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HW2^1
- Sec. 1.2: 26. A projectile is Öred straight upward with an initial velocity of 100 m/sec^2 from the top of a building 20m high and falls to the ground at the base of the building. Find (a) its maximum height above the ground; (b) when it passes the top of the building; (c) its total time in the air. Solution. Let x (t) be the distance between the projectile and the base of the building at time t. We have:
x (0) = 0; x^0 (0) = 100; x^00 = g ) x (t) = x (0) + v (0) t gt^2 =2 = 20 + 100t gt^2 = 2 :
(a) Maximum height:
x = 20+100t gt^2 = 2 ) x^0 = gt 100 ) t = 100=g = 100= 9 : 8 10 : 204 :
Then
xmax = x (10: 204) = 20 + 100 10 : 204 9 : 8 (10:204)^2 =2 = 530: 2 m:
(b) It passes the top of the building when
x (t) = 0 ) 20 + 100t 9 : 8 t^2 =2 = 20 ) t 20 : 408 :
(c) Total time in the air: 20 + 100t 9 : 8 t^2 =2 = 0 ) ft = 0 :198 08g ; ft = 20: 606 g. So, the total time in the air= 20 : 606 sec :
- Sec. 1.2: 35. Let x (t) be the height of the stone at time t. We have
x^00 = g ) x (t) = x 0 + v 0 t + gt^2 = 2 :
We are given that v 0 = 0 and x (0) = h )
x (t) = h gt^2 = 2 :
So, x = 0 when h gt^2 =2 = 0 ) t =
p 2 h=g:
v (t) = gt ) vground = g
p 2 h=g =
p 2 gh: (^1) Note to the grader. Please grade the following problems: Sec. 1.2: 26, Sec. 1.3: 9, ec. 1.3: 13, Sec. 1.3: 25, #8 (numbering is from the solutions) (20 pts for each problem)
- Sec. 1.3: 9. dy dx
= x^2 y 2 :
The direction Öeld is given on the sketch:
Sketch for y ( 2) = 2 :
y ( 1) = 2:
y (0) = 2:
y (0) = 2:
y (1) = 2:
y (1) = 2:
y (2) = 2:
y (2) = 2:
y ( 2 :5) = 2 :
y ( 0 :5) = 2.
y (0:5) = 2 :
y (0:5) = 2 :
y (1:5) = 2 :
y (1:5) = 2 :
y (2:5) = 2 :
y (2:5) = 2 :
(students do not have to include the sketch).
- Sec. 1.3: 19. y^0 = ln
1 + y^2
; y (0) = 0: It is evident that f (x; y) = ln (1 + y^2 ) is continuous in a neighborhood of (0; 0). Further, fy =
2 y 1 + y^2 is continuous in a neighborhood of (0; 0). Hence, there is a unique solution.
- Sec. 1.3: 21. y^0 = x + y; y (0) = 0; y ( 4) =? f (x; y) = x + y. Slope Öeld:
Sec. 1.3: 21.
Solution:We see from the sketch that approximately y ( 4) = 3. The following sketch is produced by the slope calculator:
- Sec. 1.3: 25. v^0 = 32 1 : 6 v f (t; v) = 32 1 : 6 v: The slope Öeld:
Sec. 1.3: 25.
From the sketch: limiting velocity=20ft/sec. 95% of the limiting ve- locity is 95 20 100
One can see from the sketch that 95% of the limiting velocity is achieved in approximately 2 seconds.
- (a) Use the method of isoclines to construct the slope Öeld for the fol- lowing di§erential equation
y^0 = x^2 y^2 :
(b) Use part (a) to sketch the solution curve for the following initial value problem: dy dx
= x^2 y^2
y (1) = 1 :
Solution. (a) Isoclines: x^2 y^2 = c:
Take c = 1; 2 ; 1 ; 3 :
The sketch below is produced by using MAPLE:
- (a)
If t = 0, then y 0 = 5 + A ) A = y 0 5 )
y (t) = 5 + (y 0 5) e t:
(d) The sketch from part (a) indicates that solutions should approach to the equilibrium as t! + 1. Indeed,
y (t) = 5 + Ae t:
Because e t^! 0 as t! + 1 , we see that
lim t!+ 1 y (t) = lim t!+ 1
5 + Ae t
- Sec. 1.4:13.
y^3
dy dx
y^4 + 1
cos x:
This is a separable equation:
y^3 y^4 + 1
dy = cos xdx:
By integrating, Z y^3 y^4 + 1
dy =
Z
cos xdx = sin x + C:
Now we calculate the integral in the left-hand side: Z y^3 y^4 + 1
dy =
Z
d (y^4 + 1) y^4 + 1
; whence
Z
y^3 y^4 + 1
dy
ln y^4 + 1 + C =
ln
y^4 + 1
+ C:
Answer (implicit solution):
ln
y^4 + 1
= 4 sin x + C:
- Sec. 1.4:17. y^0 = 1 + x + y + xy: First we factor the right-hand side:
1 + x + y + xy = (1 + x) + y (1 + x) = (1 + x) (1 + y) :
Hence we have
dy dx
= (1 + x) (1 + y) )
dy 1 + y
dx 1 + x
Z
dy 1 + y
Z
(1 + x) dx ) ln j1 + yj = x +
x^2 2
+ C.
Then
ln j1 + yj = x +
x^2 2
+ C.
Answer: (implicit solution)
ln j1 + yj = x +
x^2 2
+ C:
