Saturday, March 21, 2009

Question set 7

  1. Let A be the area of a circle with radius r at time t. If the radius changes at a rate of 2 in/sec, at what rate is the circle's area changing when r = 1?
  2. Let A be the area of a square with side s at time t. If the side changes at a rate of -4 mm/week, at what rate is the square's area changing when s = 3?
  3. Let V be the volume of a sphere with radius r at time t. If the radius changes at a rate of 3 ft/min, at what rate is the sphere's volume changing when r = 2?
  4. Let S be the surface area of a sphere with radius r at time t. If the radius changes at a rate of -5 m/hr, at what rate is the sphere's surface area changing when r = 1?
  5. Let V be the volume of a cube with side s at time t. If the side changes at a rate of 10 in/hr, at what rate is the cube's volume changing when s = 5?
  6. Let S be the surface area of a cube with side s at time t. If the side changes at a rate of 7 cm/sec, at what rate is the cube's surface area changing when s = 3?
  7. Let A be the area of a circle with radius r at time t. If the radius changes at a rate of -3 ft/sec, at what rate is the circle's area changing when r = 5?
  8. Let A be the area of a square with side s at time t. If the side changes at a rate of 2 m/day, at what rate is the square's area changing when s = 10?
  9. Let V be the volume of a sphere with radius r at time t. If the radius changes at a rate of -8 in/min, at what rate is the sphere's volume changing when r = 7?
Posted by at

26 comments:

  1. poopsiepumkinpieMarch 22, 2009 at 8:36 AM

    1)
    what we are required to find is dA/dt.
    since we know that the area,A = pi r^2

    dA/dr= 2pi r

    the change in radius is dr/dt=2 in/sec

    since we know dA/dr and dr/dt we can now find
    dA/dt

    dA/dt=dA/dr * dr/dt
    =2 pi r* 2
    =4 pi r

    dA/dt when r=1 is
    dA/dt=4 pi *(1)
    =4 pi

    ReplyDelete
  2. poopsiepumkinpieMarch 22, 2009 at 10:57 AM

    (2)

    A= s^2

    dA/ds= 2s


    ds/dt= -4 mm/week


    dA/dt=dA/ds * ds/dt

    = 2s * -4

    = -8 s

    ReplyDelete
  3. ghetto_celebMarch 22, 2009 at 8:06 PM

    number 2

    i know the area of a square is A = s^2

    rate of change of the side with respect to time is
    ds/dt = -4

    question is asking for change in area with respect to time dA/dt

    so change in area w.r.t side is
    dA/ds = 2s

    we can now find dA/dt

    dA/dt = dA/ds * ds/dt

    dA/dt = 2s * -4

    dA/dt = -8s

    therefore dA/dt when s = 3

    dA/dt = -8 * 3

    dA/dt = -24

    ReplyDelete
  4. mysticwings*March 22, 2009 at 8:46 PM

    no 3

    Vol of sphere = 4/3 pi r^3

    dv/dr = 4 pi r^2

    dr/dt = 3

    dv/dt = dv/dr * dr/dt

    = 4 pi r^2 * 3

    = 12 pi r^2

    at r = 2

    dv/dt = 12 pi (2)^2

    = 48 pi

    ReplyDelete
  5. mysticwings*March 22, 2009 at 8:54 PM

    no 4

    S = 4 pi r^2

    ds/dr = 8 pi r

    dr/dt = -5

    ds/dt = ds/dr * dr/dt

    = 8 pi r * -5

    = -40 pi r

    at r = 1

    ds/dt = -40 pi

    ReplyDelete
  6. mysticwings*March 22, 2009 at 9:00 PM

    no 5

    V = s^3

    dv/ds = 3s^2

    ds/dt = 10

    dv/dt = dv/ds * ds/dt

    = 3s^2 * 10

    = 30s^2

    at s = 5

    dv/dt = 30(5)^2

    = 750

    ReplyDelete
  7. phoenixMarch 23, 2009 at 1:12 PM

    Question #6:

    Let S be the surface area of a cube with side s at time t. If the side changes at a rate of 7 cm/sec, at what rate is the cube's surface area changing when s = 3?

