Category Archives: Engineering Mathematics Blog

Problem 1-22: Solving for the area of a circle with a given diameter

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PROBLEM:

What is the area of a circle 3.102 cm in diameter?

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SOLUTION:

The area of a circle can be computed using the formula below when the radius is given.

A=πr2A=\pi r^2

We also know that the radius is half the diameter, so the area can be calculated using the formula,

A=π(d2)2A=\pi \left(\frac{d}{2}\right)^2

So, by direct substitution

A=π(3.102 cm2)2=7.557 cm2  (Answer)A=\pi \left(\frac{3.102\:\text{cm}}{2}\right)^2=7.557\:\text{cm}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

The area of the circle is 7.557 square centimeters.

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Problem 1-21: Counting heart rate with uncertainties in number of beats and time

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PROBLEM:

A person measures his or her heart rate by counting the number of beats in 30 s. If 40±1  beats are counted in 30±0.5 s, what is the heart rate and its uncertainty in beats per minute?

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SOLUTION:

In order to compute for the heart rate in beats per minute, we need to solve for the base. The base is

A=40 beats30 sec ×60 sec1 min=80 beats/minA=\frac{40\:\text{beats}}{30\:\text{sec}\:}\times \frac{60\:\text{sec}}{1\:\text{min}}=80\:\text{beats/min}

Then we compute for the percent uncertainty by combining the uncertainties of the number of beats and time. That is

% uncertainty=(1 beat40 beats×100%)+(0.5 s30.0 s×100%)=2.5%+1.7%=4.2%\begin{align*} \text{\%\:uncertainty} & =\left( \frac{1\:\text{beat}}{40\:\text{beats}}\times 100\% \right)+ \left(\frac{0.5\:\text{s}}{30.0\:\text{s}}\times 100\% \right)\\ &=2.5\%+1.7\% \\ & =4.2\% \\ \end{align*}

Based on this percent uncertainty, we compute for the tolerance

δA=% uncertainty100 %×A=4.2%100%×80 beats/min=3.4 beats/min\begin{align*} \delta _A & =\frac{\text{\%\:uncertainty}}{100\:\%}\times A \\ & = \frac{4.2 \%}{100 \%} \times 80 \ \text{beats/min} \\ & =3.4\:\text{beats/min}\\ \end{align*}

Therefore, the heart rate is

80±3 beats/min  (Answer)\displaystyle 80\pm 3\:\text{beats/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 1-20: Solving for the percent uncertainty of a given blood pressure of 120±2 mmHg

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PROBLEM:

(a) A person’s blood pressure is measured to be 120±2120 \pm 2 mmHg. What is its percent uncertainty?

(b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of 80 mmHg?

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SOLUTION:

Part A

The percent uncertainty is computed as

% uncertainty=2 mmHg120 mmHg×100%=1.7%  (Answer)\text{\% uncertainty}=\frac{2\:\text{mmHg}}{120\:\text{mmHg}}\times 100\%=1.7\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

The uncertainty in the blood pressure is

δbp=1.7 %100 %×80 mmHg=1.3 mmHg  (Answer)\delta _{bp}=\frac{1.7\:\%}{100\:\%}\times 80\:\text{mmHg}=1.3\:\text{mmHg} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 1-19: Calculating the percent uncertainty and range of speed with the same percent uncertainty

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PROBLEM:

(a) If your speedometer has an uncertainty of 2.0 km/h at a speed of 90 km/h, what is the percent uncertainty?

(b) If it has the same percent uncertainty when it reads 60 km/ h, what is the range of speeds you could be going?

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SOLUTION:

Part a

The percent uncertainty is computed as

% uncertainty=2.0 km/hr90 km/hr×100%=2.2%  (Answer)\text{\% uncertainty}=\frac{2.0\ \text{km/hr}}{90\:\text{km/hr}}\times 100\%=2.2\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part b

The tolerance of the velocity is

δv=2.2 %100 %×60 km/hr=1 km/hr\delta _v=\frac{2.2\:\%}{100\:\%}\times 60\:\text{km/hr}=1\:\text{km/hr}

Therefore, the range of the velocity is 60±1km/h, or that is 59 to 61 km/h.   (Answer) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Problem 1-18: Significant figures, uncertainty, and accuracy of the numbers 99 and 100

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PROBLEM:

(a) How many significant figures are in the numbers 99 and 100?

(b) If the uncertainty in each number is 1, what is the percent uncertainty in each?

(c) Which is a more meaningful way to express the accuracy of these two numbers, significant figures or percent uncertainties?

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SOLUTION:

Part A

99 has 2 significant figures

100 has 3 significant figures

Part B

199×100%=1.0 %1100×100%=1.00%\begin{align*} \frac{1}{99}\times 100\% & =1.0\:\% \\ \frac{1}{100}\times 100\% & =1.00\% \end{align*}

Part C

Based on the results of parts a and b, the percent uncertainties are more meaningful.

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Problem 1-17: Stating the correct significant figures in a given calculation

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PROBLEM:

State how many significant figures are proper in the results of the following calculations:

(a) (106.7)(98.2)(46.210)(1.01)  (b) (18.7)2  (c) (1.60×10−19)(3712)\begin{align*} \left(a\right)\:\frac{\left(106.7\right)\left(98.2\right)}{\left(46.210\right)\left(1.01\right)} \ \qquad \ \left(b\right)\:\left(18.7\right)^2 \ \qquad \ \left(c\right)\:\left(1.60\times 10^{-19}\right)\left(3712\right) \end{align*}
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SOLUTION:

Part A

The answer is limited by 98.2 and 1.01. They are both 3 significant figures. So the result should be 3 significant figures.

