PLAN DEL SISTEMA DE VÍAS TRONCALES PARA BUSES

CAPÍTULO 14 Plan del Sistema de Vías Troncales

14. PLAN DEL SISTEMA DE VÍAS TRONCALES PARA BUSES

(a) Find the area between the curve y = 1

x and the x-axis, for 1 ≤ x ≤ 3, by evaluating

an appropriate integral. Then approximate the result.

(b) Estimate the area using the trapezoidal rule with three function values.

18. (a) Use the trapezoidal rule with ﬁve function values to approximate the area between the curve y = log x and the x-axis, between x = 1 and x = 5. Answer correct to four decimal places.

(b) Use Simpson’s rule with ﬁve function values to approximate the area in part (a).

C H A L L E N G E

19. (a) Sketch y = log x, for 0 ≤ x ≤ e.

(b) Evaluate the area between the curve and the y-axis, between y = 0 and y = 1. (c) Hence ﬁnd the area between the curve and thex-axis, between x = 1 and x = e.

20. (a) The area between y = √x and y = √1_x, and between x = 1 and x = 4, is rotated

about thex-axis. Find the volume of the resulting solid.

(b) Compare the volume found in part (a) with the volume of the solid generated when the area below y = √x − √1

x, also between x = 1 and x = 4, is rotated about the x-axis.

21. Consider the two curves y = 6e−x and y = ex− 1.

(a) Letu = ex. Show that thex-coordinate of the point of intersection of these two curves satisﬁesu2− u − 6 = 0.

(b) Hence ﬁnd the coordinates of the point of intersection.

(d) Find the area of the shaded region.

3 G

Calculus with Other Bases

In applications of exponential functions where calculus is required, the basee can generally be used. For example, the treatment of natural growth in Chapter Six is done entirely using basee. Nevertheless, calculus can be applied to exponential and logarithmic functions with other bases. The general principle is to express these functions in terms of the functions ex and log_ex that have base e.

This section is a little more diﬃcult than the other work on exponential and logarithmic functions and could well be left until later.

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Logarithmic Functions to Other Bases: Any logarithmic functions can be expressed

easily in terms of log_ex by using the change-of-base formula. For example, log₂x = logex

log_e2.

Thus every other logarithmic function is just a constant multiple of log_ex. This allows any other logarithmic function to be diﬀerentiated easily.

WORKEDEXERCISE:

(a) Express the functiony = log₅x in terms of the function log_ex.

(b) Hence use the calculator function labelled ln to approximate, correct to four decimal places:

(i) log530 (ii) log52 (iii) log50·07

SOLUTION:

(a) log5x =

log_ex log_e5 (b) (i) log₅30 = loge30

log_e5 =.. 2·1133

(ii) log₅2 = loge2 log_e5 =.. 0·4307

(iii) log₅0·07 = loge0·07 log_e5 =.. −1·6523

(i) 52·1133 =.. 30 (ii) 50·4307 =.. 2 (iii) 5−1·6523 =.. 0·07 Note: _{Either of the calculator functions log} _{or ln can be used to approx-} imate a number like log₅30, because the base can be changed to either 10 or e:

log530 = log1030 log₁₀5 or log530 = log_e30 log_e5 WORKEDEXERCISE:

Use the change-of-base formula to diﬀerentiate:

(a) y = log₂x (b) y = log_ax

SOLUTION:

(a) Here y = log₂x.

Using the change-of-base formula,

y = logex

log_e2. Since log_e2 is a constant,

dy dx = 1 x × 1 log_e2 = 1 x log_e2. (b) Here y = log_ax.

Using the change-of-base formula,

y = logex

log_ea. Since log_ea is a constant,

dy dx = 1 x × 1 log_ea = 1 x log_ea.

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CHAPTER3: The Logarithmic Function 3G Calculus with Other Bases 135

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Part (b) above gives the formula in the general case:

DIFFERENTIATING A LOGARITHMIC FUNCTION WITH ANOTHER BASE:

Use the change-of-base formula to write log_ax as a multiple of log_ex: log_ax = logex

log_ea.

Alternatively, remember the result as a standard form:

d dxlogax = 1 x logea. For example, d dxlog10x = 1 x log_e10 and d dxlog1·05x = 1 x log_e1·05.

A Characterisation of the Logarithmic Function: We have already seen in Section 3B

that the tangent toy = log_ex at the x-intercept has gradient exactly 1.

