19.33
Find an equation of the curve passing through the point (0,1) and having slope I2x + 1 at any point (x, y).
D x y = 12x + l. Hence, y = 6x
2 + x + C. Since (0,1) lies on the curve, 1 = C. Thus, y = 6x
2 + x + 1
is the equation of the curve.
19.34
A particle moves along the *-axis with acceleration a = 2r-3ft/s
2 . At time t = 0 it is at the origin and
moving with a speed of 4ft/s in the positive direction. Find formulas for its velocity v and position s, and
determine where it changes direction and where it is moving to the left.
D,v = a=2t- 3. So, v = J (2t-3) dt = t
2 - 3t + C. Since v = 4 when t = 0, C = 4. Thus, v =
t
2 -3t + 4. Since v = D,s, s = J (t
2 -3t + 4) dt = ^t
3 - \t
2 + 4t + C,. Since s=0 when f = 0, C,=0.
Thus, 5 = |?
3 — \t
2 + 4t. Changes of direction occur where 5 reaches a relative maximum or minimum. To
look for critical numbers for 5, we set v = 0. The quadratic formula shows that v = 0 has no real roots.
[Alternatively, note that t
2 — 3/ + 4 = (t - | )
2 + J > 0.] Hence, the particle never changes direction. Since it
is moving to the right at t = 0, it always moves to the right.
19.35
Rework Problem 19.34 when the acceleration a = t
2 - " ft/s
2 .
v = S(t
2 -¥)dt=$t
3 -%t+C. Since u=4 when t = 0, C = 4. Thus, u=|f
3 -fr + 4. Then
s = fvdt= |-|f
4 - ¥• ^
2 + 4/+C, = T^
4 -fr
2 + 4f+C,. Since s = 0 when t = 0, C\=0. Thus, s =
n/
4 -T*
2 + 4f. To find critical numbers for s, we set i> = 0, obtaining t
3 - 13^ + 12 = 0. Clearly, f = l is a
root. Dividing t
3 -13t+l2 by f-1, we obtain t
2 + t- 12 = (t + 4)(t - 3). Thus, r = 3 and f =-4
also are critical numbers. When t = \, D
2 i = a = -^<0. Hence, s has a relative maximum at t = \.
Similarly, a= " >0 when t = 3, and, therefore, 5 has a relative minimum at t = 3; a=T>0 when
t = -4, and, therefore, s has a relative minimum at t= -4. Hence, the particle changes direction at t= -4,
t = 1, and f = 3. It moves left for t< -4, where it reaches a minimum; it moves right from t= —4 to
t=l, where it reaches a maximum; it moves left from t = \ to f = 3, where it reaches a minimum; then it
moves right for f > 3.
19.36
A motorist applies the brakes on a car moving at 45 miles per hour on a straight road, and the brakes cause a
constant deceleration of 22 ft/s . In how many seconds will the car stop, and how many feet will the car have
traveled after the time the brakes were applied?
Let t = 0 be the time the brakes were applied, let the positive s direction be the direction that the car was
traveling, and let the origin 5 = 0 be the point at which the brakes were applied. Then the acceleration
a =—22. So, v = J a dt= -22t + C. The velocity at t = 0 was 45 mi/h, which is the same as
66ft/s: Hence, C = 66. Thus, v = ~22t + 66. Then s = J v dt = -llr + 66f + C,. Since . $ = 0 when
t-0, C, =0 and s = -llt
2 + 66t. The car stops when v=0, that is, when t = 3. At t = 3, s =
99. So, the car stops in 3 seconds and travels 99 feet during that time.
19.37
A particle moving on a straight line has acceleration a = 5 — 3t, and its velocity is 7 at time t = 2. lts(t)\s
the distance from the origin, find 5(2) — s(l).
v = $adt = 5t-ll
2 + C. Since the velocity is 7 when t = 2, 7 = 10-6+C, C = 3. So, v = 5t -
|<
2 + 3. Then s = \t
2 - JV
3 + 3r + C,. Hence, s(2)=12+C,, s(l) = 5 + C t , and s(2)-s(l) = 7.
19.38
Find an equation of the curve passing through the point (3, 2) and having slope 2x
2 — 5 at any point (x, y).
D :c y = 2x
2 -5. Hence, y = I*
3 - 5x + C. Since (3, 2) is on the curve, 2= |(3)
3 -5(3) + C, C=-l.
Thus, y = |jc
3 - 5x — 1 is an equation of the curve.
19.39
A particle is moving along a line with acceleration a = sin2r+ t
2 ft/s
2 . At time / = 0, its velocity is 3 ft/s.
What is the distance between the particle's location at time t = 0 and its location at time t = ir/2, and what
is the particle's speed at time / = tr/21
v = J adt= -\ cos2f + j/
3 + C. Since v = 3 when f = 0, C = 3. Thus, v = -\ cos2/ + \f + 3.
Hence, at time t=-rr/2, v = - \ cos IT + |(7r/2)3 + 3 = \ + w3/24. Its position s = J i; dt = - \ sin2r +
^- Jf4 + 3f+C, = -Jsin2;+ i(4 + 3f+C,. Then j(0) = Cl and i(w/2) = - | sin TT + A(^/2)4 +
3(7r/2)+d. Hence, 5(77/2) - s(O) = (77"+2887r)/192.
