18.01 Solutions, Problem Set 4

• §4.5 #2

A large spherical snowball is melting at the rate of 2π ft^{3}/h. At the moment when it is 30 inches in diameter, determine

a) how fast the radius is changing,

Solution: The volume is V = \frac{4}{3}πr^{3} where r is the radius. So \frac{dV}{dt} =

4πr^{2} \frac{dr}{dt}. When the diameter is 30 in, the radius is 15 in = \frac{5}{4} ft. This

gives 2π = \frac{25π}{4} \frac{dr}{dt} where \frac{dr}{dt} is in ft/h. So \frac{dr}{dt} = \frac{8}{25} ft/h.

b) and how fast the surface area is changing.

Solution: The surface area is A = 4πr^{2}, so \frac{dA}{dt} = 8πr \frac{dr}{dt}. When

r = \frac{5}{4} ft, \frac{dr}{dt} = \frac{8}{25} ft/h, and \frac{dA}{dt} = \frac{16π}{5} ft^{2}/h.

• §4.5 #3

Sand is being poured onto a conical pile at the constant rate of 50 ft^{3}/min. Frictional forces in the sand are such that the height of the pile is always equal to the radius of its base. How fast is the height of the pile increasing when the sand is 5 ft deep?

Solution: The volume of a cone is V = \frac{1}{3}πr^{2}h where r is the radius and

h is the height. If r = h, then V = \frac{1}{3}πh^{3} and \frac{dV}{dt} = πh^{2} \frac{dh}{dt}. If \frac{dV}{dt} = 50

ft^{3}/min and h = 5 ft, then \frac{dh}{dt} = 2/π ft/min.

• §4.5 #6

A light at the top of a pole is 80 ft high. A ball is dropped from the same height from a point 20 ft away from the light. (Assume that the ball falls s = 16t^{2} ft in t seconds.)

Find how fast the shadow of the ball is moving along the ground a) 1 second later;

Solution: Similar triangles (draw a picture) show that x/80 = 20/s where x is the distance between the shadow and the pole. This gives

x = 1600s^{−1} and \frac{dx}{dt} = −1600s^{−2} \frac{ds}{dt}. Since s = 16t^{2} and \frac{ds}{dt} = 32t,

this gives \frac{dx}{dt} = −200/t^{3} (where all distances are in feet, and time is in seconds). Therefore when t = 1, the shadow is moving at 200 ft/sec.

b) 2 seconds later.

Solution: When t = 2, the shadow is moving at 25 ft/sec.

• §4.5 #11

A spherical meteorite enters the earth’s atmosphere and burns up at a rate proportional to its surface area. Show that its radius decreases at a constant rate.

Solution: Since V = \frac{4}{3}πr^{3}, we have \frac{dV}{dt} = 4πr^{2} \frac{dr}{dt}. Since \frac{dV}{dt} is propor-

tional to the surface area, which is 4πr^{2}, we have \frac{dV}{dt} = 4πr^{2} · C for some

(negative) constant C. Therefore \frac{dr}{dt} = C, so the radius is decreasing at a constant rate.

• §4.6 #2

a) Show that x^{3} + 3x^{2} − 6 = 0 has only one real root, and calculate it to six decimal places of accuracy.

Solution: The function f(x) = x^{3} + 3x^{2} − 6 has derivative f′(x) = 3x^{2} + 6x, which is zero when x = −2 or x = 0. We find that f(x) is increasing on (−∞, −2), decreasing on (−2, 0) and increasing on (0, ∞). At x = −2, there is a relative maximum with value f(−2) = −2, and at x = 0, there is a relative minimum with value f(0) = −6. (Sketch the graph.) The graph therefore crosses the x-axis at most once, and since f(1) = −2 and f(2) = 14, there is exactly one root, somewhere in the interval (1, 2). We set x_{1} = 1 and repeatedly apply Newton’s method, which in this case gives the formula

x_{n+1} = x_{n} − \frac{x_{n}^{3} + 3x_{n}^{2} − 6}{3x^{2} + 6x}.

This gives x_{2} ≈ 1.222222, x_{3} ≈ 1.196215, x_{4} ≈ 1.195823, x_{5} ≈ 1.195823, . . ..

b) Show that x^{3} + 3x = 8 has only one real root, and calculate it to six decimal places of accuracy.

Solution: The function f(x) = x^{3} + 3x − 8 has derivative f′(x) = 3x^{2} + 3, which is always positive, so f(x) is always increasing, so it has at most one zero. Since f(1) = −4 and f(2) = 6, there is a zero somewhere in the interval (1, 2). Applying Newton’s Method:

x_{n+1} = x_{n} − \frac{x_{n}^{3} + 3x_{n} − 8}{3x^{2} + 3}.

with x_{1} = 1.5 gives x_{2} ≈ 1.512821, x_{3} ≈ 1.512745, x_{4} ≈ 1.512745, . . ..

