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Algebra 2 · F.BF.A.1

Radical functions in context

Water four times as deep carries a wave only twice as fast. How do you read, use and build a rule whose output grows like the square root of its input?

You need
Graphing square root functionsAlgebra 2Function notation in a real situationAlgebra 1
Where you are 3 in Radical functions
Functions13
Radical functions3
Cube root functions and roots as powersGraphing square root functions
Radical functions in context

When you reach for it

A wave rolling toward a beach slows down as the water gets shallower. For water that is shallow compared with the length of the wave, its speed in meters per second is

v(d) = \sqrt{9.8d}

where d is the depth of the water in meters. Rules like this one, with the input under a square root sign, describe many quantities: the time a pendulum takes to swing, the distance to the horizon from a height, the speed a car was going from the length of its braking marks, the side of a square from its area. You reach for this skill whenever such a rule describes a real situation and you need to say what it means, use it in both directions, or write it down in the first place.

The reading comes from Function notation in a real situation: v(20) = 14 is a sentence about water and waves, not only about numbers. The shape comes from Graphing square root functions: a half curve from an endpoint, steep at first and flattening. This lesson joins the two and adds what a situation brings to a square root rule: units, a domain set by the world rather than by the formula, the habit of running the rule backwards, and the one fact that makes square root quantities recognizable, that multiplying the input by 4 only doubles the output.

The procedure

The running example is the wave rule \displaystyle v(d) = \sqrt{9.8d}, depth d in meters, speed v in meters per second.

Read the rule: input, output and units

Before computing anything, name the two quantities and their units. The input is a depth, in meters; the output is a speed, in meters per second. Then each statement in function notation is a sentence:

v(20) = \sqrt{9.8 \cdot 20} = \sqrt{196} = 14

says that in water 20 m deep, a wave travels at 14 m/s. Write it in words with both units. A bare “14” answers nothing.

The domain the situation allows

The formula alone accepts every input that makes the radicand (the expression under the root) at least 0: 9.8d \ge 0, so d \ge 0. The situation is stricter. A depth of 0 is a dry beach with no wave at all, so the depths that make sense are d > 0. And the rule was only ever claimed for water that is shallow compared with the wave, so a sensible domain also has an upper end that depends on the waves in question. The domain of a model is always the formula’s domain cut down by what the quantities can really be.

Forward and backward

Forward means evaluate: given a depth, find the speed. Backward means solve: given a speed, find the depth. For a speed of 10 m/s,

\begin{aligned} \sqrt{9.8d} &= 10 \\ 9.8d &= 100 \\ d &= \frac{100}{9.8} \approx 10.2 \end{aligned}

Squaring both sides is safe here because 10 is not negative, so it really is the square root of 100. Doing the same with a letter v in place of 10 gives the backward rule once and for all:

d = \frac{v^2}{9.8}

the depth needed for a wave speed v. This is the inverse of the wave rule, and it is a square law: the square root rule and the square law describe the same link between depth and speed, read in opposite directions. Solving radical equations treats the backward step in general.

What multiplying the input does

Here is the fact that marks a square root quantity. Multiply the depth by any number k > 0:

v(kd) = \sqrt{9.8kd} = \sqrt{k}\,\sqrt{9.8d} = \sqrt{k}\; v(d)

The second step uses \displaystyle \sqrt{ab} = \sqrt{a}\sqrt{b}, allowed because k and 9.8d are both non-negative. So multiplying the input by k multiplies the output by \displaystyle \sqrt{k}. Four times the depth gives twice the speed, nine times the depth three times the speed, and twice the depth only about 1.41 times the speed. Read the other way, doubling the speed takes four times the depth. No calculator is needed for any of these, and none of them depends on the 9.8: the same holds for every rule of the form \displaystyle y = c\sqrt{x}.