    Answer:
    Area of a cube = 6(s^2)

    Ok, so ds/dt = 7

    And, dA/ds = 12s

    Therefore, dA/dt = 7(12s) = 84s

    But s = 3, hence dA/dt = 84(3) = 252

    ReplyDelete
  8. phoenixMarch 23, 2009 at 1:18 PM

    Question #7:
    Let A be the area of a circle with radius r at time t. If the radius changes at a rate of -3 ft/sec, at what rate is the circle's area changing when r = 5?

    Answer:
    So, Area of a circle = πr^2

    Which implies, dA/dr = 2πr

    Also, dr/dt = -3

    Therefore, dA/dt = -3(2πr) = -6πr

    But, r =5, so dA/dt = 5(-6π) = -30π

    ReplyDelete
  9. Swanky ThingMarch 24, 2009 at 3:40 PM

    2)Let A be the area of a square with side s at time t. If the side changes at a rate of -4 mm/week, at what rate is the square's area changing when s = 3?

    Area of a square = s squared

    dA/ds = 2s

    ds/dt = -4

    dA/dt = dA/ds * ds/dt

    = (2s) (-4)

    = -8s

    ReplyDelete
  10. etioncenMarch 25, 2009 at 1:50 PM

    so the value of s=3 can be substituted after solving the equation?

    ReplyDelete
  11. etioncenMarch 25, 2009 at 2:33 PM

    #8
    Area=s^2
    the differential of area with resprect to side =
    dA/ds=s^2
    =2s
    the rate of change of the area with respect to time is 10+2 because s is currently 10 and it increases by 2.....this can be represented by
    dA/dt = 10+2.
    the change of s with respect to time will be = 2
    = ds/dt=2
    THEREFOR dA/dt=dA/ds*ds/dt=
    (10+2)m* 2m
    =44m^2

    ReplyDelete
  12. Swanky ThingMarch 26, 2009 at 8:03 PM

    Yes i believe the value for s can be substituted now...

    ReplyDelete
  13. Low riderMarch 30, 2009 at 11:30 AM

    etioncen can you explain a little further from the part where you said 10+2 i don't quit understand how you came to that conclusion.

    ReplyDelete
  14. Swanky ThingApril 3, 2009 at 8:14 AM

    Queston 1

    we are required to find dA/dt.
    since we know that the area,

    A = pie r^2
    dA/dr = 2pie r

    the change in radius
    dr/dt =2 in/sec

    Therefore
    dA/dt = dA/dr * dr/dt
    = (2 pie r) x (2)
    = 4 pie r

    dA/dt when r=1 is
    dA/dt = 4 pie x (1)
    = 4 pie
    = 4 x 3.14
    = 12.57

    ReplyDelete
  15. Swanky ThingApril 3, 2009 at 8:25 AM

    Let A be the area of a square with side s at time t. If the side changes at a rate of 2 m/day, at what rate is the square's area changing when s = 10?

    What we are required to find is dA/dt when s= 10

    A = s^2
    dA/ds = 2s

    It is given that
    ds/dt = 2

    therefore
    dA/dt = dA/ds x ds/dt
    = 2s x 2
    = 4s


    The rate at which the square's area changing when s = 10?

    = 4 x 10
    = 40

    ReplyDelete
  16. Swanky ThingApril 3, 2009 at 8:42 AM

    Question 10
    Let V be the volume of a sphere with radius r at time t. If the radius changes at a rate of -8 in/min, at what rate is the sphere's volume changing when r = 7?

    We are required to find dV/dt (which is the rate of change of the sphere's volume) when r = 7.

    V = 4/3 pie r^3
    Make the equation simplier
    V = 4.19r^3
    dV/dr = 12.57r^2

    It is given that dr/dt = -8


    Therefore
    dV/dt = dV/dr x dr/dt
    = 12.57r^2 x -8
    = -100.56r^2

    When r = 7

    = -100.56 (7)^2
    = -100.56 x 49
    = -4927.44

    Am not sure how the velocity can be a minus. So if theres a problem with my answer, plez correct me..