Part B

The answer is limited by 18.7. The answer should be 3 significant figures. So the result should be 3 significant figures.

Part C

The answer is limited by 1.60 which is 3 significant figures. So the result should be 3 significant figures.

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Problem 1-16: Solving for the remaining soda after the removal of some volume

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PROBLEM:

A can contains 375 mL of soda. How much is left after 308 mL is removed?

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SOLUTION:

What is left is calculated as

Volume left=375 mL−308 mL=67 mL  (Answer)\begin{align*} \text{Volume left} & =375 \ \text{mL}-308 \ \text{mL} \\ & = 67 \ \text{mL} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, 67 mL of soda is left.

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Problem 1-15: Solving for the number of heartbeats of a person with the correct significant figure

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PROBLEM:

(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y?
(b) In 2.00 y?
(c) In 2.000 y?

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SOLUTION:

Part a

This is a problem on rounding off. Take note that 72.0 has three significant figures, 2.0 has two significant figures, 2.00 has three significant figures, and 2.000 has four significant figures.

Convert 72 beats/min to beats/year.

72 beats/min=(72.0 beats1 min)(60.0 min1.00 hr)(24.0 hr1.00 day)(365.25 days1 year)=3.7869×107 beats/year\begin{align*} 72 \ \text{beats/min} & = \left(\frac{72.0\:\text{beats}}{1\:\bcancel{\text{min}}}\right)\left(\frac{60.0\:\bcancel{\text{min}}}{1.00\:\bcancel{\text{hr}}}\right)\left(\frac{24.0\:\bcancel{\text{hr}}}{1.00\:\bcancel{\text{day}}}\right)\left(\frac{365.25\:\bcancel{\text{days}}}{1\:\text{year}}\right)\\ & =3.7869\times 10^7\:\text{beats/year} \end{align*}

In 2 years, the number of beats of an average person is

=3.7869×107 beats/year×2 years=7.5738×107 beats\begin{align*} & = 3.7869 \times 10^{7} \ \text{beats/year} \times 2 \ \text{years} \\ & = 7.5738 \times 10^7 \ \text{beats} \end{align*}

For part a, the answer is limited to 2.0 which is 2 significant figures. The answer is 7.6×107 beats (Answer)7.6 \times 10^{7} \ \text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

Part b

For item letter b), the answer is limited by 2.00, which is 3 significant figures. The answer is 7.57×107 beats (Answer)7.57\times 10^7\:\text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

Part c

For item letter c), the answer is limited by 72.0, which is 3 significant figures. The answer is  7.57×107 beats (Answer)7.57\times 10^7\:\text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

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Problem 1-14: Solving for the percent uncertainty of 130±5 beats/min

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PROBLEM:

An infant’s pulse rate is measured to be 130±5 beats/min. What is the percent uncertainty in this measurement?

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SOLUTION:

The percent uncertainty can be calculated as

% uncertainty =δAA×100%=5 beats/min130 beats/min×100%=3.85 %  (Answer)\begin{align*} \%\:\text{uncertainty}\:& =\frac{\delta _A}{A}\times 100\% \\ & =\frac{5\:\text{beats/min}}{130\:\text{beats/min}}\times 100\% \\ & =3.85\:\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Therefore, the percent uncertainty is 3.85%.

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Problem 1-13: Computing for the range of possible speeds given 90 km/h and 5% uncertainty

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PROBLEM:

(a) A car speedometer has a  5.0% uncertainty. What is the range of possible speeds when it reads 90 km/h?

(b) Convert this range to miles per hour. (1 km=0.6214 mi)

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SOLUTION:

Part A

The uncertainty in the velocity of the car is computed as

δv=5.0 %100 %×90.0 km/hr=4.5 km/hr\begin{align*} \delta _v & =\frac{5.0\:\%}{100\:\%}\times 90.0\:\text{km/hr} \\ & = 4.5\:\text{km/hr} \end{align*}

Therefore, the range of the possible speeds is 

Range=90.0 ±4.5 km/hrRange:85.8 km/hr − 94.5 km/hr  (Answer)\begin{align*} \text{Range} & =90.0\:\pm 4.5\:\text{km/hr} \\ \text{Range} & :85.8\:\text{km/hr}\:-\:94.5\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

So the range of the possible speeds is 85.5 km/hr to 94.5 km/hr.

Part B

Convert the range to mi/h

For 85.5 km/hr

85.5 km/hr=(85 kmhr)(0.6214 mi1 km)=53.13 mi/hr\begin{align*} 85.5\:\text{km/hr}=\left(85\:\frac{\text{km}}{\text{hr}}\right)\left(\frac{0.6214\:\text{mi}}{1\:\text{km}}\right)=53.13\:\text{mi/hr} \end{align*}

For 94.5 km/hr

94.5 km/hr=(94.5 kmhr)(0.6214 mi1 km)=58.72 mi/hr94.5\:\text{km/hr}=\left(94.5\:\frac{\text{km}}{\text{hr}}\right)\left(\frac{0.6214\:\text{mi}}{1\:\text{km}}\right)=58.72\:\text{mi/hr}

Therefore, the range can be represented as 53.13 mi/hr to 58.72 mi/hr.

Range:53.13 mi/hr−58.72 mi/hr  (Answer)\text{Range} : 53.13 \ \text{mi/hr} - 58.72 \ \text{mi/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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