dt - mlgibbons

advertisement
25. A police car traveling south toward Sioux Falls at 160 km/h pursues
a truck traveling east away from Sioux Falls, Iowa, at 140 km/h. At time
t = 0, the police car is 20 km north and the truck is 30 km east of Sioux
Falls. Calculate the rate at which the distance between the vehicles is
changing:
dl
l  20  30  1300
2
dl
dt
dy
dx
 160 &
 140
dt
dt
y  20  160t & x  30  140t
 l 2   30  140t    20  160t 
2
 2l
2
dt
 27.735 km/h
t 0
@ t  1/12 h, l  42.1966 km
 112.963 km/h
t 1/12
2
dl
dl
 2  30  140t 140   2  20  160t  160   l  4200  19600t  3200  25, 600t
dt
dt
dl 1000  45, 200t
 
dt
l
30. A particle moves counterclockwise around the ellipse with
equation 9x2 + 16y2 = 25.
(a)
In which of the four quadrants is dx/dt > 0? Explain.
(b) Find a relation between dx/dt and dy/dt.
(c) At what rate is the x-coordinate changing when the particle passes
the point (1, 1) if its y-coordinate is increasing at a rate of 6 m/s?
(d) Find dy/dt when the particle is at the top and bottom of the
ellipse.
dx
 x w/ respect to t
dt
dx
is positive in quadrants III & IV
dt
 x is moving to the right  .
9 x 2  16 y 2  25
 18 x
dx
dy
 32 y
0
dt
dt
dx
dy
 32 y
0
dt
dt
dy
dx
1,1 &  6  18  32  6   0
dt
dt
dx
32  6
32



m/s
dt
18
3
18 x
dy
0
dt
41. A roller coaster has the shape of the graph. Show that
when the roller coaster passes the point (x, f (x)), the
vertical velocity of the roller coaster is equal to f '(x) times
its horizontal velocity.
dy
dx
y  f  x 
 f ' x
dt
dt
In other words, vertical velocity equals
dx 

 f '  x   or f '  x   horizontal velocity.
dt 

y  f  x
Download