Rates and Changes -- what am I doing wrong?? (1 Viewer)

cssftw · Sep 3, 2011

Hi guys, I'm having a bit of a problem with this question:

Q. A sphere is expanding so that its surface area is increasing at a rate of 0.04 cm^2 /second. Find the rate of increase of its volume at the instant when the radius is 10cm.

My solution: (please someone figure out where I went wrong)

A= surface area of sphere

dA/dt = 0.04 cm^2 / second --> this is constant

we are trying to find dV/dt

dV/dt = dA/dt * dV/dA

(here's where I think I might have gone wrong in):

V = (4/3)(pi)(r^3)
A = (4)(pi)(r^2)

therefore: 3V = (A)(r)
V = Ar/3 --> r=10, therefore V = 10A/3

therefore dV/dA = 10/3

dV/dt = dA/dt * dV/dA
dV/dt = 0.04 * 10/3

dv/dt = 2/15 cm^3 / second <--- final answer

However the answers the rate of change was 0.2 cm^3 / second.

So what exactly did I do wrong, and could someone please post a proper solution, and explain WHY I was wrong? Thanks for the help

SpiralFlex · Sep 3, 2011

cssftw said:
Hi guys, I'm having a bit of a problem with this question:

Q. A sphere is expanding so that its surface area is increasing at a rate of 0.04 cm^2 /second. Find the rate of increase of its volume at the instant when the radius is 10cm.

My solution: (please someone figure out where I went wrong)

A= surface area of sphere

dA/dt = 0.04 cm^2 / second --> this is constant

we are trying to find dV/dt

dV/dt = dA/dt * dV/dA

(here's where I think I might have gone wrong in):

V = (4/3)(pi)(r^3)
A = (4)(pi)(r^2)

therefore: 3V = (A)(r)
V = Ar/3 --> r=10, therefore V = 10A/3

therefore dV/dA = 10/3

dV/dt = dA/dt * dV/dA
dV/dt = 0.04 * 10/3

dv/dt = 2/15 cm^3 / second <--- final answer

However the answers the rate of change was 0.2 cm^3 / second.

So what exactly did I do wrong, and could someone please post a proper solution, and explain WHY I was wrong? Thanks for the help

Chain rule!

We know that the surface area of a sphere is $4\pi r^2$ .

Let $A=4\pi r^2$

So $\frac{dA}{dr}=8\pi r$

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

$0.04=8 \pi (10) *\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{1}{2000\pi}$

We know that the volume of a sphere is,

$\frac{4}{3}\pi r^3$

Let $V=\frac{4}{3}\pi r^3$

So $\frac{dV}{dr}=4\pi r^2$

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

$\frac{dV}{dt}=4\pi (10)^2 * \frac{1}{2000\pi}$

$\therefore \frac{dV}{dt}=0.2 cm^3/s$

cssftw · Sep 3, 2011

SpiralFlex said:
Chain rule!

We know that the surface area of a sphere is $4\pi r^2$ .

Let $A=4\pi r^2$

So $\frac{dA}{dr}=8\pi r$

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

$0.04=8 \pi (10) *\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{1}{2000\pi}$

We know that the volume of a sphere is,

$\frac{4}{3}\pi r^3$

Let $V=\frac{4}{3}\pi r^3$

So $\frac{dV}{dr}=4\pi r^2$

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

$\frac{dV}{dt}=4\pi (10)^2 * \frac{1}{2000\pi}$

$\therefore \frac{dV}{dt}=0.2 cm^3/s$

Thanks Spiral, but I was wondering, could you please explain why my method doesn't work? I mean where exactly does it not follow the rules of mathematics and rates and changes?? e.g. like why can't I use dV/dA? thanks.

funnytomato · Sep 3, 2011

cssftw said:
V = (4/3)(pi)(r^3)
A = (4)(pi)(r^2)

therefore: 3V = (A)(r)
V = Ar/3 --> r=10, therefore V = 10A/3

therefore dV/dA = 10/3

something doesn't seem right here, hm...

funnytomato · Sep 3, 2011

V=A*r/3
here both A and r are variables

SpiralFlex · Sep 3, 2011

funnytomato said:
V=A*r/3
here both A and r are variables

I think that.

funnytomato · Sep 3, 2011

so to use ur method correctly, we need the product rule for V=A*r/3
dV/dA= r/3 + A/3*dr/dA, where A=4pi*r^2
so dA/dr=8pi*r, dr/dA= 1/8pi*r, sub this back into dV/dA equation

dV/dA= r/3 + A/3*dr/dA = r/3 + (4pi*r^2 /3) * (1/8pi*r) = r/3 + r/6 =r/2

Rates and Changes -- what am I doing wrong?? (1 Viewer)

cssftw

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SpiralFlex

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cssftw

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