The graph in context

The values v(1.25) = 3.5, v(5) = 7, v(11.25) = 10.5 and v(20) = 14 all come out exactly, since 9.8 \times 1.25 = 12.25 = 3.5^2, and so on. They show the shape.

d v 5 10 15 20 5 10 15 (1.25, 3.5) (5, 7) (11.25, 10.5) (20, 14)
Figure 1Wave speed against depth. Each step of 3.5 m/s in speed costs a longer stretch of depth than the one before, and four times the depth, from 5 m to 20 m, only doubles the speed.

The speed rises by the same 3.5 m/s from point to point, but the depth has to grow by 1.25, then 3.75, then 6.25, then 8.75 meters. That is the flattening of every square root graph, and in context it says something about the sea: near the shore, where the water is shallow, a small change in depth makes a large change in speed, while far out a few meters more or less hardly matter. The graph starts at its endpoint (0, 0) and only the part with d > 0 means anything here; nothing is drawn for negative depths.

Building a rule from a statement

Often the rule is not given. Two ways to get one are common.

  • From a relationship between the quantities. If one quantity is the square of another, or the two are tied by the The Pythagorean theorem, solve for the one you want, and a square root appears. The side of a square of area A is \displaystyle s = \sqrt{A} because s^2 = A.
  • From “proportional to the square root” and one measurement. Saying y is proportional to \displaystyle \sqrt{x} means \displaystyle y = c\sqrt{x} for some number c, in the same way as Direct variation with \displaystyle \sqrt{x} in place of x. One pair of values fixes c.

Worked examples

Example 1

Reading the wave rule

Use \displaystyle v(d) = \sqrt{9.8d} to find the wave speed in water 5 m deep and in water 2 m deep. Write each result as a sentence, and say what the statement v(d) = 3.5 asks.

Solution

Substitute each depth for d. For 5 m, 9.8 \times 5 = 49, a perfect square:

v(5) = \sqrt{49} = 7

In water 5 m deep, a wave travels at 7 meters per second. For 2 m, 9.8 \times 2 = 19.6, which is not a perfect square, so round at the end:

v(2) = \sqrt{19.6} \approx 4.43

In water 2 m deep, a wave travels at about 4.43 meters per second. Both sentences carry both units: the input is a depth in meters, the output a speed in meters per second.

The statement v(d) = 3.5 fixes the output and asks for the input: at what depth does a wave travel at 3.5 m/s? It is a question about the depth, and Example 2 answers questions of this kind.

Example 2

Running the rule backwards

(a) At what depth does a wave travel at 3.5 m/s?

(b) The same rule describes a tsunami, since its wave is hundreds of kilometers long and so even the open ocean counts as shallow for it. How fast does it travel over ocean 4000 m deep, in meters per second and in kilometers per hour? How many times faster is that than over water 100 m deep?

Solution

(a) Use the backward rule \displaystyle d = \frac{v^2}{9.8} found in the procedure:

d = \frac{3.5^2}{9.8} = \frac{12.25}{9.8} = 1.25

The depth is 1.25 m. Check: \displaystyle \sqrt{9.8 \times 1.25} = \sqrt{12.25} = 3.5.

(b) Forward this time:

v(4000) = \sqrt{39200} \approx 198

about 198 m/s. A meter per second is 3.6 kilometers per hour (there are 3600 seconds in an hour and 1000 meters in a kilometer), so the speed is about 198 \times 3.6 \approx 713 km/h, not far short of a passenger jet.

For the comparison, 4000 m is 40 times 100 m, so the speed is \displaystyle \sqrt{40} times as large, by the scaling fact:

\sqrt{40} \approx 6.32

Directly, \displaystyle v(100) = \sqrt{980} \approx 31.3 m/s, and 198 \div 31.3 \approx 6.3, which agrees. Forty times the depth gives only a little over six times the speed.

Example 3

A pendulum

A pendulum of length L meters takes

T(L) = 2\pi\sqrt{\frac{L}{9.8}}

seconds for one full swing, out and back, as long as its swings are small. (a) Give a sensible domain and find T(1). (b) How long must a pendulum be to take exactly 2 seconds? (c) A pendulum is cut to a quarter of its length. What happens to its period?

