18.01 Solutions, Problem Set 6

• §5.4 #2. Find the general solution of \frac{dy}{dx} = (3x + 1)^{3}.

Solution: Writing dy = (3x + 1)^{3} dx and integrating both sides gives:

y = ∫ dy = ∫ (3x + 1)^{3} dx = \frac{1}{3} ∫ u^{3} du = \frac{1}{12}u^{4} + C

where u = 3x + 1. So the solution is y = \frac{1}{12}(3x + 1)^{4} + C.

(Check: \frac{dy}{dx} = \frac{1}{12} · 4(3x + 1)^{3} · 3 = (3x + 1)^{3}.)

• §5.4 #4 Find the general solution of \frac{dy}{dx} = 2\sqrt{y}. Solution: Dividing by \sqrt{y}, then integrating gives:

∫ y^{−1/2}dy = ∫ 2 dx,

so 2y^{1/2} = 2x + C_{1}, or equivalently y^{1/2} = x + C. Solving for y, we get y = (x + C)^{2}. (Check: \frac{dy}{dx} = 2(x + C), and 2\sqrt{y} = 2(\sqrt{x + C})^{2} = 2|x + C|. The solution is valid for x ≥ −C.)

• §5.4 #9 Find the particular solution of \frac{dy}{dx} = 2xy^{2} satisfying the initial condition y = 1 when x = 2.

Solution: We get ∫ y^{−2} dy = ∫ 2x dx, so

−y^{−1} = x^{2} + C.

Substuting y = 1, x = 2 gives −1 = 4 + C, so C = −5 and

−y^{−1} = x^{2} − 5.

Solving for y gives y = (5 − x^{2})^{−1}. (Check: \frac{dy}{dx} = −(5 − x^{2})^{−2}(−2x) = 2x(5 − x^{2})^{−2} = 2xy^{2}, and y = 1 when x = 2.)

• §5.4 #11

Find the particular solution of y \frac{dy}{dx} = x(y^{4} +2y^{2} +1) satisfying the initial condition y = 1 when x = 4.

Solution: We get y(y^{4} +2y^{2} +1)^{−1} dy = x dx. To integrate the left side, note that y^{4} + 2y^{2} + 1 = (y^{2} + 1)^{2}, so we substitute u = y^{2} + 1. This gives

\begin{aligned}∫ \frac{1}{2}(y^{2} + 1)^{−2} · 2y dy & = ∫ \frac{1}{2}u^{−2} du \\ & = −\frac{1}{2}u^{−1} + C_{1} = −\frac{1}{2}(y^{2} + 1)^{−1} + C_{1}. \end{aligned}

Since ∫ x dx = \frac{1}{2}x^{2} + C_{2}, we get

−(y^{2} + 1)^{−1} = x^{2} + C

(multiplying through by 2 and replacing the constant 2(C_{2} − C_{1}) with C). Setting y = 1, x = 4 gives −1/2 = 16 + C, so C = −33/2. We’ll leave it as C when we solve for y:

\begin{aligned}−(y^{2} + 1)^{−1} & = x^{2} + C \\ y^{2} + 1 & = −\frac{1}{x^{2} + C} \\ y^{2} & = −\frac{x^{2} + C + 1}{x^{2} + C}. \end{aligned}

(For the right-hand-side to be positive, we need x^{2} to be between −C − 1 and −C.) We then take square-roots (positive, since the initial value of y is positive) and get

y = \sqrt{\frac{−(x^{2} + C + 1)}{x^{2} + C}} = \sqrt{\frac{2x^{2} − 31}{33 − 2x^{2}}}

(with x between \sqrt{31/2} and \sqrt{33/2}).

(Check: \frac{dy}{dx} = \frac{2x^{2} − 31}{33 − 2x^{2}}^{−1/2}· \frac{4x}{(33 − 2x^{2})^{2}} , and both sides of the differential

equation give 4x(33 − 2x^{2})^{−2}.)