The following worked exercise shows that this property distinguishes the logarithmic function base e from all other logarithmic functions.

WORKEDEXERCISE:

(a) Show that the tangent toy = log_ax at the x-intercept has gradient 1 log_ea.

(b) Show that the function y = log_ex is the only logarithmic function whose gradient at thex-intercept is exactly 1.

SOLUTION:

(a) Here y = log_ax.

When y = 0, log_ax = 0

x = 1,

and so thex-intercept is (1, 0). Using the change-of-base formula,

y = logex log_ea and diﬀerentiating, y = 1 x log_ea. Hence whenx = 1, y = 1 log_ea, as required. x y 1 2 1 −1 e y= log_ex

(b) The gradient at thex-intercept is 1 if and only if log_ea = 1,

a = e1

=e,

that is, if and only if the original basea is equal to e.

14 THE GRADIENT AT THEx-INTERCEPT: The function y = logex is the only logarithmic

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Exponential Functions with Other Bases: Before calculus can be applied to an expo-

nential function y = ax with base a diﬀerent from e, it must be written as an exponential function with basee. The important identity used to do this is

elog_ea ₌_a,

which simply expresses the fact that the functionsex and log_ex are inverse functions. Nowax can be written as

ax _{= (}_elog_ea₎x_, _replacing_{a by e}log_ea_,

=ex logea, _{using the index law (}ek₎x ₌ekx_.

Thusax has been expressed in the form ekx, wherek = log_ea is a constant.

EXPONENTIAL FUNCTIONS WITH OTHER BASES: Every number can be written as a power ofe:

a = elog_ea

Every exponential function can be written as an exponential function base e:

ax ₌_ex logea

WORKEDEXERCISE:

Express these numbers and functions as power ofe:

(a) 2 (b) 2x (c) 5−x

SOLUTION:

(a) 2 =eloge2 _{(b) 2}x ₌eloge2x

=ex loge2

=e−x loge5

Differentiating and Integrating Exponential Functions with Other Bases: Write the function as a power ofe. It can then be diﬀerentiated and integrated.

First, ax =elogeax =ex logea. Diﬀerentiating, d dxax = d dxex logea =ex logea × log ea, since _dxd ekx=k ekx,

=ax log_ea, since ex logea ₌ax_.

Integrating, axdx = ex logeadx = e x logea log_ea , since ekx = 1 kekx, = a x log_ea, since e x logea ₌ax_.

This process can be carried through every time, or the results can be remembered as standard forms.

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CHAPTER3: The Logarithmic Function 3G Calculus with Other Bases 137

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DIFFERENTIATION AND INTEGRATION WITH OTHER BASES:

dxax =axlogea

ax_{dx =} ax

log_ea+C

Note: _{The formulae for diﬀerentiating and integrating} ax _{both involve the} constant log_ea. This constant log_ea = 1 when a = e, so the formulae are simplest when the base ise. Again, this indicates that e is the appropriate base to use for the calculus of exponential functions.

WORKEDEXERCISE:

Differentiatey = 2x. Hence find the gradient ofy = 2x at they-intercept, correct to three significant figures.

SOLUTION:

Here y = 2x.

Using the standard form, y = 2xlog_e2. Hence when x = 0, y = 20× log_e2

= log_e2 =.. 0·693.

Note: _{This result should be compared with the results of physically measuring} this gradient in question 4 of Exercise 3B.

WORKEDEXERCISE:

(a) Show that the line y = x + 1 meets the curve y = 2x atA(0, 1) and B(1, 2).

(b) Sketch the two curves and shade the region contained between them.

SOLUTION:

(a) Simple substitution of x = 0 and x = 1 into both functions veriﬁes the result. (b) The graph is drawn to the right.

(upper curve− lower curve) dx = 1 0 (x + 1 − 2x)dx = 1 2x 2 +x − 2 x log_e2 1 0 = 1 2 + 1− 2 log_e2 − 0 + 0− 1 log_e2 = 11₂ − 1

log_e2 square units _x

y y x= + 1 y = 2x −1 1 1 2 =.. 0·057 30 square units.

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Exercise 3G

Note: _{Remember that log}x and ln x both mean log_ex (except on the calculator, where log means log10x and ln means logex).

In document CAPÍTULO 13 Plan del Sector de Instalaciones Viales (página 34-53)