Solution

(a) The formula needs \displaystyle \frac{L}{9.8} \ge 0, so L \ge 0. A pendulum has some length, so the domain in context is L > 0, with an upper end set by the room it hangs in. Then

T(1) = 2\pi\sqrt{\frac{1}{9.8}} \approx 2.01

A pendulum 1 m long takes about 2.01 seconds for a full swing.

(b) Set T(L) = 2 and undo the rule one operation at a time:

\begin{aligned} 2\pi\sqrt{\frac{L}{9.8}} &= 2 \\ \sqrt{\frac{L}{9.8}} &= \frac{1}{\pi} \\ \frac{L}{9.8} &= \frac{1}{\pi^2} \\ L &= \frac{9.8}{\pi^2} \approx 0.993 \end{aligned}

Dividing by 2\pi is allowed since 2\pi \ne 0, and squaring is safe because \displaystyle \frac{1}{\pi} is positive. The pendulum must be about 0.993 m long, just under a meter. This is why a clock pendulum that ticks once a second, half a swing per tick, is close to a meter long.

(c) Replacing L by \displaystyle \frac{L}{4} multiplies the period by \displaystyle \sqrt{\frac{1}{4}} = \frac{1}{2}:

T\!\left(\tfrac{L}{4}\right) = 2\pi\sqrt{\frac{L}{4 \cdot 9.8}} = \frac{1}{2} \cdot 2\pi\sqrt{\frac{L}{9.8}} = \frac{1}{2}\,T(L)

The period halves. For the 1 m pendulum, T(0.25) \approx 1.00 second, half of 2.01.

Example 4

Speed from braking distance

On a dry road, a car’s braking distance is proportional to the square of its speed, so its speed is proportional to the square root of its braking distance. A test car at 60 km/h stops in 20 m. (a) Write the speed v in km/h as a function of the braking distance d in meters. (b) A car of the same kind on the same road left braking marks 45 m long. How fast was it going? (c) What braking distance goes with 100 km/h?

Solution

(a) Proportional to \displaystyle \sqrt{d} means \displaystyle v = c\sqrt{d} for some number c. The test gives one pair of values, v = 60 when d = 20:

60 = c\sqrt{20}, \qquad c = \frac{60}{\sqrt{20}} = \frac{60}{2\sqrt{5}} = 6\sqrt{5} \approx 13.4

So \displaystyle v(d) = 6\sqrt{5}\,\sqrt{d}, with d in meters and v in km/h. Keeping c exact as \displaystyle 6\sqrt{5} makes the next part come out exactly.

(b) Substitute 45:

v(45) = 6\sqrt{5}\,\sqrt{45} = 6\sqrt{225} = 6 \cdot 15 = 90

The car was going 90 km/h. The scaling fact gives the same answer at once: 45 m is \displaystyle \frac{45}{20} = \frac{9}{4} times the test distance, so the speed is \displaystyle \sqrt{\frac{9}{4}} = \frac{3}{2} times 60 km/h.

(c) Solve \displaystyle 6\sqrt{5}\,\sqrt{d} = 100. Squaring both sides, which is safe since 100 > 0, gives 180d = 10000, so

d = \frac{10000}{180} \approx 55.6

about 55.6 m. Notice how the square law shows in the numbers: from 60 to 100 km/h is less than double the speed, yet the braking distance grows from 20 m to over 55 m.

Example 5

A shifted rule: where the endpoint is set by the situation

A bolt slips from a crane hook h meters above the ground and lands on the flat roof of a site cabin 3.2 m tall. A dropped object falls 4.9t^2 meters in its first t seconds, if air resistance is ignored. (a) Write the time of the fall as a function of h. (b) Give its domain, from the formula and from the situation. (c) How long does a fall from 22.8 m take? (d) From what height does the fall take 3 seconds?

