A.4 Practice Problems: Linear Approximations

Time to practice! This screen has a series of practice problems for linear approximations, so you can develop your skills that we introduced on the preceding screen. As you work through the questions, we'll also illustrate a few important points that we'll use as a starting point at in the next Topic.

How to use our practice problems

Note that for every problem, you can—if you wish—immediately view the solution with a single click. This is fine to do if you're feeling stuck, or wish to check a key early-step before you finish your calculation.

A word of caution, however: Some students simply read our solutions and think: Yeah, that's exactly what I would have done so I won't bother. Then later, when faced with an actual exam question, they discover they didn't actually get the approach "in their hands," and then wish that they had taken this opportunity to practice. As with any new skill, the only way to become fully competent and comfortable is by doing it for yourself; merely watching us demonstrate our skills is of little value. We want you to do well! Only you can do this part, developing the necessary neural connections in your brain and between your brain and your hands.

Since there's no penalty for getting something wrong here, go ahead and dive in. And make as many mistakes as you need to -- and you need to make some, because that's where real learning happens. (It's where your "learning edge" is!) Our goal is for you to be confident and proficient in solving these types of problems, so that they feel routine to you when you encounter them on exam. We're providing the opportunity to practice so you can get there; we sincerely hope you'll take advantage.

Practice Problem 1: Approximate

\sqrt 16.2

Consider the square-root function

𝑓 (𝑥) = \sqrt 𝑥 .

You know that $𝑓 (16) = \sqrt 16 = 4 .$ We are given that at $𝑥 = 16,$ the function changes at the rate $𝑑 𝑓 𝑑 𝑥 ∣ 𝑥 = 16 = 0.125$ Using our linear approximation method, the approximate value of $\sqrt 16.2$ is $(A) 4.02 (B) 4.0249 (C) 4.025 (D) 4.0125 (E) n o n e o f t h e s e$

A B C D E

View/Hide Solution

$𝑓 v a l u e a t 16 + 𝑑 𝑥 ⏞ ¯¯ ⏞ ¯¯ ⏞ 𝑓 (16 + 𝑑 𝑥) \approx 𝑓 v a l u e a t 16 ⏞ 𝑓 (16) + (r a t e a t 𝑥 = 16) * (𝑑 𝑥) ⏞ ¯¯¯ ⏞ ¯¯¯ ⏞ s m a l l c h a n g e 𝑑 𝑓 𝑓 (16 + 0.2) \approx 4 + (0.125) (0.2) = 4 + 0.025 = 4.025 ⟹ (C) ✓$ For reference, the actual numeric value to 6 decimal places is $\sqrt 16.2 = 4.02492236 .$

You can use the interactive Desmos graph below to explore this result graphically. Once again, visuals for dx and df will appear once have zoomed-in sufficiently. You can also try out different values of dx by using the slider beneath the graph.

So far we've considered a function for which $𝑓 = 𝑓 (𝑥)$ : our independent variable has been x, and so the relevant rate of change has been with respect to x: $𝑑 𝑓 𝑑 𝑥 ∣ a t a p a r t i c u l a r v a l u e o f 𝑥$ . Let's now consider a function that has as its independent variable $𝜃$ : $𝑔 = 𝑔 (𝜃),$ and its associated rate of change with respect to $𝜃,$ $𝑑 𝑔 𝑑 𝜃 ∣ a t a p a r t i c u l a r v a l u e o f 𝜃$ . You're still doing a practice problem on linear approximations, now just with different notation for the variables.

Practice Problem 2: Approximate

s i n (- 0.13)

Consider the function

𝑔 (𝜃) = s i n (𝜃) .

You know

s i n (0) = 0 .

