Applications of First Order Equations
4.5 APPLICATIONS TO CURVES
4.5.2. Differentiating (A) exy D cy yields (B) .xy0C y/exy D cy0. From (A), cD exy
y . Substituting this into (B) and cancelling exyyields xy0C y D y0
y, so y0 D y2 .xy 1/. 4.5.4.Solving yD x1=2Ccx for c yields c D y
x x 1=2, and differentiating yields 0D y0 x
y
x2Cx 3=2 2 , or xy0 yD x1=2
2 . 4.5.6.Rewriting yD x3C c
x as xy D x4C c and differentiating yields xy0C y D 4x3. 4.5.8. Rewriting y D exC c.1 C x2/ as y
1C x2 D ex
1C x2 C c and differentiating yields y0 1C x2 2xy
.1C x2/2 D ex 1C x2
2xex
.1C x2/2, so .1C x2/y0 2xyD .1 x/2ex.
4.5.10. If (A) y D f C cg, then (B) y0 D f0C cg0. Mutiplying (A) by g0 and (B) by g yields (C) yg0 D fg0Ccgg0and (D) y0gD f0gCcg0g, and subtracting (C) from (D) yields y0g yg0D f0g fg0. 4.5.12.Let .x0; y0/ be the center and r be the radius of a circle in the family. Since . 1; 0/ and .1; 0/ are on the circle, .x0C 1/2C y20 D .x0 1/2C y02, which implies that x0 D 0. Therefore,the equation of the circle is (A) x2C .y y0/2D r2. Since .1; 0/ is on the circle, r2 D 1 C y02. Substituting this into (A) shows that the equation of the circle is x2C y2 2yy0D 1, so 2y0D x2C y2 1
y . Differentiating y.2xC 2yy0/ y0.x2C y2 1/D 0, so y0.y2 x2C 1/ C 2xy D 0.
4.5.14. From Example 4.5.6 the equation of the line tangent to the parabola at .x0; x02/ is (A) y D x02C 2x0x.
(a) From (A), .x; y/D .5; 9/ is on the tangent line through .x0; x20/ if and only if 9D x02C 10x0, or x02 10x0C 9 D .x0 1/.x0 9/D 0. Letting x0D 1 in (A) yields the line y D 1 C 2x, tangent to the parabola at .x0; x20/D .1; 1/. Letting x0D 9 in (A) yields the line y D 81 C 18x, tangent to the parabola at .x0; x20/D .9; 81/.
(b) From (A), .x; y/D .6; 11/ is on the tangent line through .x0; x02/ if and only if 11D x02C 12x0, or x20 12x0C 11 D .x0 1/.x0 11/D 0. Letting x0D 1 in (A) yields the line y D 1 C 2x, tangent to the parabola at .x0; x20/D .1; 1/. Letting x0 D 11 in (A) yields the line y D 121 C 22x, tangent to the parabola at .x0; x02/D .11; 21/.
(c) From (A), .x; y/D . 6; 20/ is on the tangent line through .x0; x02/ if and only if 20D x02 12x0, or x02C 12x0C 20 D .x0C 2/.x0C 10/ D 0. Letting x0 D 2 in (A) yields the line y D 4 4x, tangent to the parabola at .x0; x02/D . 2; 4/. Letting x0D 10 in (A) yields the line y D 100 20x, tangent to the parabola at .x0; x02/D . 10; 100/.
(d) From (A), .x; y/D . 3; 5/ is on the tangent line through .x0; x02/ if and only if 5D x20 6x0, or x20C 6x0C 5 D .x0C 1/.x0C 5/ D 0. Letting x0D 1 in (A) yields the line y D 1 2x, tangent to the parabola at .x0; x02/D . 1; 1/. Letting x0D 5 in (A) yields the line y D 25 10x, tangent to the parabola at .x0; x02/D . 5; 25/.
Section 4.5Applications to Curves 47 4.5.15. (a) If .x0; y0/ is any point on the circle such that x0 ¤ ˙1 (and therefore y0 ¤ 0), then differentiating (A) yields 2x0C 2y0y00 D 0, so y00 D x0
y0
. Therefore,the equation of the tangent line is yD y0 x0
y0
.x x0/. Since x02C y02D 1, this is equivalent to (B).