Solution

(a) The bolt falls from height h to the roof at 3.2 m, a distance of h - 3.2 meters. So the time t satisfies 4.9t^2 = h - 3.2, and since time is not negative, take the non-negative root:

t(h) = \sqrt{\frac{h - 3.2}{4.9}}

(b) The formula needs h - 3.2 \ge 0, so h \ge 3.2. Its graph is a half curve with endpoint (3.2, 0), and the endpoint has a meaning: a bolt let go at roof level takes no time to land. The situation says the same and a little more: the hook is above the roof, so h > 3.2, and no higher than the crane can lift it.

(c) Here h - 3.2 = 19.6, and 19.6 \div 4.9 = 4:

t(22.8) = \sqrt{4} = 2

The fall takes 2 seconds.

(d) Set t(h) = 3 and square, which is safe as 3 > 0:

\begin{aligned} \frac{h - 3.2}{4.9} &= 9 \\ h - 3.2 &= 44.1 \\ h &= 47.3 \end{aligned}

From 47.3 m. Check: \displaystyle \frac{47.3 - 3.2}{4.9} = \frac{44.1}{4.9} = 9, and \displaystyle \sqrt{9} = 3. Compare (c) and (d): the extra second of falling needs 24.5 m more height, because each second of a fall covers more distance than the one before.

Example 6

How far is the horizon?

Standing at the edge of the sea with your eyes h meters above the water, you can see the sea up to a distance d where your line of sight just grazes the surface. Taking the Earth as a ball of radius 6371 km, find a rule for d as a function of h, then use it for eyes 1.6 m above the water and for a lighthouse lamp 64 m up. How high must you be to see 10 km?

Solution

The line of sight touches the sea at one point T, so it is a tangent to the circle in the figure, and a tangent is perpendicular to the radius at the point where it touches. The triangle with corners at the center O, the eye E and T therefore has a right angle at T.

O
E
T
R
h
R
d
Figure 2The sight line from an eye at height h just touches the sea at T, so it is perpendicular to the radius OT. The height is drawn far larger than it really is.

Work in kilometers, with R = 6371 and the height H in km. The hypotenuse OE is R + H, and the legs are R and d. By the The Pythagorean theorem,

\begin{aligned} (R + H)^2 &= R^2 + d^2 \\ R^2 + 2RH + H^2 &= R^2 + d^2 \\ d &= \sqrt{2RH + H^2} \end{aligned}

taking the positive root because d is a distance. That rule is exact. For any height a person can stand at, H^2 is tiny next to 2RH: for eyes at 1.6 m, H = 0.0016, so 2RH \approx 20.39 while H^2 = 0.00000256. Dropping it gives \displaystyle d \approx \sqrt{2RH}, a rule of the form \displaystyle c\sqrt{H}. With the height in meters, \displaystyle H = \frac{h}{1000}, and

d \approx \sqrt{\frac{2 \times 6371\,h}{1000}} = \sqrt{12.742\,h} \approx 3.57\sqrt{h}

kilometers, with h in meters. Now the questions:

  • Eyes at 1.6 m: \displaystyle d \approx \sqrt{12.742 \times 1.6} = \sqrt{20.3872} \approx 4.5 km.
  • A lamp at 64 m: \displaystyle d \approx 3.57\sqrt{64} = 3.57 \times 8 \approx 28.6 km.
  • To see 10 km, solve \displaystyle \sqrt{12.742\,h} = 10: \displaystyle h = \frac{100}{12.742} \approx 7.8 m.

The lamp is 40 times as high as the eyes, and it sees \displaystyle \sqrt{40} \approx 6.3 times as far, which is why a lighthouse has to be tall to be seen from far out at sea.

Boundaries

A model’s domain is smaller than its formula’s. The radicand being at least 0 is a condition on the formula. The situation adds its own conditions: a depth is positive, a pendulum has a length, the bolt starts above the roof. Answers outside the situation’s domain, such as a negative depth found by careless algebra, are rejected even when the formula accepts them.

