1}{8 \pi \times 3} = \frac {1}{24 \pi}$

$\frac {dV}{dt} = \frac {dV}{dr} \times \frac {dr}{dt} = 4 \pi r^2 \times \frac {1}{24 \pi} = \frac {4 \pi \times 3^2}{24 \pi} = \frac {3}{2} cm^3/sec$

b)

$\frac {dV}{dt} = 10 \implies \frac {dV}{dr} \times \frac {dr}{dt} \implies 4 \pi r^2 \times \frac {dr}{dt} = 10 \implies \frac {dr}{dt} = \frac {10}{4 \pi r^2} = \frac {1}{8 \pi r}$

$\therefore 4 \pi r^2 = 8 \times 10 \times \pi r \implies r = 20$

$\therefore V(20) = \frac {4}{3} \times \pi \times 20^3 = \frac {32000 \pi}{3} cm^3$

SpiralFlex · Jun 17, 2011

I feel like blue today.

a) $\frac{dA}{dt} = \frac{dA}{dr} * \frac{dr}{dt}$

$1 = 8 \pi r *\frac{dr}{dt}$

Radius of 3,

$1 = 24 \pi *\frac{dr}{dt}$

$\therefore \frac{dr}{dt}= \frac{1}{24 \pi} \; cm/s$

b) $\frac{dV}{dt} = \frac{dV}{dr} * \frac{dr}{dt}$

$\frac{dV}{dt} = 4 \pi r^2 * \frac{1}{24 \pi}$

$= 4 \pi (3)^2 * \frac{1}{24 \pi}$

$\therefore \frac{dV}{dt} = 1.5 \; cm^3/s$

c) Consider, $\frac{dA}{dt} = \frac{dA}{dr} * \frac{dr}{dt}$

and $\frac{dV}{dt} = \frac{dV}{dr} * \frac{dr}{dt}$

We can make $\frac{dr}{dt}$ the subject, of the first one.

$\frac{dr}{dt} = \frac{dA}{dt} * \frac{dr}{dA}$

Also second one,

$\frac{dr}{dt} = \frac{dV}{dt} * \frac{dr}{dV}$

Hence,

$\frac{dV}{dt} * \frac{dr}{dV} = \frac{dA}{dt} * \frac{dr}{dA}$

Substitute in our values,

$10* \frac{1}{4 \pi r^2} = 1 * \frac{1}{8 \pi r}$

Now times all sides by $8 \pi r^2$ . We get,

$r=20$

Now substitute into our volume formula,

$V=\frac{4}{3} \pi (20)^3$

$\therefore V = \frac{32000 \pi}{3} \; cm^3$

SpiralFlex · Jun 17, 2011

Drongoski said:
$V(r) = \frac {4}{3}\pi r^3 \implies \frac {dV}{dt} = 4 \pi r^2$

$A(r) = 4 \pi r^2 \implies \frac {dA}{dt} = 8 \pi r$

$\frac {dA}{dt} = \frac {dA}{dr} \times \frac {dr}{dt} = 1 \implies \frac {dr}{dt} = \left ( { \frac {dA}{dr} \right ) ^{-1} = \frac {1}{8 \pi r} = \frac {1}{8 \pi \times 3} = \frac {1}{24 \pi}$

$\frac {dV}{dt} = \frac {dV}{dr} \times \frac {dr}{dt} = 4 \pi r^2 \times \frac {1}{24 \pi} = \frac {4 \pi \times 3^2}{24 \pi} = \frac {3}{2} cm^3/sec$

$\frac {dV}{dt} = 10 \implies \frac {dV}{dr} \times \frac {dr}{dt} = 10 \implies 4 \pi r^2 \times \frac {dr}{dt} = 10 \implies \frac {dr}{dt} = \frac {10}{4 \pi r^2} = \frac {1}{8 \pi r}$

$\therefore 4 \pi r^2 = 8 \times 10 \times \pi r \implies r = 20$

So many arrows.

Drongoski · Jun 17, 2011

SpiralFlex said:
So many arrows.

More arrows $\implies$ less lines.

JAult · Jun 17, 2011

Thank you so much <3

Rates of Change... (1 Viewer)

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