Linear Approximation
Calculus 1 Lia Vas
Linear Approximation The differential. Consider a function y = f (x) and the two points (x, f (x)) and (x+h, f (x+h)) on its graph. Recall that ∆x is sometimes used to denote the difference h between the x-values and ∆y is used for the difference f (x + h) − f (x) of the y-values. In this notation, the derivative can be calculated as follows. ∆y dy = lim = f 0 (x) ⇒ dy = f 0 (x)dx dx ∆x→0 ∆x In this last formula, the quantity dy measures the rise (or decline) of the tangent line when x-values change by dx = ∆x. The quantity dy = f 0 (x)dx is known as the differential of the function y = f (x). Consider the geometric meaning of the change in y-values ∆y = f (x + h) − f (x) and the differential dy = f 0 (x)dx on the figure on the right. dy = f 0 (x)dx This also agrees with the fact that the quody is the tangent of the angle adjacent to dx tient dx which is the is the slope of the hypotenuse in the triangle on the right. This exactly corresponds to dy the derivative dx = f 0 (x). Linear Approximation. Although different in general, the change in y-values ∆y = f (x + h) − f (x) and the differential dy = f 0 (x)dx are similar in size when dx is small. In this case, ∆y and dy are approximately equal and we write ∆y ≈ dy ⇒ f (x + h) − f (x) ≈ f 0 (x)h ⇒ f (x + h) ≈ f (x) + f 0 (x)h. The last formula on the right represents the y-value of the tangent line at point x+h. This formula represents the point-slope equation if we denote h + x by x and x by a, in which case dx = x − a, and consider ∆y as f (x) − f (a). The equation of line passing (a, f (a)) tangent to the graph of f (x) has the slope f 0 (a) so its point-slope equation is y − f (a) = f 0 (a)(x − a) ⇒ y = f (a) + f 0 (a)(x − a).
1
The expression f (a)+f 0 (a)(x−a) is called the linear approximation of f (x) at x = a. Thus, approximating ∆y by the differential dy amounts to approximating f (x) by its linear approximation f (a) + f 0 (a)(x − a). ∆y ≈ dy ⇔ f (x) − f (a) ≈ f 0 (a)(x − a) ⇔ f (x) ≈ f (a) + f 0 (a)(x − a) In applications, you can think of the value f (a) as of the present value, the value f (x) then represents the future value, (x − a) the time lapsed and f 0 (a) the change rate. Thus f (x)
≈
f (a)
+
future ≈ present value value
+
f 0 (a)
(x − a)
change time rate elapsed
The linear approximation is particularly useful when certain phenomena is modeled by a function which is either • too complex to be manipulated or • such that its exact formulas is not known. but its value and the value of its derivative are known at a point. We illustrate both scenarios in the next two examples. Example 1. Find the linear approximation of ex at 0 and use it to approximate e0.01 by a rational number. Solution. Let f (x) = ex so that f 0 (x) = ex . Thus f (0) = e0 = 1 and f 0 (0) = e0 = 1. The linear approximation f (x) ≈ f (a) + f 0 (a)(x − a) for f (x) = ex and a = 0 gives you that ex ≈ 1 + 1(x − 0) = 1 + x. Note that this answer also indicates that the line 1 + x is tangent to the graph of ex at x = 0. When x = 0.01 this approximation gives you e0.01 ≈ 1 + 0.01 = 1.01 = 101 . Comparing with 100 0.01 the calculator value e ≈ 1.01005 . . . we can see that our approximation is accurate to first four decimal places. Example 2. Approximate f (3.06) and f (2.9) given that f (3) = 1 and f 0 (3) = 0.5. Solution. The linear approximation formula for f (x) at 3 gives us that f (x) ≈ f (3)+f 0 (3)(x−3). Thus f (3.06) ≈ f (3) + f 0 (3)(3.01 − 3) = 1 + 0.5(0.06) = 1.03 and f (2.9) ≈ f (3) + f 0 (3)(2.9 − 3) = 1 + 0.5(−0.1) = 0.95. 2
Let us also consider a more applied example. Example 3. The profit P of a company depends on the number of items x produced. The production level x depends on the time t (measured in years). Assume that the profit increases by $200 with each new item produced and that the production level increases by 150 items each year. (a) Determine the rate of increase of the profit per year. (b) If the company is presently making a profit of $800,000, approximate the profit in four year time. Solution. (a) dP = 200 dollars per item and dx = 150 items per year. By the chain rule, dx dt dP dP dx = dx dt = 200 · 150 = 30, 000 dollars per year. dt (b) Using the linear approximation, P (4) ≈ P (0) + P 0 (0)(4 − 0) = 800, 000 + 30, 000(4) = 920, 000 dollars. do at least these... the rest of this handout is optional Practice problems. 1. Find the linear approximation of the following functions at the given x-value and use them to approximate given y-values. (a) f (x) = sin x, x = 0, approximate sin 0.1. (b) f (x) = ln(1 + 3x), x = 0, approximate ln 0.7 = ln(1 − 3(0.1)). 2. If f (2) = 5 and f 0 (2) = 3, approximate f (2.1) and f (1.9). 3. If f (1) = 1 and f 0 (1) = −2, approximate f (1.01). √ 4. Use the linear √ approximation to estimate 3 26 with a rational number. Compare to the calculator value of 3 26. √ 4 5. Use the linear approximation to estimate 16.2 with a rational number. Compare to the √ calculator value of 4 16.2. 6. The cost of producing 10 items is $200 and the cost of each new item produced is $15. Approximate the cost of producing 12 items. 7. The number of bacteria five hours after the start of experiment is 2000 and the number is increasing by 100 bacteria per hour. Approximate the number of bacteria five and a half hours after the start of experiment. Solutions. 1. (a) Let f (x) = sin x so that f 0 (x) = cos x. Thus f (0) = 0 and f 0 (0) = cos 0 = 1. The linear approximation f (x) ≈ f (a) + f 0 (a)(x − a) for a = 0 gives you that sin x ≈ 0 + 1(x − 0) = x. Note that this answer also indicates that the line x is tangent to the graph of sin x at x = 0.