Each rule holds only where its physics holds. The wave rule is for water shallow compared with the wavelength; in deep water waves follow a different law. The pendulum rule is for small swings. The falling rule ignores air resistance, and the braking rule assumes the same car on the same road. Outside those conditions the formula still gives numbers, but they are no longer speeds or times of anything real.

The scaling fact needs a pure root, \displaystyle y = c\sqrt{x}. For the shifted rule \displaystyle t(h) = \sqrt{\frac{h - 3.2}{4.9}}, multiplying h by 4 does not double t: t(22.8) = 2 but t(91.2) \approx 4.24, not 4. It is the radicand, h - 3.2, that must be multiplied by 4.

Squaring to solve needs a non-negative target. The step from \displaystyle \sqrt{9.8d} = v to 9.8d = v^2 is reversible only when v \ge 0, since a square root is never negative. A negative target means no input works, and squaring would hide that. In context a negative speed or time does not arise, but a result should still be checked in the original rule.

Rounded constants give rounded answers. The 9.8 in these rules is the pull of gravity, rounded, and 3.57 in the horizon rule is rounded too. Answers carry no more accuracy than that: 28.6 km is honest, 28.5600 km is not.

Where this goes next

The backward step in every example, squaring to free the input, is the subject of Solving radical equations, where an equation may have its root anywhere and squaring can create an answer that does not belong. That a square root rule and a square law are the same link read in two directions is the idea behind Inverse functions, and Restricting a domain to get an inverse explains why the square law has to be cut to non-negative inputs to be undone by a root. Quantities that grow like a cube root rather than a square root, such as the side of a cube from its volume, belong to Cube root functions and roots as powers, and all of these sit in the family of Radical functions.

Exercises

Reading a rule in context

  1. Use \displaystyle v(d) = \sqrt{9.8d} to find the wave speed in water 10 m deep, to two decimal places, and write the result as a sentence.
  2. Explain what v(2.45) = 4.9 says about waves, and check that it is true.
  3. Use \displaystyle T(L) = 2\pi\sqrt{\frac{L}{9.8}} to find the period of a pendulum 0.5 m long, to two decimal places.
  4. Use the horizon rule \displaystyle d(h) = 3.57\sqrt{h} (distance in km, height in m) to find how far out to sea you can see from a cliff top 25 m above the water.
  5. A circular pond has area A square meters. Write its radius r as a function of A, and find the radius of a pond of area 200 m², to two decimal places.

The domain in context

  1. A harbor is 0 to 3 m deep, and the waves crossing it are long compared with that depth. Give the domain of the formula \displaystyle v(d) = \sqrt{9.8d}, a domain that fits waves in this harbor, and the speeds the rule then allows, to two decimal places.
  2. A ladder 6.5 m long leans against a wall with its foot x m from the wall. Explain why the height of its top is \displaystyle h(x) = \sqrt{42.25 - x^2}, give the formula’s domain and the situation’s, and find h(2.5) and h(1.6).
  3. A conker falls from a branch h m above the ground onto a garden wall 1.7 m tall, so it falls for \displaystyle t(h) = \sqrt{\frac{h - 1.7}{4.9}} seconds. Give the domain, say what the endpoint of the graph means, and find t(6.6).

Running the rule backwards

  1. How long must a pendulum be to take 3 seconds for a full swing? Give the length to two decimal places.
  2. Using \displaystyle d(h) = 3.57\sqrt{h}, how high must a lookout stand to see 20 km out to sea? Give the height to one decimal place.
  3. Using your rule from Exercise 5, what area does a circular pond of radius 5 m have, to one decimal place?
  4. At what depth does a wave travel at 20 m/s? Give the depth to one decimal place.

