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# Logarithmic equations

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Skydivers use an instrument called an altimeter to determine their height above Earths surface. An altimeter measures atmospheric pressure and converts it to altitude based on the relationship between pressure and altitude. One model for atmospheric pressure P (in kilo pascals, kPa) as a function of altitude a (in kilometers) is P=100e^(-a/8).

A. Since an altimeter measures pressure directly, pressure is the independent variable for an altimeter.   Rewrite the model P=100e^(-a/8) so that it gives altitude as a function of pressure.

B. To check the function in part a, use the fact that atmospheric pressure at Earths surface is about 100kPa.

C. Suppose a skydiver deploys the parachute when the altimeter measures 87kPa. Use the function in part a to determine the skydivers altitude. Give your answer in both kilometers and feet (1 kilometer=3281 feet).

Guest Dec 8, 2015
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Skydivers use an instrument called an altimeter to determine their height above Earths surface. An altimeter measures atmospheric pressure and converts it to altitude based on the relationship between pressure and altitude. One model for atmospheric pressure P (in kilo pascals, kPa) as a function of altitude a (in kilometers) is P=100e^(-a/8).

A. Since an altimeter measures pressure directly, pressure is the independent variable for an altimeter.   Rewrite the model P=100e^(-a/8) so that it gives altitude as a function of pressure.

$$\small{ \begin{array}{rcll} P &=& 100 e^{\frac{-a}{8}} \qquad & | \qquad : 100 \\\\ \frac{P}{100} &=& e^{\frac{-a}{8}} \qquad & | \qquad \ln{} \\\\ \ln{ ( \frac{P}{100} ) } &=& \ln{ ( e^{\frac{-a}{8}} ) } \\ \ln{ ( \frac{P}{100} ) } &=& \frac{-a}{8} \qquad & | \qquad \cdot (-1) \\\\ -\ln{ ( \frac{P}{100} ) } &=& \frac{a}{8} \\\\ \frac{a}{8} &=& -\ln{ ( \frac{P}{100} ) } \\\\ \frac{a}{8} &=& 0 -\ln{ ( \frac{P}{100} ) } \qquad & | \qquad \ln{(1)} = 0 \\\\ \frac{a}{8} &=& \ln{(1)} -\ln{ ( \frac{P}{100} ) } \\\\ \frac{a}{8} &=& \ln{ \left( \frac{1} { \frac{P}{100} } \right) } \\\\ \frac{a}{8} &=& \ln{ ( \frac{100}{P} ) } \qquad & | \qquad \cdot 8 \\\\ \mathbf{ a } &\mathbf{=}& \mathbf{ 8\cdot \ln{ \left( \frac{100}{P} \right) } }\\ \end{array} }$$

B. To check the function in part a, use the fact that atmospheric pressure at Earths surface is about 100kPa.

$$\small{ \begin{array}{rcll} a & = & 8\cdot \ln{ \left( \frac{100}{P} \right) } \qquad P = 100\ \mathrm{kPa} \\\\ a & = & 8\cdot \ln{ \left( \frac{100}{100} \right) } \\\\ a & = & 8\cdot \ln{ ( 1 ) } \qquad & | \qquad \ln{(1)} = 0 \\\\ a & = & 8\cdot 0 \\\\ \mathbf{ a } & \mathbf{=} & \mathbf{0\ \mathrm{km} } \end{array} }$$

C. Suppose a skydiver deploys the parachute when the altimeter measures 87kPa. Use the function in part a to determine the skydivers altitude. Give your answer in both kilometers and feet (1 kilometer=3281 feet).

$$\small{ \begin{array}{rcll} a & = & 8\cdot \ln{ \left( \frac{100}{P} \right) } \qquad P = 87\ \mathrm{kPa} \\\\ a & = & 8\cdot \ln{ \left( \frac{100}{87} \right) } \\\\ a & = & 8\cdot \ln{ ( 1.14942528736 ) } \\\\ a & = & 8\cdot 0.13926206733 \\\\ \mathbf{ a } & \mathbf{=} & \mathbf{1.11409653867\ \mathrm{km} } \\\\ a & = & 1.11409653867\ \mathrm{km}\cdot \frac{3281\ \mathrm{ft} }{ 1\ \mathrm{km}}\\\\ \mathbf{ a } & \mathbf{=} & \mathbf{3655.35074337\ \mathrm{feet} } \\\\ \end{array} }$$

heureka  Dec 8, 2015

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