3
When x = 0.1 this approximation gives you sin 0.1 ≈ 0.1 = can compare with the calculator value sin 0.1 ≈ 0.998 . . .
1 . 10
To check out the accuracy, you
3 (b) Let f (x) = ln(1 + 3x) so that f 0 (x) = 1+3x . Thus f (0) = 0 and f 0 (0) = 3. The linear 0 approximation f (x) ≈ f (a)+f (a)(x−a) for a = 0 gives you that ln(1+3x) ≈ 0+3(x−0) = 3x. So 3x is tangent to the graph of ln(1 + 3x) at x = 0.
Note that when y = ln 0.7 = ln(1 − 3(0.1)), x = −0.1. The linear approximation with x = −0.1 is ln 0.7 ≈ 3(−0.1) = −0.3 = −3 . To check out the accuracy, you can compare with the 10 calculator value ln 0.7 = −0.367 . . . 2. f (2) = 5 and f 0 (2) = 3 ⇒ f (x) ≈ f (2) + f 0 (2)(x − 2) = 5 + 3(x − 2) or 3x − 1. Thus f (2.1) ≈ 5 + 3(2.1 − 2) = 5 + 0.3 = 5.3 and f (1.9) ≈ 5 + 3(1.9 − 2) = 5 − 0.3 = 4.7. 3. f (1) = 1 and f 0 (1) = −2 ⇒ f (x) ≈ f (1) + f 0 (1)(x − 1) = 1 − 2(x − 1) or 2x + 3. Thus f (1.01) ≈ 1 − 2(1.01 − 1) = 1 − 0.02 = 0.98. √ √ 3 consider the function f (x) = x = x1/3 and its linear approximation 4. To approximate 3 26, √ at a = 27 (since 3 27 = 3 is easy to determine and 26 is relatively close to 27. In this 1 case, f (x) = x1/3 ⇒ f 0 (x) = 13 x−2/3 . Thus f (27) = 3 and f 0 (27) = 31 312 = 27 so that the 1 1/3 linear approximation is x ≈ 3 + 27 (x − 27). When x = 26, the linear approximation is 1 261/3 ≈ 3 − 27 = 80 ≈ 2.962963. The calculator value is 2.9624960 . . . 27 √ √ 5. To approximate 4 √ 16.2, consider the function f (x) = 4 x = x1/4 and its linear approximation at a = 16 (since 4 16 = 2 is easy to determine and 16.2 is relatively close to 16. In this 1 case, f (x) = x1/4 ⇒ f 0 (x) = 14 x−3/4 . Thus f (16) = 2 and f 0 (16) = 41 213 = 32 so that the 1 linear approximation is x1/4 ≈ 2 + 32 (x − 16). When x = 16.2, the linear approximation is 1 1 (0.2) = 2 + 160 = 321 = 2.00625. The calculator value is 2.00622 . . . 16.21/4 ≈ 2 + 32 160 6. If x denotes the number of items produced and f (x) denotes the cost of producing x items, then f (10) = 200 and f 0 (10) = 15. Thus, the linear approximation is f (x) ≈ f (10)+f 0 (10)(x−10) = 200 + 15(x − 10). When x = 12 this approximation gives you f (12) ≈ 200 + 15(2) = 230 dollars. 7. If x denotes the number of hours after the start of experiment and f (x) denotes the number of bacteria at that time, then f (5) = 2000 and f 0 (5) = 100. Thus, the linear approximation is f (x) ≈ f (5) + f 0 (5)(x − 5) = 2000 + 100(x − 5). When x = 5.5 this approximation gives you f (5.5) ≈ 2000 + 100(0.5) = 2050 bacteria.
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Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
Linear Approximations and Differentials PROBLEM: Approximate the number
√ 4
1.1.
IDEA: We have seen that a curve lies very close to its tangent line near the point of tangency. This observation is the basis for a method of finding approximate values of functions. The idea is that it√might be easy to calculate a value f (a) of a function (in our case 4 1), but difficult (or even impossible) to compute √ 4 nearby values of f (in our case 1.1). So we settle for the easily computed values of the linear function L whose graph is the tangent line of f at (a, f (a)). In other words, we use the tangent line at (a, f (a)) as an approximation to the curve y = f (x) when x is near a. Solution: The point-slope equation of the tangent line is y = f (a) + f ′ (a)(x − a) therefore f (x) ≈ f (a) + f ′ (a)(x − a) if x is close to a √ In particular, if f (x) = 4 x, then √ 4
x≈
√ 4
a+
1 (x − a) 4a3/4
1 . Plugging in x = 1.1 and a = 1, we get 4a3/4 √ √ 1 1 4 4 (1.1 − 1) = 1 + (0.1) = 1.025 (the true value is 1.024113689...) 1.1 ≈ 1 + 3/4 4·1 4
since f ′ (a) =
The approximation f (x) ≈ f (a) + f ′ (a)(x − a)
(1)
is called the linear approximation or tangent line approximation of f at a. The linear function whose graph is this tangent line, that is, L(x) = f (a) + f ′ (a)(x − a) is called the linearization of f at a.
1
(2)
Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
√ EXAMPLE: Find the linearization of the function f (x) = x + 3 at a = 1 and use it to √ √ approximate the numbers 3.98 and 4.05. Are these approximations overestimates or underestimates? √ Solution: The derivative of f (x) = x + 3 is f ′ (x) = (x + 3)1/2
′
1 1 1 = (x + 3)1/2−1 · (x + 3)′ = (x + 3)−1/2 · 1 = √ 2 2 2 x+3
and so we have f (1) = 2 and f ′ (1) = 14 . Putting these values into (2), we see that the linearization is 7 x 1 L(x) = f (1) + f ′ (1)(x − 1) = 2 + (x − 1) = + 4 4 4 The corresponding linear approximation (1) is √
x+3 ≈
7 x + 4 4
(when x is near 1)
In particular, we have √
3.98 ≈
√ 7 0.98 7 1.05 4.05 ≈ + + = 1.995 and = 2.0125 4 4 4 4
(3)
The linear approximation is illustrated in the figure below.
We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 1. We also see that our approximations are overestimates because the tangent line lies above the curve. √ √ Of course, a calculator could give us approximations for 3.98 and 4.05, but the linear approximation gives an approximation over an entire interval. In the table above we compare estimates from the obtained linear approximation with the true values. Notice from this table, and also from the figure above, that the tangent line approximation gives good estimates when x is close to 1 but the accuracy of the approximation deteriorates when x is farther away from 1. √ EXAMPLE: Find of the function f (x) = x at a = 4 and use it to approximate √ √ the linearization the numbers 3.98 and 4.05. Are these approximations overestimates or underestimates?
2
Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
√ EXAMPLE: Find the linearization of the function f (x) = x at a = 4 and use it to approximate √ √ the numbers 3.98 and 4.05. Are these approximations overestimates or underestimates? √ Solution: The derivative of f (x) = x is f ′ (x) = x1/2
′
1 1 1 = x1/2−1 = x−1/2 = √ 2 2 2 x
and so we have f (4) = 2 and f ′ (4) = 14 . Putting these values into (2), we see that the linearization is x 1 L(x) = f (4) + f ′ (4)(x − 4) = 2 + (x − 4) = 1 + 4 4 The corresponding linear approximation (1) is √
x≈1+
x 4
(when x is near 4)
In particular, we have √
3.98 ≈ 1 +
√ 3.98 4.05 = 1.995 and = 2.0125 4.05 ≈ 1 + 4 4
Note that we got the same results as in (3) since 1+
3.98 3 + 0.98 3 0.98 7 0.98 =1+ =1+ + = + 4 4 4 4 4 4
1+
3 + 1.05 3 1.05 7 1.05 4.05 =1+ =1+ + = + 4 4 4 4 4 4
and or, in general,
x 3+x−3 3 x−3 7 x−3 =1+ =1+ + = + 4 4 4 4 4 4 Our approximations are overestimates because the tangent line lies above the curve. 1+
√ 3 EXAMPLE: Find the linearization of the function f (x) = 1 + x at a = 0 and use it to √ √ approximate the numbers 3 0.95 and 3 1.1. Are these approximations overestimates or underestimates?
3
Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
√ 3 EXAMPLE: Find the linearization of the function f (x) = 1 + x at a = 0 and use it to √ √ approximate the numbers 3 0.95 and 3 1.1. Are these approximations overestimates or underestimates? √ Solution: The derivative of f (x) = 3 1 + x is ′ 1 1 1 f ′ (x) = (1 + x)1/3 = (1 + x)1/3−1 · (1 + x)′ = (1 + x)−2/3 · 1 = p 3 3 3 3 (1 + x)2 and so we have f (0) = 1 and f ′ (0) = 31 . Putting these values into (2), we see that the linearization is x L(x) = f (0) + f ′ (0)(x − 0) = 1 + 3 The corresponding linear approximation (1) is √ x 3 (when x is near 0) 1+x≈1+ 3 In particular, we have √ −0.05 3 0.95 ≈ 1 + = 0.9833... (the true value is 0.9830475725...) 3 and √ 0.1 3 = 1.0333... (the true value is 1.032280115...) 1.1 ≈ 1 + 3 Our approximations are overestimates because the tangent line lies above the curve. √ EXAMPLE: Find the linearization of the function f (x) = 3 x at a = 1 and use it to approximate √ √ the numbers 3 0.95 and 3 1.1. Are these approximations overestimates or underestimates? √ Solution: The derivative of f (x) = 3 x is ′ 1 1 1 f ′ (x) = x1/3 = x1/3−1 = x−2/3 = √ 3 3 3 3 x2 and so we have f (1) = 1 and f ′ (1) = 13 . Putting these values into (2), we see that the linearization is x−1 2 x L(x) = f (1) + f ′ (1)(x − 1) = 1 + = + 3 3 3 The corresponding linear approximation (1) is √ 2 x 3 x≈ + (when x is near 1) 3 3 In particular, we have √ 2 0.95 3 = 0.9833... (the true value is 0.9830475725...) 0.95 ≈ + 3 3 and √ 2 1.1 3 1.1 ≈ + = 1.0333... (the true value is 1.032280115...) 3 3 EXAMPLE: Find the linearization of the function f (x) = sin x at a = 0 and use it to approximate the numbers sin(−0.1) and sin(0.1). Are these approximations overestimates or underestimates? 4
Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
EXAMPLE: Find the linearization of the function f (x) = sin x at a = 0 and use it to approximate the numbers sin(−0.1) and sin(0.1). Are these approximations overestimates or underestimates? Solution: The derivative of f (x) = sin x is f ′ (x) = (sin x)′ = cos x and so we have f (0) = 0 and f ′ (0) = 1. Putting these values into (2), we see that the linearization is L(x) = f (0) + f ′ (0)(x − 0) = 0 + 1 · (x − 0) = x The corresponding linear approximation (1) is sin x ≈ x (when x is near 0)
In particular, we have sin(−0.1) ≈ −0.1 (the true value is -0.09983341665...) and sin(0.1) ≈ 0.1 (the true value is 0.09983341665...) The first approximation is an underestimate because the tangent line lies below the curve when x is near 0 from the left. The second approximation is an overestimate because the tangent line lies above the curve when x is near 0 from the right. EXAMPLE: For what values of x is the linear approximation sin x ≈ x accurate to within 0.1? Solution: Accuracy to within 0.1 means that the functions should differ by less than 0.1: | sin x − x| < 0.1
⇐⇒
−0.1 < sin x − x < 0.1
Using a graphing calculator we can conclude that the approximation sin x ≈ x is accurate to within 0.1 when −0.86 < x < 0.86. 5
Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
Differentials The ideas behind linear approximations are sometimes formulated in the terminology and notation of differentials. If y = f (x), where f is a differentiable function, then the differential dx is an independent variable; that is, dx can be given the value of any real number. The differential dy is then defined in terms of dx by the equation dy = f ′ (x)dx So dy is a dependent variable; it depends on the values of x and dx. If dx is given a specific value and x is taken to be some specific number in the domain of f, then the numerical value of dy is determined. Let P (x, f (x)) and Q(x + ∆x, f (x + ∆x)) be points on the graph of f and let dx = ∆x. The corresponding change in y is ∆y = f (x + ∆x) − f (x)
The slope of the tangent line P R is the derivative f ′ (x). Thus the directed distance from S to R is f ′ (x)dx = dy. Therefore, dy represents the amount that the tangent line rises or falls (the change in the linearization), whereas ∆y represents the amount that the curve y = f (x) rises or falls when x changes by an amount dx. Notice that the approximation ∆y ≈ dy becomes better as ∆x becomes smaller.
If we let dx = x − a, then x = a + dx and we can rewrite the linear approximation (1) f (x) ≈ f (a) + f ′ (a)(x − a) in the notation of differentials: f (a + dx) ≈ f (a) + dy √ For instance, for the function f (x) = x + 3 in Example 1, we have dx dy = f ′ (x)dx = √ 2 x+3 If a = 1 and dx = ∆x = 0.05, then 0.05 = 0.0125 dy = √ 2 1+3 and
√
4.05 = f (1.05) ≈ f (1) + dy = 2.0125
just as we found in Example 1.
