Question

A house cost $120,000 when it was purchased. The value of the house increases by 10% each year. Find the rate of growth each month.

Answer 1

FIRST MODEL: 

Well the model for the value of the house is:

$V={ \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000$

V = Value

t = Years passed {t≥0}

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When t=0, V=120000

When t=1, V=132000

When t=2, V=145200

etc... etc...

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Now, this model is actually curved so there is no constant rate of growth each month. We can only calculate what the rate of growth is at a particular time. If we want to find out the rate of growth at a particular time, we must differentiate the formula (model) above.

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$V={ \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000\\ \\ \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000 \right) } }$

$\\ \\ \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ t } \right) } } +\ln { \left( 120000 \right) } \\ \\ \ln { V=t\ln { \left( \frac { 11 }{ 10 } \right) } } +\ln { \left( 120000 \right) }$

$\\ \\ \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \\ \\ V\cdot \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \cdot V$

$\\ \\ \therefore \quad \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \cdot { \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000$

Plug any value of (t) that is greater than 0 into the formula above to find out how quickly the investment is growing. If you want to find out how quickly the investment was growing after 1 month had passed, transform t into 1/12.

The rate of growth is being measured in years, not months. So when t=1/12, the rate of growth turns out to be 11528.42 per annum.

SECOND MODEL (What you are ultimately looking for):

$V={ \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000$

V = Value of house

t = months that have gone by {t≥0}

Formula above differentiated:

$V={ \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000\\ \\ \ln { V } =\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000 \right) }$

$\\ \\ \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } } \right) } } +\ln { \left( 120000 \right) }$

$\\ \\ \ln { V=\frac { t }{ 12 } } \ln { \left( \frac { 11 }{ 10 } \right) } +\ln { \left( 120000 \right) }$

$\\ \\ \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) }$

$\\ \\ V\cdot \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) } \cdot V$

$\\ \\ \therefore \quad \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) } \cdot { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000$

When t=1, dV/dt = 960.70 (2dp)

dV/dt in this case will measure the rate of growth monthly. As more money is accumulated, this rate of growth will rise. The rate of growth is constantly increasing as the graph of V is actually a curve. You can only find out the rate at which the house value is growing monthly at a particular time.