(b) Since y0D x0
y0
on the tangent line, we can rewrite (B) as y xy0D 1 y0
. Hence (F) 1 .y xy0/2 D y02 and (G) x02 D 1 y20 D .y xy0/2 1
.y xy0/2 . Since .y0/2 D x02
y02, (F) and (G) imply that .y0/2 D .y xy0/2 1, which implies (C).
(c) Using the quadratic formula to solve (C) for y0yields
y0D xy˙px2C y2 1
x2 1 .H/
if .x; y/ is on a tangent line with slope y0. If yD 1 x0x y0
, then x2C y2 1D x2C 1 x0x y0
2
1D
x x0
y0
2
(since x02C y02 D 1). Since y0 D x0
y0
, this implies that (H) is equivalent to x0
y0 D 1
x2 1
x.1 x0x/
y0 ˙
ˇ ˇ ˇ ˇ
x x0
y0
ˇ ˇ ˇ ˇ
, which holds if and only if we choose the “˙" so that ˙ ˇ ˇ ˇ ˇ
x x0
y0
ˇ ˇ ˇ ˇD
x x0
y0
. Therefore,we must choose˙ D if x x0
y0
> 0, so (H) reduces to (D), or˙ D C if x x0
y0
< 0, so (H) reduces to (E).
(d) Differentiating (A) yields 2xC 2yy0 D 0, so y0 D x
y on either semicircle. Since (D) and (E) both reduce to y0 D xy
1 x2 D x
y (since x2C y2D 1) on both semicircles, the conclusion follows.
(e) From (D) and (E) the slopes of tangent lines from (5,5) tangent to the circle are y0D 25˙p 49
24 D
3 4;4
3. Therefore, tangent lines are yD 5 C3
4.x 5/D 1C 3x=5
4=5 and yD 5 C4
3.x 5/D 1 4x=5 3=5 , which intersect the circle at . 3=5; 4=5/ .4=5; 3=5/, respectively. (See (B)).
4.5.16. (a) If .x0; y0/ is any point on the parabola such that x0 > 0 (and therefore y0 ¤ 0), then differentiating (A) yields 1 D 2y0y00, so y00 D 1
2y0
. Therefore,the equation of the tangent line is y D y0C 1
2y0
.x x0/. Since x0D y20, this is equivalent to (B).
(b) Since y0 D 1 2y0
on the tangent line, we can rewrite (B) as y0
2 D y xy0. Substituting this into (B) yields yD .y xy0/C x
4.y xy0/, which implies (C).
(c) Using the quadratic formula to solve (C) for y0yields
y0D y˙py2 x
2x .F/
if .x; y/ is on a tangent line with slope y0. If y D y0
2 C x 2y0
, then y2 x D 1 4
y0
x y0
2
so (F)
is equivalent to 1 2y0 D
y0C x y0 ˙
ˇ ˇ ˇ ˇ
y0
x y0
ˇ ˇ ˇ ˇ
4x which holds if and only if we choose the “˙" so that
˙ ˇ ˇ ˇ ˇ
y0
x y0
ˇ ˇ ˇ ˇD
y0
x y0
. Therefore,we must choose˙ D C if x > y02D x0, so (F) reduces to (D), or˙ D if x < y02D x0, so (F) reduces to (E).
(d) Differentiating (A) yields 1D 2yy0, so y0 D 1
2y on either half of the parabola. Since (D) and (E) both reduce to this if xD y2, the conclusion follows.
4.5.18. The equation of the line tangent to the curve at .x0; y.x0// is y D y.x0/C y0.x0/.x x0/.
Since y.x0=2/D 0, y.x0/ y0.x0/x0
2 D 0. Since x0is arbitrary, it follows that y0 D 2y x , so y0
y D 2 x, lnjyj D 2 ln jxj C k, and y D cx2. Since .1; 2/ is on the curve, c D 2. Therefore,y D 2x2.
4.5.20. The equation of the line tangent to the curve at .x0; y.x0// is y D y.x0/C y0.x0/.x x0/.
Since .x1; y1/ is on the line, y.x0/C y0.x0/.x1 x0/ D y1. Since x0 is arbitrary, it follows that yC y0.x1 x/D y1, so y0
y y1 D 1 x x1
, lnjy y1j D ln jx x1j C k, and y y1 D c.x x1/.