• §5.4 #16 Verify that y = x + A\sqrt{x^{2} + 1} is a solution of the differential equation

(x^{2} + 1)\frac{dy}{dx} = xy + 1 for all choices of the constant A.

Solution: We have \frac{dy}{dx} = 1 + Ax(x^{2} + 1)^{−1/2}, so the left-hand side is

x^{2} + 1 + Ax(x^{2} + 1)^{1/2},

which agrees with the right-hand side.

• §5.5 #4

A ball is thrown upward from the top of a cliff 96 ft high with an initial velocity of 64 ft/s. Find the maximum height of the ball above the ground below. Assuming that the ball misses the cliff on its way down, how long does it take to hit the ground?

Solution: Since the acceleration is constant (-32 ft/s^{2}), the velocity v and height y are given by:

\begin{aligned}v & = −32t + v_{0} & = −32t + 64 \\ y & = −16t^{2} + 64t + y_{0} & = −16t^{2} + 64t + 96, \end{aligned}

where t is in seconds, y is in feet and v is in ft/s. The maximum height is attained when v = 0, so t = 2 and y = 160 (feet). The ball hits the ground when y = 0. Solving for t gives t^{2} − 4t − 6 = 0, so

t = \frac{4 ± \sqrt{40}}{2} = 2 ± \sqrt{10}.

Ruling out the negative solution, we get t ≈ 5.16 (seconds).

• §5.5 #10

How much time does a train traveling 144 km/h take to stop if it has a constant negative acceleration of 4 m/s^{2}? How far does the train travel in this time?

Solution: Converting km/h to m/s, we find the train’s initial velocity is 40 m/s. So the velocity v and distance traveled x are given by

v = 40 − 4t, x = 40t − 2t^{2},

where t is in seconds, x is in meters and v is in m/s. The train takes 10 seconds to stop, during which time it travels 200 meters.

• §5.5 #12

On the surface of the moon the acceleration due to gravity is approximately \frac{1}{6} that at the surface of the earth, and on the surface of the sun it is approximately 29 times as great as at the surface of the earth. If a person on earth can jump with enough initial velocity to rise 5 ft, how high will the same initial velocity carry that person (a) on the moon? (b) on the sun?

Solution: Let a be the downward constant of acceleration due to gravity on one of these bodies. The height after jumping with initial velocity v_{0} is given by s = −\frac{1}{2}at^{2} +v_{0}t. The maximum height is attained when t = v_{0}/a,

so s = \frac{v_{0}^{2}}{2a}. If this is 5 ft on the earth, then it will be 30 ft on the moon and just over 2 inches on the sun.

• §8.5 #2

If the rate of increase of a population of a country is 3 percent per year, by what factor does it increase every 10 years? What percentage increase will double the population every 10 years?

Solution: The population satisfies P = P_{0}e^{kt} where P_{0} is the initial popoluation, t is in years and k is a constant. When t = 1, P = 1.03 · P_{0}, so k = ln(1.03). This gives P = P_{0}e^{t ln(1.03)} (or one could argue more directly that P = P_{0}(1.03)^{t}). When t = 10, we have P ≈ 1.34P_{0}, so the population increases 34%. To determine the constant k which would give P = 2P_{0} when t = 10, we solve 2 = e^{10k}, so k = (ln 2)/10 ≈ 0.0693. So after 1 year, the population would be about 1.07 · P_{0}, an increase of about 7%. (Or more directly, the tenth root of 2 is ≈ 1.07.)

• §8.5 #5

The half-life of radium is 1620 years. What percentage of a given quantity of radium will remain after 100 years?

Solution: The quantity remaining after t years is x = x_{0}e^{−t(ln 2)/1620}. When t = 10, we have x ≈ 0.96 · x_{0}, so about 96% remains.

• §8.5 #6

Cobalt 60, with a half-life of 5.3 years, is used extensively in medical radiology. How long does it take for 90% of a given quantity to decay?

Solution: The quantity remaining after t years is x = x_{0}e^{−t(ln 2)/5.3}. Solving 0.1 = e^{−t(ln 2)/5.3} for t gives ln 0.1 = −t(ln 2)/5.3, so

t = 5.03 ln(10)/ ln 2 ≈ 17.6 (years).