What multiplying the input does

  1. Without a calculator, how many times faster is a wave in water 18 m deep than in water 2 m deep?
  2. A pendulum takes 1.5 seconds for a full swing. How many times as long must a pendulum be to take 3 seconds?
  3. A lookout climbs from 30 m to 60 m above the sea. By what factor does the distance to the horizon grow? Check with \displaystyle d(h) = 3.57\sqrt{h} at both heights.
  4. Use the braking rule of Example 4 for braking marks 5 m long.

Building a rule

  1. The top speed of a boat that pushes through the water, rather than skimming over it, is roughly proportional to the square root of its length at the waterline. A boat 9 m long at the waterline has a top speed of about 7.2 knots. Write the rule, and estimate the top speed of a boat 16 m long.
  2. Write the perimeter P of a square as a function of its area A, and find the perimeter of a square field of area 30 m², to one decimal place.
  3. A drone rises straight up from a point 30 m from where you stand. Write its distance D from you as a function of its height h, find D(40), and find the height at which it is 60 m from you, to one decimal place.

Reasoning

  1. Show that for any rule \displaystyle y = c\sqrt{x} with c > 0, multiplying the input by k > 0 multiplies the output by \displaystyle \sqrt{k}. Then show that this fails for \displaystyle y = \sqrt{x + 1} by trying x = 3 and k = 4.
  2. Explain, from the graph of \displaystyle v(d) = \sqrt{9.8d}, why the speed of a wave changes much more between depths of 1 m and 2 m than between 11 m and 12 m.
  3. A classmate solves \displaystyle \sqrt{9.8d} = -3 by squaring and gets \displaystyle d = \frac{9}{9.8}. Explain what is wrong, both with the algebra and with the question.
  4. The horizon rule \displaystyle d \approx \sqrt{2RH} (in km, with R = 6371) dropped the H^2 term. For a spacecraft 400 km up, compare \displaystyle \sqrt{2RH} with the exact \displaystyle \sqrt{2RH + H^2}, and say whether dropping the term is still reasonable.
Answers
  1. \displaystyle v(10) = \sqrt{98} \approx 9.90. In water 10 m deep, a wave travels at about 9.90 meters per second.
  2. In water 2.45 m deep, a wave travels at 4.9 meters per second. Check: 9.8 \times 2.45 = 24.01 and \displaystyle \sqrt{24.01} = 4.9.
  3. \displaystyle T(0.5) = 2\pi\sqrt{\frac{0.5}{9.8}} \approx 1.42 seconds.
  4. \displaystyle d(25) = 3.57\sqrt{25} = 3.57 \times 5 = 17.85, so about 18 km. From 25 m up you see about 18 km out to sea.
  5. \pi r^2 = A, so \displaystyle r(A) = \sqrt{\frac{A}{\pi}}, taking the positive root as r is a length. \displaystyle r(200) = \sqrt{\frac{200}{\pi}} \approx 7.98 m.
  6. The formula’s domain is d \ge 0. For this harbor, 0 < d \le 3. The speeds are \displaystyle 0 < v \le \sqrt{29.4}, that is up to about 5.42 m/s, since the rule rises as d rises.
  7. The wall, the ground and the ladder make a right triangle with hypotenuse 6.5, so x^2 + h^2 = 6.5^2 = 42.25 and \displaystyle h = \sqrt{42.25 - x^2}, positive as it is a height. The formula needs 42.25 - x^2 \ge 0, so -6.5 \le x \le 6.5; the situation needs 0 < x < 6.5. \displaystyle h(2.5) = \sqrt{36} = 6 m and \displaystyle h(1.6) = \sqrt{39.69} = 6.3 m.
