19.74
If a particle starts from rest, what constant acceleration is required to move the particle 50 mm in 5 s along a
straight line?
v = J a dt = at + v a . Since the particle starts from rest, i> 0 = 0, v = at. Then s = J v dt = \at
2 . [We
set s = 0 at t = 0.] So, 50= |a(25), a = 4m/s
2 .
ANTIDERIVATIVES (INDEFINITE INTEGRALS)
149
19.75
What constant deceleration is needed to slow a particle from a velocity of 45 ft/s to a dead stop in 15 ft?
Since v g = 45, v = at + 45. When the particle stops, v = 0, that is, t = -45 la.
Now,
Hence,
at
Then
[We let
19.76
A ball is rolled in a straight line over a level lawn, with an initial velocity of 10 ft/s. If, because of friction, the
velocity decreases at the rate of 4 ft/s
2 , how far will the ball roll?
The deceleration a = -4. v = J a dt = -4t + v a . When f = 0, v = 10; hence, v 0 = 10, v =-4t +
10. Then the position s = J v dt = — 2t
2 + Wt [We assume s = 0 when f = 0.] The ball stops when
u = 0, that is, when t = 2.5. Hence, the distance rolled is -2(2.5)
2 + 10(2.5) = 12.5 ft.
19.77
Find the equation of the family of curves whose tangent line at any point (x, y) has slope equal to — 3x
2 .
Thus, the family is a family of cubic curves y = — x
3 + C.
19.78
Find
Let w = 1 + tan x, du = sec
2 x dx. Then
19.79
Evaluate J x
2 esc
2 x
3 dx.
Let u = cot x
3 , du - (-esc
2 Jt
3 )(3jt
2 dx). Then
19.80
Find J (tan B + cot 0)
2 rffl.
19.81
Evaluate
Hence,
19.82
Evaluate
substitute x = Tr/2 — i;
Hence,
and use Problem 19.81.]
19.83
Find the family of curves for which the slope of the tangent line at(x, y) is (1+x)/(1-y).
Then
y' = -3x
2 . y = J y' dx = -x
3 + C.
v = / a dt = at + V Q .
dx.
dy/dx = (l+x)/(l-y).
C = 0, (x + l)2 + (y-l)2 = CWith C, > 0, this is a family of circles with center at (—1, 1).
l .
f(l-y)dy = f(l + x)dx, y - {y
2 = x + {x
2 + C, x
2 + y
2 +2x-2y +
(sec" x — tan x sec x) dx = tan x — sec x + C [Or:
s=0
t=0.
/ (tan 0 + cot 0)
2 d6 = J (tan
2 6 + 2 + cot
2 0) d6 = J (sec
2 0 - 1 + 2 + esc
2 0 - 1) d6 = J (sec
2 0 + esc
2 6) d6
= tan 0 - cot 0 + C.
If a particle starts from rest, what constant acceleration is required to move the particle 50 mm in 5 s along a
straight line?
v = J a dt = at + v a . Since the particle starts from rest, i> 0 = 0, v = at. Then s = J v dt = \at
2 . [We
set s = 0 at t = 0.] So, 50= |a(25), a = 4m/s
2 .
ANTIDERIVATIVES (INDEFINITE INTEGRALS)
149
19.75
What constant deceleration is needed to slow a particle from a velocity of 45 ft/s to a dead stop in 15 ft?
Since v g = 45, v = at + 45. When the particle stops, v = 0, that is, t = -45 la.
Now,
Hence,
at
Then
[We let
19.76
A ball is rolled in a straight line over a level lawn, with an initial velocity of 10 ft/s. If, because of friction, the
velocity decreases at the rate of 4 ft/s
2 , how far will the ball roll?
The deceleration a = -4. v = J a dt = -4t + v a . When f = 0, v = 10; hence, v 0 = 10, v =-4t +
10. Then the position s = J v dt = — 2t
2 + Wt [We assume s = 0 when f = 0.] The ball stops when
u = 0, that is, when t = 2.5. Hence, the distance rolled is -2(2.5)
2 + 10(2.5) = 12.5 ft.
19.77
Find the equation of the family of curves whose tangent line at any point (x, y) has slope equal to — 3x
2 .
Thus, the family is a family of cubic curves y = — x
3 + C.
19.78
Find
Let w = 1 + tan x, du = sec
2 x dx. Then
19.79
Evaluate J x
2 esc
2 x
3 dx.
Let u = cot x
3 , du - (-esc
2 Jt
3 )(3jt
2 dx). Then
19.80
Find J (tan B + cot 0)
2 rffl.
19.81
Evaluate
Hence,
19.82
Evaluate
substitute x = Tr/2 — i;
Hence,
and use Problem 19.81.]
19.83
Find the family of curves for which the slope of the tangent line at(x, y) is (1+x)/(1-y).
Then
y' = -3x
2 . y = J y' dx = -x
3 + C.
v = / a dt = at + V Q .
dx.
dy/dx = (l+x)/(l-y).
C = 0, (x + l)2 + (y-l)2 = CWith C, > 0, this is a family of circles with center at (—1, 1).
l .
f(l-y)dy = f(l + x)dx, y - {y
2 = x + {x
2 + C, x
2 + y
2 +2x-2y +
(sec" x — tan x sec x) dx = tan x — sec x + C [Or:
s=0
t=0.
/ (tan 0 + cot 0)
2 d6 = J (tan
2 6 + 2 + cot
2 0) d6 = J (sec
2 0 - 1 + 2 + esc
2 0 - 1) d6 = J (sec
2 0 + esc
2 6) d6
= tan 0 - cot 0 + C.