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Linear Approximation The differential. Consider a function y = f (x) and the two points (x, f (x)) and (x+h, f (x+h)) on its graph. Recall that ∆x is sometimes used to denote the difference h between the x-values and ∆y is used for the difference f (x + h) − f (x) of the y-values. In this notation, the derivative can be calculated as follows. ∆y dy = lim = f 0 (x) ⇒ dy = f 0 (x)dx dx ∆x→0 ∆x In this last formula, the quantity dy measures the rise (or decline) of the tangent line when x-values change by dx = ∆x. The quantity dy = f 0 (x)dx is known as the differential of the function y = f (x). Consider the geometric meaning of the change in y-values ∆y = f (x + h) − f (x) and the differential dy = f 0 (x)dx on the figure on the right. dy = f 0 (x)dx This also agrees with the fact that the quody is the tangent of the angle adjacent to dx tient dx which is the is the slope of the hypotenuse in the triangle on the right. This exactly corresponds to dy the derivative dx = f 0 (x). Linear Approximation. Although different in general, the change in y-values ∆y = f (x + h) − f (x) and the differential dy = f 0 (x)dx are similar in size when dx is small. In this case, ∆y and dy are approximately equal and we write ∆y ≈ dy ⇒ f (x + h) − f (x) ≈ f 0 (x)h ⇒ f (x + h) ≈ f (x) + f 0 (x)h. The last formula on the right represents the y-value of the tangent line at point x+h. This formula represents the point-slope equation if we denote h + x by x and x by a, in which case dx = x − a, and consider ∆y as f (x) − f (a). The equation of line passing (a, f (a)) tangent to the graph of f (x) has the slope f 0 (a) so its point-slope equation is y − f (a) = f 0 (a)(x − a) ⇒ y = f (a) + f 0 (a)(x − a).
1
The expression f (a)+f 0 (a)(x−a) is called the linear approximation of f (x) at x = a. Thus, approximating ∆y by the differential dy amounts to approximating f (x) by its linear approximation f (a) + f 0 (a)(x − a). ∆y ≈ dy ⇔ f (x) − f (a) ≈ f 0 (a)(x − a) ⇔ f (x) ≈ f (a) + f 0 (a)(x − a) In applications, you can think of the value f (a) as of the present value, the value f (x) then represents the future value, (x − a) the time lapsed and f 0 (a) the change rate. Thus f (x)
≈
f (a)
+
future ≈ present value value
+
f 0 (a)
(x − a)
change time rate elapsed
The linear approximation is particularly useful when certain phenomena is modeled by a function which is either • too complex to be manipulated or • such that its exact formulas is not known. but its value and the value of its derivative are known at a point. We illustrate both scenarios in the next two examples. Example 1. Find the linear approximation of ex at 0 and use it to approximate e0.01 by a rational number. Solution. Let f (x) = ex so that f 0 (x) = ex . Thus f (0) = e0 = 1 and f 0 (0) = e0 = 1. The linear approximation f (x) ≈ f (a) + f 0 (a)(x − a) for f (x) = ex and a = 0 gives you that ex ≈ 1 + 1(x − 0) = 1 + x. Note that this answer also indicates that the line 1 + x is tangent to the graph of ex at x = 0. When x = 0.01 this approximation gives you e0.01 ≈ 1 + 0.01 = 1.01 = 101 . Comparing with 100 0.01 the calculator value e ≈ 1.01005 . . . we can see that our approximation is accurate to first four decimal places. Example 2. Approximate f (3.06) and f (2.9) given that f (3) = 1 and f 0 (3) = 0.5. Solution. The linear approximation formula for f (x) at 3 gives us that f (x) ≈ f (3)+f 0 (3)(x−3). Thus f (3.06) ≈ f (3) + f 0 (3)(3.01 − 3) = 1 + 0.5(0.06) = 1.03 and f (2.9) ≈ f (3) + f 0 (3)(2.9 − 3) = 1 + 0.5(−0.1) = 0.95. 2
Let us also consider a more applied example. Example 3. The profit P of a company depends on the number of items x produced. The production level x depends on the time t (measured in years). Assume that the profit increases by $200 with each new item produced and that the production level increases by 150 items each year. (a) Determine the rate of increase of the profit per year. (b) If the company is presently making a profit of $800,000, approximate the profit in four year time. Solution. (a) dP = 200 dollars per item and dx = 150 items per year. By the chain rule, dx dt dP dP dx = dx dt = 200 · 150 = 30, 000 dollars per year. dt (b) Using the linear approximation, P (4) ≈ P (0) + P 0 (0)(4 − 0) = 800, 000 + 30, 000(4) = 920, 000 dollars. do at least these... the rest of this handout is optional Practice problems. 1. Find the linear approximation of the following functions at the given x-value and use them to approximate given y-values. (a) f (x) = sin x, x = 0, approximate sin 0.1. (b) f (x) = ln(1 + 3x), x = 0, approximate ln 0.7 = ln(1 − 3(0.1)). 2. If f (2) = 5 and f 0 (2) = 3, approximate f (2.1) and f (1.9). 3. If f (1) = 1 and f 0 (1) = −2, approximate f (1.01). √ 4. Use the linear √ approximation to estimate 3 26 with a rational number. Compare to the calculator value of 3 26. √ 4 5. Use the linear approximation to estimate 16.2 with a rational number. Compare to the √ calculator value of 4 16.2. 6. The cost of producing 10 items is $200 and the cost of each new item produced is $15. Approximate the cost of producing 12 items. 7. The number of bacteria five hours after the start of experiment is 2000 and the number is increasing by 100 bacteria per hour. Approximate the number of bacteria five and a half hours after the start of experiment. Solutions. 1. (a) Let f (x) = sin x so that f 0 (x) = cos x. Thus f (0) = 0 and f 0 (0) = cos 0 = 1. The linear approximation f (x) ≈ f (a) + f 0 (a)(x − a) for a = 0 gives you that sin x ≈ 0 + 1(x − 0) = x. Note that this answer also indicates that the line x is tangent to the graph of sin x at x = 0.