• §8.5 #11

According to Newton’s law of cooling, a body at temperature T cools at a rate proportional to the difference between T and the temperature of the surrounding air. A vat of boiling soup at 100^{∘}C is brought into a room where the air is 20^{∘}C, and is left to cool. After 1 hour its temperature is 60^{∘}C. How much additional time is required for it to cool to 30^{∘}C?

Solution: The temperature T in ^{∘}C satisfies the differential equation

\frac{dT}{dt} = k(T − 20)

where k is a (negative) constant. So

∫ \frac{dT}{T − 20} = ∫ k dt,

which gives

ln(T − 20) = kt + C.

Exponentiating both sides gives

T − 20 = e^{C}e^{kt},

so the constant e^{C} is T_{0} − 20 where T_{0} is the initial temperature. Since T_{0} − 20 = 80 and T − 20 = 40 after 1 hour, we get k = − ln 2 (with t in hours). The formula for T is therefore

T = 20 + 80e^{−t ln 2} = 20 + 80/2^{t}.

Setting T = 30 and solving for t gives t = 3, so the tempurature is 30^{∘}C after 2 additional hours (a total of 3 hours of cooling).

• §7.2 #2, 8, 12

Sketch the curves and find the areas of the regions they bound:

#2 y = x^{2}, x = y^{2}.

Solution: The curves intersect when x = 0 and x = 1. We need to compute the area of the region below y = \sqrt{x} and above y = x^{2} for x between 0 and 1. This is given by

∫_{0}^{1}(x^{1/2} − x^{2}) dx = (\frac{2}{3}x^{3/2} − \frac{1}{3}x^{3})]_{0}^{1} = \frac{1}{3}.

#8 y = x^{3} − 4x, y = 5x (x ≥ 0).

Solution: We find the points of intersection by solving x^{3}−4x = 5x, so

x^{3} − 9x = x(x^{2} − 9) = x(x − 3)(x + 3) = 0,

giving x = 0 and x = 3 (we assumed x ≥ 0). The area is given by

∫_{0}^{3}(5x − (x^{3} − 4x)) dx = ∫_{0}^{3}(9x − x^{3}) dx = (\frac{9}{2}x^{2} − \frac{1}{4}x^{4})]_{0}^{3} = \frac{81}{4}.

#12 y = x^{3}, y = 2x − x^{2}.

Solution: Solving x^{3} = 2x − x^{2} gives

x^{3} + x^{2} − 2x = x(x^{2} + x − 2) = x(x − 1)(x + 2) = 0,

so the curves intersect when x = −2, 0 and 1. If −2 ≤ x ≤ 0, then x^{3} ≥ 2x − x^{2}, but if 0 ≤ x ≤ 1, then 2x − x^{2} ≥ x^{3}. So the total area of the two regions is

∫_{−2}^{0}(x^{3} − 2x + x^{2}) dx + ∫_{0}^{1}(2x − x^{2} − x^{3}) dx.

Since ∫ (x^{3} −2x+x^{2}) dx = F (x)+C, where F (x) = \frac{1}{4}x^{4} + \frac{1}{3}x^{3} −x^{2},

the area is

F (x)]_{−2}^{0} − F (x)]_{0}^{1} = F (0) − F (−2) − F (1) + F (0) = \frac{37}{12}.

• §7.2 #25

Find the area between y = x^{3} and its tangent at x = 1.

Solution: The slope of the tangent line is the value of \frac{dy}{dx} = 3x^{2} at x = 1, so the equation of the tangent line is y − 1 = 3(x − 1), or y = 3x − 2. We know the line intersects the curve when x = 1, but we need to find the other points of intersection, so we need to solve the equation x^{3} − 3x + 2 = 0. Since x = 1 is a root, we can factor out x − 1 to get

x^{3} − 3x + 2 = (x − 1)(x^{2} + x − 2) = (x − 1)^{2}(x + 2) = 0.