We are given that at

𝜃 = 0,

the function changes at the rate

𝑑 𝑔 𝑑 𝜃 ∣ 𝜃 = 0 = 1

when

𝜃

is measured in radians. Using our linear approximation method, the approximate value of

s i n (- 0.13)

(A) 0 (B) 1 (C) 0.13 (D) - 0.13 (E) n o n e o f t h e s e

A B C D E

View/Hide Solution

$𝑔 v a l u e a t 0 + 𝑑 𝜃 ⏞ 𝑔 (0 + 𝑑 𝜃) \approx 𝑔 v a l u e a t 0 ⏞ 𝑔 (0) + (r a t e a t 𝜃 = 0) * (𝑑 𝜃) ⏞ ¯¯¯ ⏞ ¯¯¯ ⏞ s m a l l c h a n g e 𝑑 𝑔 𝑔 (0 + (- 0.13)) \approx 0 + (1) (- 0.13) = 0 - 0.13 = - 0.13 ⟹ (D) ✓$ For reference, the actual numeric value to 6 decimal places is $s i n (- 0.13) = - 0.129634 .$

As an aside, this problem illustrates one of the most common approximations used in Physics and other fields: $s i n (𝑑 𝜃) \approx 𝑑 𝜃$ for small values of $𝑑 𝜃$ when $𝑑 𝜃$ is measured in radians. This is more commonly expressed as $s i n (𝜃) \approx 𝜃 f o r s m a l l v a l u e s o f 𝜃$ when $𝜃$ is expressed in radians.

You can use the interactive Desmos graph below to explore this result graphically. Once again, visuals for $𝑑 𝜃$ and dg will appear once have zoomed-in sufficiently. You can also try out different values of $𝑑 𝜃$ by using the slider beneath the graph.

As you'll see, replacing the curve $𝑦 = s i n (𝜃)$ with the line $𝑦 = 𝜃$ works quite well for small values of $𝜃 .$

As we saw, the preceding problem illustrates one of the most frequently used approximations used in Physics and other fields: $s i n (𝑑 𝜃) \approx 𝑑 𝜃$ for small values of $𝑑 𝜃$ when $𝑑 𝜃$ is measured in radians. This is more commonly expressed as

Small Angle Approximation of $𝐬 𝐢 𝐧 (𝜽)$

$s i n (𝜃) \approx 𝜃 f o r s m a l l v a l u e s o f 𝜃$ when $𝜃$ is expressed in radians.

IMPORTANT NOTE:
This approximation works only for small values of $𝜃$ that are close to $𝜃 = 0,$ because (I) at our "base point" of $𝜃 = 0,$ $s i n (0) = 0,$ and (II) that's where the rate $𝑑 𝑔 𝑑 𝜃 ∣ 𝜃 = 0 = 1$ is valid.

At other locations on the sine curve, the function's value and its rate of change are different. To illustrate, let's consider the same function, $𝑔 (𝜃) = s i n (𝜃),$ but at a different value of $𝜃 .$ As we'll see, at this new point of interest the function changes half-as-quickly as it did in Problem 2.

Practice Problem 3: Approximate

s i n (𝜋 / 3 + 0.018)

Consider the function $𝑔 (𝜃) = s i n (𝜃) .$ You know $s i n (𝜋 3) = \sqrt 3 2 \approx 0.866 .$ We are given that at $𝜃 = 𝜋 3,$ the function changes at the rate $𝑑 𝑔 𝑑 𝜃 ∣ 𝜃 = 𝜋 / 3 = 12$ when $𝜃$ is measured in radians.

Using our linear approximation method, the approximate value of $s i n (𝜋 / 3 + 0.018)$ is $(A) 0.366 (B) 0.946 (C) 0.857 (D) 0.875 (E) n o n e o f t h e s e$

A B C D E

View/Hide Solution

$𝑔 v a l u e a t 𝜋 / 3 + 𝑑 𝜃 ⏞ ¯¯ ⏞ ¯¯ ⏞ 𝑔 (𝜋 / 3 + 𝑑 𝜃) \approx 𝑔 v a l u e a t 𝜋 / 3 ⏞ 𝑔 (𝜋 / 3) + (r a t e a t 𝜃 = 𝜋 / 3) * (𝑑 𝜃) ⏞ ¯¯¯ ⏞ ¯¯¯ ⏞ s m a l l c h a n g e 𝑑 𝑔 𝑔 (𝜋 / 3 + 0.018) \approx 0.866 + (12) (0.018) = 0.866 + 0.009 = 0.875 ⟹ (D) ✓$ For reference, the actual numeric value to 6 decimal places is $s i n (𝜋 / 3 + 0.018) = 0.874885 .$

Let's consider the function $𝑔 (𝜃) = s i n (𝜃)$ again as a practice problem for linear approximations, now at $𝜃 = 𝜋$ where the function's rate of change is negative.