ANTIDERIVATIVES (INDEFINITE INTEGRALS)
145
Find an equation of the curve passing through the point (0,1) and having slope I2x + 1 at any point (x, y).
D x y = 12x + l. Hence, y = 6x
2 + x + C. Since (0,1) lies on the curve, 1 = C. Thus, y = 6x
2 + x + 1
is the equation of the curve.
19.34
A particle moves along the *-axis with acceleration a = 2r-3ft/s
2 . At time t = 0 it is at the origin and
moving with a speed of 4ft/s in the positive direction. Find formulas for its velocity v and position s, and
determine where it changes direction and where it is moving to the left.
D,v = a=2t- 3. So, v = J (2t-3) dt = t
2 - 3t + C. Since v = 4 when t = 0, C = 4. Thus, v =
t
2 -3t + 4. Since v = D,s, s = J (t
2 -3t + 4) dt = ^t
3 - \t
2 + 4t + C,. Since s=0 when f = 0, C,=0.
Thus, 5 = |?
3 — \t
2 + 4t. Changes of direction occur where 5 reaches a relative maximum or minimum. To
look for critical numbers for 5, we set v = 0. The quadratic formula shows that v = 0 has no real roots.
[Alternatively, note that t
2 — 3/ + 4 = (t - | )
2 + J > 0.] Hence, the particle never changes direction. Since it
is moving to the right at t = 0, it always moves to the right.
19.35
Rework Problem 19.34 when the acceleration a = t
2 - " ft/s
2 .
v = S(t
2 -¥)dt=$t
3 -%t+C. Since u=4 when t = 0, C = 4. Thus, u=|f
3 -fr + 4. Then
s = fvdt= |-|f
4 - ¥• ^
2 + 4/+C, = T^
4 -fr
2 + 4f+C,. Since s = 0 when t = 0, C\=0. Thus, s =
n/
4 -T*
2 + 4f. To find critical numbers for s, we set i> = 0, obtaining t
3 - 13^ + 12 = 0. Clearly, f = l is a
root. Dividing t
3 -13t+l2 by f-1, we obtain t
2 + t- 12 = (t + 4)(t - 3). Thus, r = 3 and f =-4
also are critical numbers. When t = \, D
2 i = a = -^<0. Hence, s has a relative maximum at t = \.
Similarly, a= " >0 when t = 3, and, therefore, 5 has a relative minimum at t = 3; a=T>0 when
t = -4, and, therefore, s has a relative minimum at t= -4. Hence, the particle changes direction at t= -4,
t = 1, and f = 3. It moves left for t< -4, where it reaches a minimum; it moves right from t= —4 to
t=l, where it reaches a maximum; it moves left from t = \ to f = 3, where it reaches a minimum; then it
moves right for f > 3.
19.36
A motorist applies the brakes on a car moving at 45 miles per hour on a straight road, and the brakes cause a
constant deceleration of 22 ft/s . In how many seconds will the car stop, and how many feet will the car have
traveled after the time the brakes were applied?
Let t = 0 be the time the brakes were applied, let the positive s direction be the direction that the car was
traveling, and let the origin 5 = 0 be the point at which the brakes were applied. Then the acceleration
a =—22. So, v = J a dt= -22t + C. The velocity at t = 0 was 45 mi/h, which is the same as
66ft/s: Hence, C = 66. Thus, v = ~22t + 66. Then s = J v dt = -llr + 66f + C,. Since . $ = 0 when
t-0, C, =0 and s = -llt
2 + 66t. The car stops when v=0, that is, when t = 3. At t = 3, s =
99. So, the car stops in 3 seconds and travels 99 feet during that time.
19.37
A particle moving on a straight line has acceleration a = 5 — 3t, and its velocity is 7 at time t = 2. lts(t)\s
the distance from the origin, find 5(2) — s(l).
v = $adt = 5t-ll
2 + C. Since the velocity is 7 when t = 2, 7 = 10-6+C, C = 3. So, v = 5t -
|<
2 + 3. Then s = \t
2 - JV
3 + 3r + C,. Hence, s(2)=12+C,, s(l) = 5 + C t , and s(2)-s(l) = 7.
19.38
Find an equation of the curve passing through the point (3, 2) and having slope 2x
2 — 5 at any point (x, y).
D :c y = 2x
2 -5. Hence, y = I*
3 - 5x + C. Since (3, 2) is on the curve, 2= |(3)
3 -5(3) + C, C=-l.
Thus, y = |jc
3 - 5x — 1 is an equation of the curve.
19.39
A particle is moving along a line with acceleration a = sin2r+ t
2 ft/s
2 . At time / = 0, its velocity is 3 ft/s.
What is the distance between the particle's location at time t = 0 and its location at time t = ir/2, and what
is the particle's speed at time / = tr/21
v = J adt= -\ cos2f + j/
3 + C. Since v = 3 when f = 0, C = 3. Thus, v = -\ cos2/ + \f + 3.
Hence, at time t=-rr/2, v = - \ cos IT + |(7r/2)3 + 3 = \ + w3/24. Its position s = J i; dt = - \ sin2r +
^- Jf4 + 3f+C, = -Jsin2;+ i(4 + 3f+C,. Then j(0) = Cl and i(w/2) = - | sin TT + A(^/2)4 +
3(7r/2)+d. Hence, 5(77/2) - s(O) = (77"+2887r)/192.
ANTIDERIVATIVES (INDEFINITE INTEGRALS)
145