• §4.6 #6 Consider a spherical shell 1 ft thick whose volume equals the volume of the hollow space inside it. Use Newton’s method to calculate the shell’s outer radius to six decimal places of accuracy.

Solution: Let r denote the outer radius in feet. The outer sphere is supposed to have twice the volume of the inner sphere, which has radius r − 1. Therefore, we have to solve the equation

\frac{4}{3}πr^{3} = 2 · \frac{4}{3}π(r − 1)^{3},

or equivalently r^{3} − 2(r − 1)^{3} = 0. So we are looking for a zero of f(x) = x^{3} − 2(x − 1)^{2}. A good starting point is x_{1} = 5, since 5^{3} is a little smaller than 2 · 4^{3}. Newton’s Method gives the formula:

x_{n+1} = x_{n} − \frac{x_{n}^{3} − 2(x_{n} − 1)^{3}}{3x_{n}^{2} − 6(x_{n} − 1)^{2}} ,

so x_{2} ≈ 4.857143, x_{2} ≈ 4.847367, x_{3} ≈ 4.847322, x_{3} ≈ 4.847322, . . .. (It’s not hard to solve the above equation algebraically; the solution is r = ^{3} \sqrt{2}/( ^{3} \sqrt{2} − 1).)

• §4.6 #13 Find the smallest positive solution of each of the following equations, correct to six decimal places.

a) 4(x − 1) = sin x;

Solution: Let f(x) = 4(x − 1) − sin x. Then f′(x) = 4 − cos x is always positve, so f(x) is always increasing, so there is at most one solution of f(x) = 0. Since f(1) = − sin 1 is negative and f(2) = 4 − sin 2 is positive, there is a solution somewhere in the interval (1, 2). We start with x_{1} = 1.5 and apply Newton’s Method, which gives

x_{n+1} = x_{n} − \frac{4(x_{n} − 1) − sin x_{n}}{4 − cos x_{n}} .

So x_{2} ≈ 1.244862, x_{3} ≈ 1.236140, x_{4} ≈ 1.236130, x_{5} ≈ 1.236130, . . .. b) x^{2} = sin x;

Solution: Let f(x) = x^{2} − sin x. From the discussion in class, we are looking for a zero of f(x) in the interval (0.5, 1), so we might start with x_{1} = 0.75 for example. Newton’s method gives the formula

x_{n+1} = x_{n} − \frac{x_{n}^{2} − sin x_{n}}{2x_{n} − cos x_{n}} ,

so x_{2} ≈ 0.905066, x_{3} ≈ 0.877663, x_{3} ≈ 0.876727, x_{4} ≈ 0.876726, x_{5} ≈ 0.876726, . . ..

• §12.2 Find the following limits.

#2 lim_{x→1} \frac{ln x}{x − 1}.

Solution: Letting f(x) = ln x and g(x) = x − 1, the hypotheses for l’Hôpital’s Rule are satisfied (f(x) and g(x) are differentiable at x = 1 and f(1) = g(1) = 0.) Therefore

lim_{x→1} \frac{ln x}{x − 1} = lim_{x→1} \frac{1/x}{1} = 1.

#4 lim_{x→0} \frac{e^{x} − e^{−x}}{sin 5x}.

Solution: The hypotheses for l’Hôpital’s Rule are again satisfied, so

lim_{x→0} \frac{e^{x} − e^{−x}}{sin 5x} = lim_{x→0} \frac{e^{x} + e^{−x}}{5 cos 5x} = \frac{2}{5}.

#9 lim_{x→0} \frac{e^{x} − 1 − x}{x^{2}}.

Solution: Applying l’Hôpital’s Rule again gives

lim_{x→0} \frac{e^{x} − 1 − x}{x^{2}} = lim_{x→0} \frac{e^{x} − 1}{2x},

to which we can again apply the rule to get

lim_{x→0} \frac{e^{x} − 1}{2x} = lim_{x→0} \frac{e^{x}}{2} = \frac{1}{2}.

#11 lim_{x→0} \frac{x^{3}}{sin x − x}.

Solution: L’Hôpital’s Rule applies again to give

lim_{x→0} \frac{x^{3}}{sin x − x} = lim_{x→0} \frac{3x^{2}}{cos x − 1},

and again to give

lim_{x→0} \frac{3x^{2}}{cos x − 1} = lim_{x→0} \frac{6x}{− sin x},

and again to give

lim_{x→0} \frac{6x}{− sin x} = lim_{x→0} \frac{6}{− cos x} = −6.

#22 lim_{x→0} \frac{sin x^{3}}{sin^{3} x}.

