150
CHAPTER 19
19.84
Find the family of curves for which the slope of the tangent line at (x, y) is (1 + x)/(l + y).
Then
This consists of two families of hyperbolas with center at (—1, — 1).
19.85
Find the family of curves for which the slope of the tangent line at(x,y) isx
Then
19.86
If y" = 24/x
3 at all points of a curve, and, at the point (1,0), the tangent line is I2x + y = 12, find the
equation of the curve.
Since the slope of 12* + y = 12 is -12, when *=1. y'=
Since the curve passes through (1,0).
So, the equation of the curve is y = 12
Then y = J y' dx = Ux~
l + C,.
or xy = 12(1 - x).
19.87
A rocket is shot from the top of a tower at an angle of 45° above the horizontal (Fig. 19-1). It hits the ground in 5
seconds at a horizontal distance from the foot of the tower equal to three times the height of the tower. Find the
height of the tower.
Let h be the height, in feet, of the tower. The height of the rocket s=h+v0t-16t2, where v0 is the
vertical component of the initial velocity. When ( = 5, s = 0. So, 0 = h + 5v 0 -400, v 0 = (400- h)/5.
Since the angle of projection is 45°, the horizontal component of the velocity has the initial value v 0 , and this value
is maintained. Hence, the horizontal distance covered in 5 seconds is 5v 0 . Thus, 5[(400- h)/5] = 3h,
h = 100 ft.
In Problems 19.88-19.94, find the general solution of the indicated differential equation.
19.88
19.89
19.90
19.91
y = $(24x
3 + I8x
2 -8x + 3)dx = 6x' + 6x
3 -4x
2 + 3x+C.
= 6x
2 + 4x-5.
= (3* + l)
3
.
= 24x3 + I8x2 - 8x + 3.
y = / (3* + I)
3 dx. Let u = 3x + l, du = 2>dx. Then y = £J u
3 du = I • J • w
4 + C= A(3x + I)
4 + C.
Fig. 19-1
y" = 24jT
3
. v'= f y"d;c = -12x"
2 + C.
-12. Thus, C = 0, y' = -12^;"
2
.
C, = -12.
C = 0, (x + l)
2 -(y + l)
2 + C 1 =0.
dy/dx = (l + x)/(l+y).
I(l + y)dy = f(l + x)dx, y + \y
2 = x+ {x
2 + C, x
2 -y
2 +2x-2y +
y = J (6X + 4x-5)dx = 2x3 + 2x2 -5x+C.
dy/dx = xVy- Then I y'1'2 dy = f x dx, 2//2 = |*2 + C, //2=J*2 + C,.
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