Equal sign or approximation sign?

[shuxue]

The authors of a physics textbook want to determine the number of grains, N in a beach of 500 m long, 100 m wide, and 3 m deep. They assumed that each grain is 1-mm-diameter sphere. They also assumed that the grains are so tightly packed that the volume of the space between the grains is negligible compared to the volume of the sand itself. The authors found the answer as follows:

Volume of the beach = N X Volume of each grain
N = (Volume of the beach) / (Volume of each grain)
$$N=\frac{(500)(100)(3)}{(\frac{4}{3}\pi)(0.5\times10^{-3})^3}=2.9\times10^{14}\approx3\times10^{14}$$

Why equal sign is used between $$\frac{(500)(100)(3)}{(\frac{4}{3}\pi)(0.5\times10^{-3})^3}$$ and $$2.9\times10^{14}$$ instead of approximation sign? Why approximation sign is used between $$2.9\times10^{14}$$ and $$3\times10^{14}$$ instead of equal sign?

 

[Orodruin]

You are giving too much significance to the particular sign used. The computation in itself is approximative.

 

[Mark44]

$$N=\frac{(500)(100)(3)}{(\frac{4}{3}\pi)(0.5\times10^{-3})^3}=2.9\times10^{14}\approx3\times10^{14}$$

The first equals sign should really be ≈, IMO, since the 2.9 x 10¹⁴ is an approximation. This value, in turn, is approximately equal to 3.0 x 10¹⁴.

 

[shuxue]

The example is taken from "Physics: For Scientists and Engineers", by Paul A. Tipler and Gene Mosca.

I found another similar example in another textbook, "Fundamental of physics" by Halliday, Resnick, and Walker. The example is as follows:

A pirate ship is 560 m from a fort defending a harbor entrance. A defense cannon, located at sea level in front of the fort fires balls at initial speed of v = 82 m/s. Thus the maximum range, R of the cannonballs is

$$R=\frac{v^2}{g}=((82 \enspace m/s)^2)/(9.8\enspace m \enspace s^{-2})=686 \enspace m\approx690 \enspace m$$

Again, why equal sign is used between ((82 m/s)^2)/(9.8 m s^(-2)) and 686 m instead of approximation sign? Why approximation sign is used between 686 m and 690 m instead of equal sign?

 

[Mark44]

The example is taken from "Physics: For Scientists and Engineers", by Paul A. Tipler and Gene Mosca.

I found another similar example in another textbook, "Fundamental of physics" by Halliday, Resnick, and Walker. The example is as follows:

A pirate ship is 560 m from a fort defending a harbor entrance. A defense cannon, located at sea level in front of the fort fires balls at initial speed of v = 82 m/s. Thus the maximum range, R of the cannonballs is

$$R=\frac{v^2}{g}=((82 \enspace m/s)^2)/(9.8\enspace m \enspace s^{-2})=686 \enspace m\approx690 \enspace m$$

Again, why equal sign is used between ((82 m/s)^2)/(9.8 m s^(-2)) and 686 m instead of approximation sign? Why approximation sign is used between 686 m and 690 m instead of equal sign?

The = isn't appropriate because $\frac{82^2}{9.8}$ is not equal to 686. A better approximation is 686.122.

I would have written the calculation this way:
$$\frac{82^2}{9.8} \doteq 686 \approx 690$$

For your second question, since 686 is not equal to 690 (obviously), it would be incorrect to write 686 = 690.

 

[sjb-2812]

For your second question, since 686 is not equal to 690 (obviously), it would be incorrect to write 686 = 690.

But then g is not equal to 9.8 ms^-2 either, unless this has been specifically mentioned earlier, even with the restraint of "at sea level"

 

[shuxue]

I want to solve the equation $2 \sin^2\theta+2 \sin \theta-1=0$, $0\leq\theta<2\pi$. By using the quadratic formula. I found one of the solution to be $\theta=\sin^{-1}( \frac{-1+\sqrt{3}}{2})$. Which of the following ways of writing the numerical value of $\theta$ is correct? (i) or (ii)?

(i) $\theta=0.3747$, (ii) $\theta\approx 0.3747$

Are both of them acceptable? I noticed that, frequently, in most trigonometric textbooks the numerical solution of a trigonometric equation is written using the equal sign (as in (ii)) even the numerical answer has been rounded. Why equal sign is used instead of approximation sign?

 

[Mark44]

I want to solve the equation $2 \sin^2\theta+2 \sin \theta-1=0$, $0\leq\theta<2\pi$. By using the quadratic formula. I found one of the solution to be $\theta=\sin^{-1}( \frac{-1+\sqrt{3}}{2})$. Which of the following ways of writing the numerical value of $\theta$ is correct? (i) or (ii)?

(i) $\theta=0.3747$, (ii) $\theta\approx 0.3747$

Are both of them acceptable?

Not in my opinion. The first is only an approximation, but using = doesn't indicate that.

I noticed that, frequently, in most trigonometric textbooks the numerical solution of a trigonometric equation is written using the equal sign (as in (ii)) even the numerical answer has been rounded.

As I recall, most of the trig books I've used (either as a student or as a teacher), the authors took pains to distinguish between exact answers (using = ) and answers that were rounded (using ≈ or $\doteq$).

Why equal sign is used instead of approximation sign?

You keep asking this, so apparently you aren't understanding. The = sign should be used for exact answers, and the ≈ should be used if you are writing only an approximate value.