3
When x = 0.1 this approximation gives you sin 0.1 ≈ 0.1 = can compare with the calculator value sin 0.1 ≈ 0.998 . . .
1 . 10
To check out the accuracy, you
3 (b) Let f (x) = ln(1 + 3x) so that f 0 (x) = 1+3x . Thus f (0) = 0 and f 0 (0) = 3. The linear 0 approximation f (x) ≈ f (a)+f (a)(x−a) for a = 0 gives you that ln(1+3x) ≈ 0+3(x−0) = 3x. So 3x is tangent to the graph of ln(1 + 3x) at x = 0.
Note that when y = ln 0.7 = ln(1 − 3(0.1)), x = −0.1. The linear approximation with x = −0.1 is ln 0.7 ≈ 3(−0.1) = −0.3 = −3 . To check out the accuracy, you can compare with the 10 calculator value ln 0.7 = −0.367 . . . 2. f (2) = 5 and f 0 (2) = 3 ⇒ f (x) ≈ f (2) + f 0 (2)(x − 2) = 5 + 3(x − 2) or 3x − 1. Thus f (2.1) ≈ 5 + 3(2.1 − 2) = 5 + 0.3 = 5.3 and f (1.9) ≈ 5 + 3(1.9 − 2) = 5 − 0.3 = 4.7. 3. f (1) = 1 and f 0 (1) = −2 ⇒ f (x) ≈ f (1) + f 0 (1)(x − 1) = 1 − 2(x − 1) or 2x + 3. Thus f (1.01) ≈ 1 − 2(1.01 − 1) = 1 − 0.02 = 0.98. √ √ 3 consider the function f (x) = x = x1/3 and its linear approximation 4. To approximate 3 26, √ at a = 27 (since 3 27 = 3 is easy to determine and 26 is relatively close to 27. In this 1 case, f (x) = x1/3 ⇒ f 0 (x) = 13 x−2/3 . Thus f (27) = 3 and f 0 (27) = 31 312 = 27 so that the 1 1/3 linear approximation is x ≈ 3 + 27 (x − 27). When x = 26, the linear approximation is 1 261/3 ≈ 3 − 27 = 80 ≈ 2.962963. The calculator value is 2.9624960 . . . 27 √ √ 5. To approximate 4 √ 16.2, consider the function f (x) = 4 x = x1/4 and its linear approximation at a = 16 (since 4 16 = 2 is easy to determine and 16.2 is relatively close to 16. In this 1 case, f (x) = x1/4 ⇒ f 0 (x) = 14 x−3/4 . Thus f (16) = 2 and f 0 (16) = 41 213 = 32 so that the 1 linear approximation is x1/4 ≈ 2 + 32 (x − 16). When x = 16.2, the linear approximation is 1 1 (0.2) = 2 + 160 = 321 = 2.00625. The calculator value is 2.00622 . . . 16.21/4 ≈ 2 + 32 160 6. If x denotes the number of items produced and f (x) denotes the cost of producing x items, then f (10) = 200 and f 0 (10) = 15. Thus, the linear approximation is f (x) ≈ f (10)+f 0 (10)(x−10) = 200 + 15(x − 10). When x = 12 this approximation gives you f (12) ≈ 200 + 15(2) = 230 dollars. 7. If x denotes the number of hours after the start of experiment and f (x) denotes the number of bacteria at that time, then f (5) = 2000 and f 0 (5) = 100. Thus, the linear approximation is f (x) ≈ f (5) + f 0 (5)(x − 5) = 2000 + 100(x − 5). When x = 5.5 this approximation gives you f (5.5) ≈ 2000 + 100(0.5) = 2050 bacteria.
4
Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
Linear Approximations and Differentials PROBLEM: Approximate the number
√ 4
1.1.
