Approximations

On this screen we'll develop a straightforward strategy for computing linear approximations, perhaps the most frequently used Calculsu tool in science and engineering. As always, free practice problems are below, each with a complete solution immediately available.

You might recall that back in our "Introductory Ideas" chapter, we introduced linear approximations for just this reason. Back on those screens we had to provide the value of the derivative at the point of interest because you didn't know how to calculate it yet – but now you do, and so we can pull all of the pieces together and you can complete the calculation from start to finish on your own (at least after some practice). You might want to revisit that screen since each problem there has a Desmos calculator so you can see how the approximation works, whereas the problems on this screen do not (yet).

The following Problem Solving Strategy illustrates the approach, and then we'll dive straight into problems.

PROBLEM SOLVING STRATEGY: Approximations

Problems of this type will ask you to approximate the value of a function that's close to a value you immediately know. For instance, you might be asked to approximate $\sqrt 4.02$ , which is close to the value you know of $\sqrt 4 = 2$ . Or you could be asked to approximate $s i n (0.1)$ , since you know $s i n (0) = 0$ .

$Curve with red dot at (x_0, f(x_0)), and blue dot on the curve a horizontal distance delta-x away$

We'll call the point you know about $𝑥 0$ . For instance, if you want to approximate $\sqrt 4.02$ , then $𝑥 0 = 4$ . And if you want to approximate $s i n (0.1)$ , then $𝑥 0 = 0$ . We represent this point that we know about, $(𝑥 𝑜, 𝑓 (𝑥 0))$ , as the red dot in the top figure.

We'll call the horizontal distance to the point you're interested in $Δ 𝑥$ , so you're looking for an approximation to $𝑓 (𝑥 0 + Δ 𝑥)$ . We show this point as the green dot in the lower figure. So in the examples we're considering: $\sqrt 4.02 = \sqrt 4 + 0.02, w e h a v e 𝑥 0 = 4 a n d Δ 𝑥 = 0.02; a n d f o r s i n (0.1) = s i n (0 + 0.1), w e h a v e 𝑥 0 = 0 a n d Δ 𝑥 = 0.1$

$Top figure shows the tangent line at the point (x_0, f(x_0). The lower figure shows the horizontal distance delta-x, and the associated change delta-y equals m * delta-x.$

To make the approximation, we replace the function's actual growth (or decrease) with linear growth (or decrease), which is the same thing as pretending that the function follows the tangent line as you move a small distance away from $𝑥 0$ . (See the upper figure.) Said differently, if we imagine walking from the point we know about to the point of interest in order to compute the change, instead of walking from the point along the function's actual curve (shown in blue), we're going to walk along the tangent line (shown in green) to get an approximation.

The method works because we can easily compute the slope $𝑚$ of the line tangent to the curve at the point you know about, $𝑥 0$ , since the slope equals the derivative at that point:

𝑚 = 𝑑 𝑓 𝑑 𝑥 ∣ 𝑥 = 𝑥 0

Then computing the approximate growth (or decrease) $Δ 𝑦$ from the value we know about is straightforward (see the lower figure):

Δ 𝑦 = 𝑚 Δ 𝑥 = 𝑑 𝑓 𝑑 𝑥 ∣ 𝑥 = 𝑥 0 Δ 𝑥

Putting all the pieces together,

𝑓 (𝑥 0 + Δ 𝑥) \approx 𝑓 (𝑥 0) + Δ 𝑦 \approx 𝑓 (𝑥 0) + 𝑑 𝑓 𝑑 𝑥 ∣ 𝑥 = 𝑥 0 Δ 𝑥

This approach to estimation is known as a linear approximation since we are replacing the function around the point of interest with a line whose slope equals the tangent to the curve there.

Many students who find the discussion above rather abstract find the method straightforward after practicing the concrete problems below.

Practice Problem 1: Square roots

Without using a calculator, estimate:

(a)

\sqrt 4.04

(b)

\sqrt 3.96

Practice Problem 2: sin

(a)

s i n (0.1)

(b)

s i n (- 0.1)

Practice Problem 3:

c o s (𝜋 + 1 / 100)

(based on an actual exam question)

Without using a calculator, estimate to two decimal places:

c o s (𝜋 + 1 / 100)

Practice Problem 4:

\sqrt 99

Without using a calculator, estimate to two decimal places:

\sqrt 99

Practice Problem 5:

(1 + Δ 𝑥) 𝑛

(a)

Estimate

(1.0003) 100

(b)

Show that

(1 + Δ 𝑥) 𝑛 \approx 1 + 𝑛 Δ 𝑥

, if

Δ 𝑥

is small compared to 1.

Practice Problem 6: Adding a coat of paint

You're going to add a coat of paint of thickness 0.02 cm to a cube of edge-length 10 cm. Approximately how many cubic centimeters of paint will you use?
(This question could state instead, "Use differentials to estimate how much paint you will use.")

Practice Problem 7: Increasing a circle's radius

You increase a circle's radius by 1%. By approximately what percentage does its area change?
(This question could state instead, "Use differentials to estimate the percentage change in area.")

Practice Problem 8: Differentials of a circle's area, and of a sphere's volume

In this question we're going to try to make geometric sense of the differentials associated with a circle's area, and a sphere's volume.

(a)

The formula for the area of a circle is

𝐴 c i r c l e = 𝜋 𝑟 2

. (i) Find

𝑑 𝐴 c i r c l e 𝑑 𝑟

. (ii) The result from (i) should look familiar. What does

𝑑 𝐴 c i r c l e 𝑑 𝑟

represent geometrically? Hint: Look at the result in the form

𝑑 𝐴 c i r c l e =___𝑑 𝑟

(b)

The formula for the surface area of a sphere is

𝑉 s p h e r e = 43 𝜋 𝑟 3

. (i) Find

𝑑 𝑉 s p h e r e 𝑑 𝑟

. (ii) The result from (i) should look familiar. What does

𝑑 𝑉 s p h e r e 𝑑 𝑟

represent geometrically? Hint: Look at the result in the form

𝑑 𝑉 s p h e r e =___𝑑 𝑟

(a) (i) $2 𝜋 𝑟$ ; (ii) See detailed solution.
(b) (i) $4 𝜋 𝑟 2$ ; (ii) See detailed solution.

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