So the only other point of intersection is x = −2. The area is given by

∫_{−2}^{1}(x^{3} − 3x + 2) dx = (\frac{1}{4}x^{4} − \frac{3}{2}x^{2} + 2x)]_{−2}^{1} = \frac{27}{4}.

• §7.2 #30

Sketch the graph of y = sin^{4} x cos x on the interval 0 ≤ x ≤ π/2, and find the area of the region between the curve and the x-axis.

Solution: Since cos x ≥ 0 for x between 0 and π/2, the graph lies above the axis. It meets the axis when x = 0 (so sin x = 0) and x = π/2 (so cos x = 0). The first and second derivatives are

\frac{dy}{dx} = 4 sin^{3} x cos^{2} x − sin^{5} x  = sin^{3} x(4 cos^{2} x − sin^{2} x),
\frac{d^{2}y}{dx^{2}} = 12 sin^{2} x cos^{3} x − 13 sin^{4} x cos x = cos x sin^{2} x(12 cos^{2} x − 13 sin^{2} x).

Ignoring the endpoints, the only relative extremum is the maximum when 4 cos^{2} x = sin^{2} x, so tan x = 2 (since tan x ≥ 0 on the interval), which gives x ≈ 1.107. The only inflection point (again ignoring endpoints) occurs when tan x = \sqrt{13/12}, so x ≈ 0.825.

The area of the region is given by ∫_{0}^{π/2} sin^{4} x cos x dx. Substituting u =

sin x to find the indefinite integral:

∫ sin^{4} x cos x dx = ∫ u^{4} du = \frac{1}{5}u^{5} + C = \frac{1}{5} sin^{5} x + C,

we get

∫_{0}^{π/2} sin^{4} x cos x dx = \frac{1}{5} sin^{5} x]_{0}^{π/2} = \frac{1}{5}.

• §7.3 #1a

Find the volume of the solid of revolution generated when the area bounded by y = \sqrt{x}, y = 0, x = 4 is revolved about the x-axis.

Solution: We integrate dV = πy^{2} dx = x dx from x = 0 to x = 4. This gives

V = ∫_{0}^{4} πx dx = \frac{π}{2} x^{2}]_{0}^{4} = 8π.

• §7.3 #1f

Find the volume of the solid of revolution generated when the area bounded by x^{2/3} + y^{2/3} = a^{2/3} (first quadrant) is revolved about the x-axis.

Solution: The curve meets the axis when y = 0, so x = a. We integrate dV = πy^{2} dx from x = 0 to x = a. Since y^{2} = (a^{2/3} − x^{2/3})^{3}, the volume is

V = ∫_{0}^{a} π(a^{2/3} − x^{2/3})^{3} dx = ∫_{0}^{a} π(a^{2} − 3a^{4/3}x^{2/3} + 3a^{2/3}x^{4/3} − x^{2}) dx
= π (a^{2}x − \frac{9}{5}a^{4/3}x^{5/3} + \frac{9}{7}a^{2/3}x^{7/3} − \frac{1}{3}x^{3})]_{0}^{a}
= (a^{3} − \frac{9}{5}a^{3} + \frac{9}{7}a^{3} − \frac{1}{3}a^{3}) π
= \frac{16πa^{3}}{105} .

• §7.3 #4

A horn-shaped solid is generated by a moving circle perpendicular to the y-axis whose diameter lies in the xy-plane and extends from y = 27x^{3} to y = x^{3}. Find the volume of this solid between y = 0 and y = 8.

Solution: The cross-section in a plane perpendicular to the y-axis is a disk with diameter running from x = \frac{1}{3}y^{1/3} to x = y^{1/3}, so its radius is \frac{1}{3}y^{1/3}. The area of a circular cross-section is therefore πr^{2} = \frac{1}{9}πy^{2/3}. To get the volume, we integrate dV = \frac{1}{9}πy^{2/3} dy for y = 0 to y = 8, giving

V = ∫_{0}^{8} \frac{1}{9}πy^{2/3} dy = \frac{π}{15}y^{5/3}]_{0}^{8} = \frac{32π}{15} .