Practice Problem 4: Approximate

s i n (𝜋 + 0.07)

Consider the function $𝑔 (𝜃) = s i n (𝜃) .$ You know $s i n (𝜋) = 0 .$ We are given that at $𝜃 = 𝜋,$ the function changes at the rate $𝑑 𝑔 𝑑 𝜃 ∣ 𝜃 = 𝜋 = - 1$ when $𝜃$ is measured in radians.

Using our linear approximation method, the approximate value of $s i n (𝜋 + 0.07)$ is $(A) 0.35 (B) . 07 (C) - 0.07 (D) - 1 (E) n o n e o f t h e s e$

A B C D E

View/Hide Solution

$𝑔 v a l u e a t 𝜋 + 𝑑 𝜃 ⏞ 𝑔 (𝜋 + 𝑑 𝜃) \approx 𝑔 v a l u e a t 𝜋 ⏞ 𝑔 (𝜋)) + (r a t e a t 𝜃 = 𝜋)) * (𝑑 𝜃) ⏞ ¯¯¯ ⏞ ¯¯¯ ⏞ s m a l l c h a n g e 𝑑 𝑔 𝑔 (𝜋 + 0.07) \approx 0 + (- 1) (0.07) = 0 - 0.07 = - 0.07 ⟹ (C) ✓$ For reference, the actual numeric value to 6 decimal places is $s i n (𝜋 + 0.07) = - 0.069943 .$

Let's finally consider the sin function one last time, now at a location where its rate of change is zero.

Practice Problem 5: Approximate

s i n (𝜋 / 2 + 0.04)

Consider the function $𝑔 (𝜃) = s i n (𝜃) .$ You know $s i n (𝜋 2) = 1 .$ We are given that at $𝜃 = 𝜋 2,$ the function changes at the rate $𝑑 𝑔 𝑑 𝜃 ∣ 𝜃 = 𝜋 / 2 = 0$ when $𝜃$ is measured in radians.

Using our linear approximation method, the approximate value of $s i n (𝜋 2 + 0.04)$ is $(A) 0.04 (B) 0 (C) 1.61 (D) 1 (E) n o n e o f t h e s e$

A B C D E

View/Hide Solution

$𝑔 v a l u e a t 𝜋 / 2 + 𝑑 𝜃 ⏞ ¯¯ ⏞ ¯¯ ⏞ 𝑔 (𝜋 2 + 𝑑 𝜃) \approx 𝑔 v a l u e a t 𝜋 / 2 ⏞ 𝑔 (𝜋 2) + (r a t e a t 𝜃 = 𝜋 / 2)) * (𝑑 𝜃) ⏞ ¯¯¯ ⏞ ¯¯¯ ⏞ s m a l l c h a n g e 𝑑 𝑔 𝑔 (𝜋 2 + 0.04) \approx 1 + (0) (0.04) = 1 + 0 = 1 ⟹ (D) ✓$ For reference, the actual numeric value to 6 decimal places is $s i n (𝜋 2 + 0.04) = 0.999550 .$ The error between our approximate value and the true value is 0.04%.

The result of Problem 5 is often initially surprising to students: since $𝑑 𝑔 𝑑 𝜃 ∣ 𝜃 = 𝜋 / 2 = 0,$ the function's value doesn't change—to first order—if you move a tiny bit away from $𝑥 = 𝜋 2$ . As usual, the further away from $𝑥 = 𝜋 2$ we go, the worse our approximation becomes. But as you can see from the interactive Desmos graph, to first order if you walk a little bit to the right or a little bit to the left from $𝑥 = 𝜋 2,$ you remain quite close to $𝑦 = 1 .$

In the next Topic we'll pull together all of the results we've developed so far, and start to draw some generalizations.

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