Solution: Applying l’Hôpital’s Rule gives

lim_{x→0} \frac{sin x^{3}}{sin^{3} x} = lim_{x→0} \frac{3x^{2} cos x^{3}}{3 sin^{2} x cos x} = ( lim_{x→0} \frac{x}{sin x})^{2} lim_{x→0} \frac{cos x^{3}}{cos x} = 1, since cos 0^{3} = cos 0 = 1 and lim_{x→0} \frac{sin x}{x} = 1.

• §12.3 Evaluate the following limits by any method.

#2 lim_{x→∞} \frac{ln(ln x)}{ln x}. Solution: L’Hôpital’s Rule applies to the “indeterminate form” \frac{∞}{∞}, giving

lim_{x→∞} \frac{ln(ln x)}{ln x} = lim_{x→∞} \frac{1/(x ln x)}{1/x} = lim_{x→∞} \frac{1}{ln x},

which is 0 since the denominator → ∞ as x → ∞.

#10 lim_{x→0} x cot x.

Solution: Substituting x = 0, we find this has the indeterminate form 0 · ∞. We write it instead as lim_{x→0} \frac{x cos x}{sin x} and apply L’Hôpital’s Rule, which gives

lim_{x→0} \frac{x cos x}{sin x} = lim_{x→0} \frac{cos x − x sin x}{cos x},

which is 1 since the numerator and denominator both → 1 as x → 0.

#20 lim_{x→0} (\frac{1}{x} − \frac{1}{sin x}).

Solution: This has the indeterminate form ∞ − ∞. We write it as lim_{x→0} \frac{sin x − x}{x sin x} and apply L’Hôpital’s Rule, which gives

lim_{x→0} \frac{sin x − x}{x sin x} = lim_{x→0} \frac{cos x − 1}{sin x + x cos x}.

L’Hôpital’s Rule applies again, giving lim_{x→0} \frac{cos x − 1}{sin x + x cos x} = lim_{x→0} \frac{− sin x}{2 cos x − x sin x} = \frac{0}{2} = 0.

#27 lim_{x→0^{+}}(e^{x} − 1)^{x}. The expression has the indeterminate form 0^{0} at x = 0, so we consider its logarithm instead: ln(e^{x}−1)^{x} = x ln(e^{x}−1). This has the indeterminate form 0 · ∞, so we write it as \frac{ln(e^{x} − 1)}{1/x} instead (indeterminate form \frac{∞}{∞}) and apply l’Hôpital’s Rule. We get

lim_{x→0^{+}} \frac{ln(e^{x} − 1)}{1/x} = lim_{x→0^{+}} \frac{e^{x}/(e^{x} − 1)}{−1/x^{2}} = lim_{x→0^{+}} \frac{−x^{2}e^{x}}{e^{x} − 1} .

This has the indeterminate form \frac{0}{0}; applying l’Hôpital’s Rule again gives

lim_{x→0^{+}} \frac{−x^{2}e^{x}}{e^{x} − 1} = lim_{x→0^{+}} \frac{−(x^{2}e^{x} + 2xe^{x})}{e^{x}} = lim_{x→0^{+}} −(x^{2} + 2x) = 0.

Therefore the logarithm of lim_{x→0^{+}}(e^{x} − 1)^{x} is 0, so

lim_{x→0^{+}}(e^{x} − 1)^{x} = e^{0} = 1.

• Find the linearization of ^{3} \sqrt{1 + x} at x = 0.

Solution: with f(x) = ^{3} \sqrt{1 + x} = (1 + x)^{1/3}, we have f′(x) = \frac{1}{3}(1 + x)^{−2/3}. Recall the linearization of f(x) at x = a is just the tangent line approximation

f(x) ≈ f(a) + f′(a)(x − a).

Since f(0) = 1 and f′(0) = \frac{1}{3}, the linearization of ^{3} \sqrt{1 + x} at x = 0 is

3 \sqrt{1 + x} ≈ 1 + \frac{x}{3}.

• Use a linear approximation to estimate

a) \sqrt{9.01}, Solution: Setting f(x) = \sqrt{x} and a = 9, we have f′(x) = \frac{1}{2}x^{−1/2}, so f(a) = 3 and f′(a) = \frac{1}{6}. The linear approximation is therefore

\sqrt{9 + 0.01} ≈ 3 + \frac{1}{6} · 0.01 = 3 + \frac{1}{600} ≈ 3.00167.

b) ln 1.01. Solution: Setting f(x) = ln x and a = 1, we have f′(x) = 1/x, so f(a) = 0 and f′(a) = 1. The linear approximation is therefore

ln 1 + 0.01 ≈ 0 + 1 · 0.01 = 0.01.