148
CHAPTER 1!
19.63
Find J (2- *
3 )V dx.
19.64
Evaluate
Let w = r
2 + 3, du = 2tdt. Then
19.65
Find
19.66
Find the equation of the curve passing through (1, 5) and whose tangent line at (x, y) has slope 4x.
Substituting (1,5) for (x, y), we have 5 = 2(1)
2 + C, C = 3.
Hence,
Hence, y = 2x
2 + 3.
19.67
Find the equation of the curve passing through (9,18) and whose tangent line at (x, y) has slope Vx.
Substituting (9,18) for (x, y), we obtain
Hence,
19.68
Find the equation of the curve passing through (4, 2) and whose tangent line at (x, y) has slope x/y.
Substituting (4,2) for (x, y), we have
(a hyperbola).
19.69
Find the equation of the curve passing through (3, 2) and whose tangent line at (x, y) has slope *
2
/y
3
.
Substituting (3,2) for (x, y), we obtain 4 = 9+C,
19.70
Find the equation of a curve such that y" is always 2 and, at the point (2,6), the slope of the tangent line is 10.
y' = \y"dx = $2dx = 2x + C. When x = 2, y'= 10; so, 10 = 4 + C, C = 6, y' = 2x + 6. Hence,
y = / y' dx = x
2 + 6x + C,. When x = 2, y = 6; thus, 6 = 16 + C,, C, = -10, y = x
2 + 6x - 10.
19.71
Find the equation of a curve such that y" = 6x - 8 and, at the point (1, 0), y'=4.
/ = / y"dx = $(6x-8)dx = 3x
2 -8x+C. When x = l, y'=4; so, 4=-5+C, C = 9. Thus,
/ = 3x
2 ~ &x + 9. Then y = J y' dx = *
3 -4;c
2 + 9* + C,. When x = l, y = Q. Hence, 0 = 6+C U
Cj = ~6. So, y = x
3 - 4x
2 + 9x - 6.
19.72
A car is slowing down at the rate of 0.8 ft/s
2 . How far will the car move before it stops if its speed was initially
15mi/h?
o = -0.8. Hence, v = J a dt = -Q.&t + C. When t = 0, v=22ft/s [15 mi/h = (15 • 5280)73600 ft/s =
22ft/s.] So, u = -0.8f + 22. The car stops when v=0, that is, when r = 27.5. The position 5 =
J u rff = -0.4r
2 + 22f + s u , where s a is the initial position. Hence, the distance traveled in 27.5 seconds is
-0.4(27.5)
2 + 22(27.5) = 302.5 ft.
19.73
A block of ice slides down a 60-meter chute with an acceleration of 4 m/s
2 . What was the initial velocity of the
block if it reaches the bottom in 5 s?
v = J a dt = J4 dt = 4t+ v 0 , where v 0 is the initial velocity. Then the position s = J v dt = 2f + v 0 t. [We
let s = 0 at the top of the chute.] Then 60 = 2(25) + 5u 0 , i; 0 =2m/s.
Let u = 2 - x
3 , du = -3x
2 dx. Then
So,
y' = 4x.
y = $4xdx = 2x
2 + C.
dy/dx = xly. J y afy = J x dx, \ y
2 = {x'- + C, y
2 = x
2 + C,.
4 = 16+C,, C, = -12, / = ^
2 -12, or ^r
2 - y
2 = 12
4y/dx = x
2 iy\ J y
3 rf>- = | x
2 dx, \ y
4 = |x
3 + C.
C=-5, \y=\x-5, y = \x -20.
c = o.
CHAPTER 1!
19.63
Find J (2- *
3 )V dx.
19.64
Evaluate
Let w = r
2 + 3, du = 2tdt. Then
19.65
Find
19.66
Find the equation of the curve passing through (1, 5) and whose tangent line at (x, y) has slope 4x.
Substituting (1,5) for (x, y), we have 5 = 2(1)
2 + C, C = 3.
Hence,
Hence, y = 2x
2 + 3.
19.67
Find the equation of the curve passing through (9,18) and whose tangent line at (x, y) has slope Vx.
Substituting (9,18) for (x, y), we obtain
Hence,
19.68
Find the equation of the curve passing through (4, 2) and whose tangent line at (x, y) has slope x/y.
Substituting (4,2) for (x, y), we have
(a hyperbola).
19.69
Find the equation of the curve passing through (3, 2) and whose tangent line at (x, y) has slope *
2
/y
3
.
Substituting (3,2) for (x, y), we obtain 4 = 9+C,
19.70
Find the equation of a curve such that y" is always 2 and, at the point (2,6), the slope of the tangent line is 10.
y' = \y"dx = $2dx = 2x + C. When x = 2, y'= 10; so, 10 = 4 + C, C = 6, y' = 2x + 6. Hence,
y = / y' dx = x
2 + 6x + C,. When x = 2, y = 6; thus, 6 = 16 + C,, C, = -10, y = x
2 + 6x - 10.
19.71
Find the equation of a curve such that y" = 6x - 8 and, at the point (1, 0), y'=4.
/ = / y"dx = $(6x-8)dx = 3x
2 -8x+C. When x = l, y'=4; so, 4=-5+C, C = 9. Thus,
/ = 3x
2 ~ &x + 9. Then y = J y' dx = *
3 -4;c
2 + 9* + C,. When x = l, y = Q. Hence, 0 = 6+C U
Cj = ~6. So, y = x
3 - 4x
2 + 9x - 6.
19.72
A car is slowing down at the rate of 0.8 ft/s
2 . How far will the car move before it stops if its speed was initially
15mi/h?
o = -0.8. Hence, v = J a dt = -Q.&t + C. When t = 0, v=22ft/s [15 mi/h = (15 • 5280)73600 ft/s =
22ft/s.] So, u = -0.8f + 22. The car stops when v=0, that is, when r = 27.5. The position 5 =
J u rff = -0.4r
2 + 22f + s u , where s a is the initial position. Hence, the distance traveled in 27.5 seconds is
-0.4(27.5)
2 + 22(27.5) = 302.5 ft.
19.73
A block of ice slides down a 60-meter chute with an acceleration of 4 m/s
2 . What was the initial velocity of the
block if it reaches the bottom in 5 s?
v = J a dt = J4 dt = 4t+ v 0 , where v 0 is the initial velocity. Then the position s = J v dt = 2f + v 0 t. [We
let s = 0 at the top of the chute.] Then 60 = 2(25) + 5u 0 , i; 0 =2m/s.
Let u = 2 - x
3 , du = -3x
2 dx. Then
So,
y' = 4x.
y = $4xdx = 2x
2 + C.
dy/dx = xly. J y afy = J x dx, \ y
2 = {x'- + C, y
2 = x
2 + C,.
4 = 16+C,, C, = -12, / = ^
2 -12, or ^r
2 - y
2 = 12
4y/dx = x
2 iy\ J y
3 rf>- = | x
2 dx, \ y
4 = |x
3 + C.
C=-5, \y=\x-5, y = \x -20.
c = o.