IDEA: We have seen that a curve lies very close to its tangent line near the point of tangency. This observation is the basis for a method of finding approximate values of functions. The idea is that it√might be easy to calculate a value f (a) of a function (in our case 4 1), but difficult (or even impossible) to compute √ 4 nearby values of f (in our case 1.1). So we settle for the easily computed values of the linear function L whose graph is the tangent line of f at (a, f (a)). In other words, we use the tangent line at (a, f (a)) as an approximation to the curve y = f (x) when x is near a. Solution: The point-slope equation of the tangent line is y = f (a) + f ′ (a)(x − a) therefore f (x) ≈ f (a) + f ′ (a)(x − a) if x is close to a √ In particular, if f (x) = 4 x, then √ 4
x≈
√ 4
a+
1 (x − a) 4a3/4
1 . Plugging in x = 1.1 and a = 1, we get 4a3/4 √ √ 1 1 4 4 (1.1 − 1) = 1 + (0.1) = 1.025 (the true value is 1.024113689...) 1.1 ≈ 1 + 3/4 4·1 4
since f ′ (a) =
The approximation f (x) ≈ f (a) + f ′ (a)(x − a)
(1)
is called the linear approximation or tangent line approximation of f at a. The linear function whose graph is this tangent line, that is, L(x) = f (a) + f ′ (a)(x − a) is called the linearization of f at a.
1
(2)
Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
√ EXAMPLE: Find the linearization of the function f (x) = x + 3 at a = 1 and use it to √ √ approximate the numbers 3.98 and 4.05. Are these approximations overestimates or underestimates? √ Solution: The derivative of f (x) = x + 3 is f ′ (x) = (x + 3)1/2
′
1 1 1 = (x + 3)1/2−1 · (x + 3)′ = (x + 3)−1/2 · 1 = √ 2 2 2 x+3
and so we have f (1) = 2 and f ′ (1) = 14 . Putting these values into (2), we see that the linearization is 7 x 1 L(x) = f (1) + f ′ (1)(x − 1) = 2 + (x − 1) = + 4 4 4 The corresponding linear approximation (1) is √
x+3 ≈
7 x + 4 4
(when x is near 1)
In particular, we have √
3.98 ≈
√ 7 0.98 7 1.05 4.05 ≈ + + = 1.995 and = 2.0125 4 4 4 4
(3)
The linear approximation is illustrated in the figure below.
We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 1. We also see that our approximations are overestimates because the tangent line lies above the curve. √ √ Of course, a calculator could give us approximations for 3.98 and 4.05, but the linear approximation gives an approximation over an entire interval. In the table above we compare estimates from the obtained linear approximation with the true values. Notice from this table, and also from the figure above, that the tangent line approximation gives good estimates when x is close to 1 but the accuracy of the approximation deteriorates when x is farther away from 1. √ EXAMPLE: Find of the function f (x) = x at a = 4 and use it to approximate √ √ the linearization the numbers 3.98 and 4.05. Are these approximations overestimates or underestimates?
2
Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
√ EXAMPLE: Find the linearization of the function f (x) = x at a = 4 and use it to approximate √ √ the numbers 3.98 and 4.05. Are these approximations overestimates or underestimates? √ Solution: The derivative of f (x) = x is f ′ (x) = x1/2
′
1 1 1 = x1/2−1 = x−1/2 = √ 2 2 2 x
and so we have f (4) = 2 and f ′ (4) = 14 . Putting these values into (2), we see that the linearization is x 1 L(x) = f (4) + f ′ (4)(x − 4) = 2 + (x − 4) = 1 + 4 4 The corresponding linear approximation (1) is √
x≈1+
x 4
(when x is near 4)
In particular, we have √
3.98 ≈ 1 +
√ 3.98 4.05 = 1.995 and = 2.0125 4.05 ≈ 1 + 4 4
Note that we got the same results as in (3) since 1+
3.98 3 + 0.98 3 0.98 7 0.98 =1+ =1+ + = + 4 4 4 4 4 4
1+
3 + 1.05 3 1.05 7 1.05 4.05 =1+ =1+ + = + 4 4 4 4 4 4
and or, in general,
x 3+x−3 3 x−3 7 x−3 =1+ =1+ + = + 4 4 4 4 4 4 Our approximations are overestimates because the tangent line lies above the curve. 1+
√ 3 EXAMPLE: Find the linearization of the function f (x) = 1 + x at a = 0 and use it to √ √ approximate the numbers 3 0.95 and 3 1.1. Are these approximations overestimates or underestimates?