  8. Domain h \ge 1.7 from the formula, and h > 1.7 in the situation, since the branch is above the wall. The endpoint (1.7, 0) says a conker starting at the top of the wall takes no time to land on it. \displaystyle t(6.6) = \sqrt{\frac{4.9}{4.9}} = 1 second.
  9. \displaystyle \sqrt{\frac{L}{9.8}} = \frac{3}{2\pi}, so \displaystyle L = 9.8\left(\frac{3}{2\pi}\right)^2 \approx 2.23 m.
  10. \displaystyle \sqrt{h} = \frac{20}{3.57}, so \displaystyle h = \left(\frac{20}{3.57}\right)^2 \approx 31.4 m.
  11. From \displaystyle r = \sqrt{\frac{A}{\pi}}, squaring gives A = \pi r^2 = 25\pi \approx 78.5 m².
  12. \displaystyle d = \frac{20^2}{9.8} = \frac{400}{9.8} \approx 40.8 m.
  13. 3 times, because 18 m is 9 times 2 m and \displaystyle \sqrt{9} = 3. (Directly, \displaystyle \sqrt{176.4} \approx 13.28 and \displaystyle \sqrt{19.6} \approx 4.43.)
  14. 4 times as long. Doubling the period needs the length multiplied by k with \displaystyle \sqrt{k} = 2.
  15. By \displaystyle \sqrt{2} \approx 1.41. Check: \displaystyle 3.57\sqrt{30} \approx 19.6 km and \displaystyle 3.57\sqrt{60} \approx 27.7 km, and 27.7 \div 19.6 \approx 1.41.
  16. \displaystyle v(5) = 6\sqrt{5}\,\sqrt{5} = 6 \cdot 5 = 30 km/h. As a check, 5 m is a quarter of 20 m, so the speed is half of 60 km/h.
  17. \displaystyle v = c\sqrt{L} with \displaystyle 7.2 = c\sqrt{9} = 3c, so c = 2.4 and \displaystyle v(L) = 2.4\sqrt{L} knots. v(16) = 2.4 \times 4 = 9.6 knots.
  18. The side is \displaystyle \sqrt{A}, so \displaystyle P(A) = 4\sqrt{A}. \displaystyle P(30) = 4\sqrt{30} \approx 21.9 m.
  19. The ground distance 30, the height h and the distance D form a right triangle, so \displaystyle D(h) = \sqrt{900 + h^2}. \displaystyle D(40) = \sqrt{2500} = 50 m. For D = 60: 900 + h^2 = 3600, h^2 = 2700, \displaystyle h = \sqrt{2700} \approx 52.0 m.
  20. \displaystyle c\sqrt{kx} = c\sqrt{k}\sqrt{x} = \sqrt{k}\,(c\sqrt{x}), using \displaystyle \sqrt{ab} = \sqrt{a}\sqrt{b} for a, b \ge 0. For \displaystyle \sqrt{x + 1}: at x = 3 the output is 2, and at x = 12 it is \displaystyle \sqrt{13} \approx 3.61, not 2 \times 2 = 4. The +1 inside means multiplying x does not multiply the radicand.
  21. The graph is steep near d = 0 and flattens as d grows. Equal steps of depth therefore give big rises in speed near the start and small ones further along: from 1 m to 2 m the speed goes from about 3.13 to 4.43 m/s, a rise of about 1.3, while from 11 m to 12 m it goes from about 10.38 to 10.84 m/s, a rise of under 0.5.
  22. A square root is never negative, so \displaystyle \sqrt{9.8d} = -3 has no solution. Squaring turned it into 9.8d = 9, which has one, and \displaystyle d = \frac{9}{9.8} gives \displaystyle \sqrt{9} = 3, not -3. In context the question asks for a depth with a negative speed, and the rule gives speeds as sizes, never negative.
  23. 2RH = 2 \times 6371 \times 400 = 5096800 and H^2 = 160000. So \displaystyle \sqrt{2RH} \approx 2257.6 km while the exact value is \displaystyle \sqrt{5256800} \approx 2292.8 km, about 35 km more. The error is now about 1.5\%: tolerable for a rough answer, but no longer negligible as it was for a person on a beach, because H is no longer tiny next to R.
kg1128 · updated 2026-10-19 · CC-BY-SA-4.0