• §7.3 #9

The base of a certain solid is the circle x^{2} + y^{2} = a^{2}. Each plane perpendicular to the x-axis intersects the solid in a square cross-section with one side in the base of the solid. Find its volume.

Solution: The cross-sections perpendicular to the x-axis have area

s^{2} = (2y)^{2} = 4y^{2} = 4(a^{2} − x^{2}).

Integrating dV = 4(a^{2} − x^{2}) dx from x = −a to x = a gives

V = ∫_{−a}^{a} 4(a^{2} − x^{2}) dx = 4 (a^{2}x − \frac{1}{3}x^{3})]_{−a}^{a} = \frac{16a^{3}}{3} .

• §7.3 #12

Find the volume of the solid formed by revolving the area inside the curve x^{2} + y^{4} = 1 about (a) the x-axis; (b) the y-axis.

Solution: Revolving about the x-axis, we get dV = πy^{2} dx = π(1 − x^{2})^{1/2} dx, so

V = ∫_{−1}^{1} π(1 − x^{2})^{1/2} dx = π ∫_{−1}^{1}(1 − x^{2})^{1/2} dx
= π · (areaofhalftheunitcircle)
= π \frac{π}{2}
= \frac{π^{2}}{2} .

Revolving about the y-axis, we get dV = πx^{2} dy = π(1 − y^{4}) dy, so

V = ∫_{−1}^{1} π(1 − y^{4}) dx = π (y − \frac{1}{5}y^{5})]_{−1}^{1} = \frac{8}{5}π.

• §7.4 #3

Sketch the region bounded by the curves y = \sqrt{x}, x = 4, y = 0, and use the shell method to find the volume of the solid generated by revolving this region about the y-axis.

Solution: The volume of a shell of radius r = x is dV = 2πxy dx = 2πx^{3/2} dx, so

V = ∫_{0}^{4} 2πx^{3/2} dx = \frac{4π}{5} x^{5/2}]_{0}^{4} = \frac{128}{5}π.

• §7.4 #7

Sketch the region bounded by the curves y = \frac{1}{x}, x = a, x = b (0 < a < b), y = 0, and use the shell method to find the volume of the solid generated by revolving this region about the y-axis.

Solution: The volume of a shell of radius r = x is dV = 2πxy dx = 2π dx, so

V = ∫_{a}^{b} 2π dx = 2π(b − a).

• §7.4 #10

A hole of radius \sqrt{3} is bored through the center of a sphere of radius 2. Find the volume removed.

Solution: Think of the hole as a “solid” of revolution gotten by revolving the the part of the circle, x^{2} + y^{2} = 4, with 0 ≤ x ≤ \sqrt{3}, about the y-axis. Then dV = 2πx(2y) dx where y = \sqrt{4 − x^{2}}, so

V = ∫_{0}^{√\overline{3}} 4πx(4 − x^{2})^{1/2} dx.

The subsitution u = 4 − x^{2}, du = −2x dx gives

∫ 4πx(4−x^{2})^{1/2} dx = ∫ −2πu^{1/2} du = −\frac{4}{3}πu^{3/2}+C = −\frac{4}{3}π(4−x^{2})^{3/2}+C.

So the volume is

V = −\frac{4}{3}π(4 − x^{2})^{3/2}]_{0}^{√\overline{3}} = −\frac{4}{3}π + \frac{32}{3}π = \frac{28}{3}π.

• §7.4 #12

Let r and h be positive numbers. The region bounded by the line x/r + y/h = 1 and the axes is revolved about the y-axis. Use the shell method to obtain the standard formula for the volume of a cone.

Solution: The volume of a shell of radius r = x is dV = 2πxy dx = 2πxh(1 − x/r) dx, so

V = ∫_{0}^{r} 2πh(x − x^{2}/r) dx = 2πh (\frac{1}{2}x^{2} − \frac{1}{3}x^{3}/r)]_{0}^{r} = \frac{1}{3}πhr^{2}.