3
Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
√ 3 EXAMPLE: Find the linearization of the function f (x) = 1 + x at a = 0 and use it to √ √ approximate the numbers 3 0.95 and 3 1.1. Are these approximations overestimates or underestimates? √ Solution: The derivative of f (x) = 3 1 + x is ′ 1 1 1 f ′ (x) = (1 + x)1/3 = (1 + x)1/3−1 · (1 + x)′ = (1 + x)−2/3 · 1 = p 3 3 3 3 (1 + x)2 and so we have f (0) = 1 and f ′ (0) = 31 . Putting these values into (2), we see that the linearization is x L(x) = f (0) + f ′ (0)(x − 0) = 1 + 3 The corresponding linear approximation (1) is √ x 3 (when x is near 0) 1+x≈1+ 3 In particular, we have √ −0.05 3 0.95 ≈ 1 + = 0.9833... (the true value is 0.9830475725...) 3 and √ 0.1 3 = 1.0333... (the true value is 1.032280115...) 1.1 ≈ 1 + 3 Our approximations are overestimates because the tangent line lies above the curve. √ EXAMPLE: Find the linearization of the function f (x) = 3 x at a = 1 and use it to approximate √ √ the numbers 3 0.95 and 3 1.1. Are these approximations overestimates or underestimates? √ Solution: The derivative of f (x) = 3 x is ′ 1 1 1 f ′ (x) = x1/3 = x1/3−1 = x−2/3 = √ 3 3 3 3 x2 and so we have f (1) = 1 and f ′ (1) = 13 . Putting these values into (2), we see that the linearization is x−1 2 x L(x) = f (1) + f ′ (1)(x − 1) = 1 + = + 3 3 3 The corresponding linear approximation (1) is √ 2 x 3 x≈ + (when x is near 1) 3 3 In particular, we have √ 2 0.95 3 = 0.9833... (the true value is 0.9830475725...) 0.95 ≈ + 3 3 and √ 2 1.1 3 1.1 ≈ + = 1.0333... (the true value is 1.032280115...) 3 3 EXAMPLE: Find the linearization of the function f (x) = sin x at a = 0 and use it to approximate the numbers sin(−0.1) and sin(0.1). Are these approximations overestimates or underestimates? 4
Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
EXAMPLE: Find the linearization of the function f (x) = sin x at a = 0 and use it to approximate the numbers sin(−0.1) and sin(0.1). Are these approximations overestimates or underestimates? Solution: The derivative of f (x) = sin x is f ′ (x) = (sin x)′ = cos x and so we have f (0) = 0 and f ′ (0) = 1. Putting these values into (2), we see that the linearization is L(x) = f (0) + f ′ (0)(x − 0) = 0 + 1 · (x − 0) = x The corresponding linear approximation (1) is sin x ≈ x (when x is near 0)
In particular, we have sin(−0.1) ≈ −0.1 (the true value is -0.09983341665...) and sin(0.1) ≈ 0.1 (the true value is 0.09983341665...) The first approximation is an underestimate because the tangent line lies below the curve when x is near 0 from the left. The second approximation is an overestimate because the tangent line lies above the curve when x is near 0 from the right. EXAMPLE: For what values of x is the linear approximation sin x ≈ x accurate to within 0.1? Solution: Accuracy to within 0.1 means that the functions should differ by less than 0.1: | sin x − x| < 0.1
⇐⇒
−0.1 < sin x − x < 0.1
Using a graphing calculator we can conclude that the approximation sin x ≈ x is accurate to within 0.1 when −0.86 < x < 0.86. 5
Section 2.8 Linear Approximations and Differentials
2010 Kiryl Tsishchanka
Differentials The ideas behind linear approximations are sometimes formulated in the terminology and notation of differentials. If y = f (x), where f is a differentiable function, then the differential dx is an independent variable; that is, dx can be given the value of any real number. The differential dy is then defined in terms of dx by the equation dy = f ′ (x)dx So dy is a dependent variable; it depends on the values of x and dx. If dx is given a specific value and x is taken to be some specific number in the domain of f, then the numerical value of dy is determined. Let P (x, f (x)) and Q(x + ∆x, f (x + ∆x)) be points on the graph of f and let dx = ∆x. The corresponding change in y is ∆y = f (x + ∆x) − f (x)
The slope of the tangent line P R is the derivative f ′ (x). Thus the directed distance from S to R is f ′ (x)dx = dy. Therefore, dy represents the amount that the tangent line rises or falls (the change in the linearization), whereas ∆y represents the amount that the curve y = f (x) rises or falls when x changes by an amount dx. Notice that the approximation ∆y ≈ dy becomes better as ∆x becomes smaller.
If we let dx = x − a, then x = a + dx and we can rewrite the linear approximation (1) f (x) ≈ f (a) + f ′ (a)(x − a) in the notation of differentials: f (a + dx) ≈ f (a) + dy √ For instance, for the function f (x) = x + 3 in Example 1, we have dx dy = f ′ (x)dx = √ 2 x+3 If a = 1 and dx = ∆x = 0.05, then 0.05 = 0.0125 dy = √ 2 1+3 and
√
4.05 = f (1.05) ≈ f (1) + dy = 2.0125
just as we found in Example 1.
6