4.5.22.The equation of the line tangent to the curve at .x0; y.x0// is yD y.x0/C y0.x0/.x x0/. Since y.0/D x0, x0D y.x0/ y0.x0/x0. Since x0is arbitrary, it follows that xD y xy0, so (A) y0 y
x D 1.
The solutions of (A) are of the form yD ux, where u0xD 1, so u0D 1
x. Therefore,uD lnjxj C c and yD x ln jxj C cx.
4.5.24. The equation of the line normal to the curve at .x0; y0/ is y D y.x0/ x x0
y0.x0/. Since y.0/ D 2y.x0/, y.x0/C x0
y0.x0/ D 2y.x0/. Since x0is arbitrary, it follows that y0yD x, so (A) y2 2 D x2
2 C c 2 and y2D x2C c. Now y.2/ D 1 $ c D 3. Therefore,y Dp
x2 3.
4.5.26. Differentiating the given equation yields 2xC 4y C 4xy0C 2yy0 D 0, so y0 D xC 2y 2xC y is a differential equation for the given family, and (A) y0 D 2xC y
xC 2y is a differential equation for the orthogonal trajectories. Substituting y D ux in (A) yields u0xCu D 2C u
1C 2u, so u0xD 2.u2 1/
1C 2u and 1C 2u
.u 1/.uC 1/u0D 2 x, or
3
u 1 C 1
uC 1
u0D 4
x. Therefore, 3 lnju 1jCln juC1j D 4 ln jxjC K, so .u 1/3.uC 1/ D k
x4. Substituting uD y
x yields the orthogonal trajectories .y x/3.yC x/ D k.
4.5.28.Differentiating yields yex2.1C2x2/Cxex2y0D 0, so y0D y.1C 2x2/
x is a differential equation for the given family. Therefore,(A) y0 D x
y.1C 2x2/ is a differential equation for the orthogonal trajectories. From (A), yy0D x
1C 2x2, soy2 2 D 1
4ln.1C 2x2/Ck
2, and the orthogonal trajectories are given by y2D 1
2ln.1C 2x2/C k.
Section 4.5Applications to Curves 49
4.5.30. Differentiating (A) y D 1 C cx2yields (B) y0 D 2cx. From (C), c D y 1
x2 . Substituting this into (B) yields the differential equation y0D 2.y 1/
x for the given family of parabolas. Therefore,y0 D x
2.y 1/ is a differential equation for the orthogonal trajectories. Separating variables yields 2.y 1/y0D x, so .y 1/2 D x2
2 Ck. Now y. 1/ D 3 $ k D 9
2, so .y 1/2D x2 2 C9
2. Therefore,(D) y D 1 C
r9 x2
2 . This curve interesects the parabola (A) if and only if the equation (C) cx2 D r9 x2
2 has a solution x2in .0; 9/. Therefore,c > 0 is a necessaary condition for intersection. We will show that it is also sufficient. Squaring both sides of (C) and simplifying yields 2c2x4Cx2 9D 0. Using the quadratic formula to solve this for x2yields x2D 1Cp
1C 72c2
4c2 . The condition x2< 9 holds if and only if 1Cp
1C 72c2< 36c4, which is equivalent to 1C72c2< .1C36c2/2D 1C72c2C1296c4, which holds for all c > 0.
4.5.32. The angles and 1 from the x-axis to the tangents to C and C1satisfy tan D f .x0; y0/ and tan 1D f .x0; y0/C tan ˛
1 f .x0; y0/ tan ˛ D tan C tan ˛
1 tan tan ˛ D tan. C ˛/. Therefore, assuming and 1are both in Œ0; 2 /, 1D C ˛.
4.5.34.Circles centered at the origin are given by x2C y2D r2. Differentiating yields 2xC 2yy0D 0, so y0 D x
y is a differential equation for the given family, and y0 D .x=y/C tan ˛
1C .x=y/ tan ˛ is a differential equation for the desired family. Substituting yD ux yields u0xCu D 1=uC tan ˛
1C .1=u/ tan ˛ D 1C u tan ˛ uC tan ˛ . Therefore,u0x D 1C u2
uC tan ˛, uC tan ˛
1C u2 u0 D 1 x and 1
2ln.1C u2/C tan ˛ tan 1u D lnjxj C k.
Substituting uD y
x yields1
2ln.x2C y2/C .tan ˛/ tan 1y x D k.