ANTIDERIVATIVES (INDEFINITE INTEGRALS)
19.95
Find the escape velocity for an object shot vertically upward from the surface of a sphere of radius R and mass M.
[Assume the inverse square law for gravitational attraction F = — G(m l m 2 /s
2 ), where G is a positive constant
(dependent on the units used for force, mass, and distance), and m l and m 2 are two masses at a distance s.
Assume also Newton's law F = ma.]
19.96
Find the escape velocity for an object shot vertically upward from the surface of the Earth. (Let —g be the
acceleration due to the Earth's gravity at the surface of the Earth; g = 32 ft/s
2 .)
19.97
The equation xy = c represents the family of all equilateral hyperbolas with center at the origin. Find the
equation of the family of curves that intersect the curves of the given family at right angles.
For the given equation, xy'+y = 0, y' = — y/x. Hence, for the orthogonal family, dy/dx = x/y,
J y dy = J x dx, \ y
2 = \x
2 + C, y
2 — x
2 = C,. This is a family of hyperbolas with axes of symmetry on the xor y-axis.
19.98
Compare the values of J 2 cos xsinxdx obtained by the substitutions (a) u = sin x and (b) u = cos x,
and reconcile the results.
(a) Let M = sin x, du = cosxdx. Then J 2 sin x cos x dx = 2 J u du = 2 • \ u2 + C = sin2 x + C. (b)
Let M = COSJC, du = -sin x dx. Then J2sin xcosxdx = — 2 J u du = — 2- \u
2 + C= — cos
2 x + C. There
is no contradiction between the results of (a) and (b). sin
2 x and —cos
2 x differ by a constant, since
sin
2 x + cos
2 x = \. So, it is not surprising that they have the same derivative.
19.99
Compute J cos
2 x dx.
Remember the trigonometric identity cos
2 x = (1 + cos 2x) 12. Hence, J cos
2 x dx = \ J (1 + cos 2x) dx =
!(* + 2 sin2x) + C= \(x + sin x -cos*) + C. For the last equation, we used the trigonometric identity
sin 2x = 2 sin x cos x.
19.100 Compute Jsin
2 *d.x.
J sin2 x dx = J (1 - cos2 x) dx = x - \ (x + sin x cos x) + C [by Problem 19.99] = | (x - sin x cos x) + C.
By Problem 19.95, -g = a =-GMAR
2 , g=GM/R
2 , GM = gR
2 . Hence, 2GM/R = 2gR. There[f we approximate the radius of the
Now,
fore, the escape velocity is
Earth by 4000 miles, then the escape velocity is about
Let m be the mass of the object, ma = — G(mM/s ),
of the sphere. Hence, Now, a=
where s is the distance of the object from the center
Thus,
Hence
the initial velocity.
When
GMIR. Thus,
In order for the object never to return to the surface of the
sphere, v must never be 0. Since GM/s approaches 0 as s—»+<», we must have
or
is the escape velocity.
Thus,
19.94
19.92
19.93
151
a=-GM/s.
s = R, v = v 0 ,
- HGMIs2) ds,
v.
v
=-GM/s2,
- GM/R.
j
y-ll*dy = Sx-"3dx, \y^=\x^ + C, y2'3-*2'3 = Q.
19.95
Find the escape velocity for an object shot vertically upward from the surface of a sphere of radius R and mass M.
[Assume the inverse square law for gravitational attraction F = — G(m l m 2 /s
2 ), where G is a positive constant
(dependent on the units used for force, mass, and distance), and m l and m 2 are two masses at a distance s.
Assume also Newton's law F = ma.]
19.96
Find the escape velocity for an object shot vertically upward from the surface of the Earth. (Let —g be the
acceleration due to the Earth's gravity at the surface of the Earth; g = 32 ft/s
2 .)
19.97
The equation xy = c represents the family of all equilateral hyperbolas with center at the origin. Find the
equation of the family of curves that intersect the curves of the given family at right angles.
For the given equation, xy'+y = 0, y' = — y/x. Hence, for the orthogonal family, dy/dx = x/y,
J y dy = J x dx, \ y
2 = \x
2 + C, y
2 — x
2 = C,. This is a family of hyperbolas with axes of symmetry on the xor y-axis.
19.98
Compare the values of J 2 cos xsinxdx obtained by the substitutions (a) u = sin x and (b) u = cos x,
and reconcile the results.
(a) Let M = sin x, du = cosxdx. Then J 2 sin x cos x dx = 2 J u du = 2 • \ u2 + C = sin2 x + C. (b)
Let M = COSJC, du = -sin x dx. Then J2sin xcosxdx = — 2 J u du = — 2- \u
2 + C= — cos
2 x + C. There
is no contradiction between the results of (a) and (b). sin
2 x and —cos
2 x differ by a constant, since
sin
2 x + cos
2 x = \. So, it is not surprising that they have the same derivative.
19.99
Compute J cos
2 x dx.
Remember the trigonometric identity cos
2 x = (1 + cos 2x) 12. Hence, J cos
2 x dx = \ J (1 + cos 2x) dx =
!(* + 2 sin2x) + C= \(x + sin x -cos*) + C. For the last equation, we used the trigonometric identity
sin 2x = 2 sin x cos x.
19.100 Compute Jsin
2 *d.x.
J sin2 x dx = J (1 - cos2 x) dx = x - \ (x + sin x cos x) + C [by Problem 19.99] = | (x - sin x cos x) + C.
By Problem 19.95, -g = a =-GMAR
2 , g=GM/R
2 , GM = gR
2 . Hence, 2GM/R = 2gR. There[f we approximate the radius of the
Now,
fore, the escape velocity is
Earth by 4000 miles, then the escape velocity is about
Let m be the mass of the object, ma = — G(mM/s ),
of the sphere. Hence, Now, a=
where s is the distance of the object from the center
Thus,
Hence
the initial velocity.
When
GMIR. Thus,
In order for the object never to return to the surface of the
sphere, v must never be 0. Since GM/s approaches 0 as s—»+<», we must have
or
is the escape velocity.
Thus,
19.94
19.92
19.93
151
a=-GM/s.
s = R, v = v 0 ,
- HGMIs2) ds,
v.
v
=-GM/s2,
- GM/R.
j
y-ll*dy = Sx-"3dx, \y^=\x^ + C, y2'3-*2'3 = Q.