    ReplyDelete
  17. the best oneApril 12, 2009 at 7:03 PM

    QUESTION1
    A = pi r^2

    dA/dr= 2pi r
    dA/dt=dA/dr * dr/dt
    =2 pi r* 2
    =4 pi r

    dA/dt when r=1 is
    dA/dt=4 pi *(1)
    =4 pi

    ReplyDelete
  18. the best oneApril 12, 2009 at 7:05 PM

    Question2
    A= s^2
    dA/ds= 2s
    ds/dt= -4 mm/week
    dA/dt=dA/ds * ds/dt
    = 2s * -4
    = -8 s
    therefore dA/dt when s = 3

    dA/dt = -8 * 3
    dA/dt = -24

    ReplyDelete
  19. the best oneApril 12, 2009 at 7:07 PM

    Question3
    Vol of sphere = 4/3 pi r^3
    dv/dr = 4 pi r^2
    dr/dt = 3
    dv/dt = dv/dr * dr/dt
    = 4 pi r^2 * 3
    = 12 pi r^2
    at r = 2
    dv/dt = 12 pi (2)^2
    = 48 pi

    ReplyDelete
  20. the best oneApril 12, 2009 at 7:09 PM

    Question5

    V = s^3
    dv/ds = 3s^2
    ds/dt = 10
    dv/dt = dv/ds * ds/dt
    = 3s^2 * 10
    = 30s^2
    at s = 5
    dv/dt = 30(5)^2
    = 750

    ReplyDelete
  21. saeedApril 14, 2009 at 1:05 PM

    1,A=PIR^2
    DA/DY=2PIR=2*2PIR=-4PIR (R=-2)
    @r=1 ,da/dr=-4pi

    2.A=s^2
    da/dy=2s =2*-4s=-8s
    @s=3 da/ds=-8*3=-24mm

    3,v=4/3pir^2
    dv/dy=4pir^2=3*4pi3^2=12pir
    dv/dr@r=2=12*pi*2^2=150.72

    4.SA=4pir^2
    ds/dy=8pir=8*-5*pir=-40pir
    @r=-5dr/dy=-40pi*-5=-628km/hr

    5.v=s^3
    dv/dy=3s^2=10*3s^2=30s^2
    @s=5ds/dt=30*5^2
    30*25=750in/hr

    6.S=6s^2
    ds/dy=12s=12*7s=84s
    @s=5 ds/dy=84*5=420

    7.A=pir^2
    da/dy=2pi=2*-3pir=-6pir
    @r=-3dr/dt=-6pi*5=94.2ft/sec

    8.A=s^2
    da/dy=2s=2*2s=4s
    @s=10 ds/dt=4*10=40m/day

    9.V=4/3pir^3
    dv/dy=4pir^2=-8*4pir^2=-32pir^2
    @r=7 dr/dt=-32pir^@=-4923.52

    ReplyDelete
  22. computationApril 26, 2009 at 7:31 AM

    question 1.

    Let A be the area of a circle with radius r at time t. If the radius changes at a rate of 2 in/sec, at what rate is the circle's area changing when r = 1?

    A = radius = pie r^2
    dA/dr=2pie r

    change in radius=dr/dt=2in/sec
    dA/dt = dA/dr * dr/dt
    = 2 pie x 2
    = 4 pie

    when r=1
    dA/dt= 4 pie x r
    =12.66

    ReplyDelete
  23. computationApril 26, 2009 at 8:19 AM

    qusetion 2

    Let A be the area of a square with side s at time t. If the side changes at a rate of -4 mm/week, at what rate is the square's area changing when s = 3?

    A = s^2
    dA/ds=2s
    ds/dt=-4 /week

    hence dA/dt=dA/ds*ds/dt
    dA/dt=2s*-4
    = -8s

    therefore dA/dt when s=3
    = -8*3

    dA/dt=-24

    ReplyDelete
  24. computationApril 26, 2009 at 8:26 AM

    question 8

    Let A be the area of a square with side s at time t. If the side changes at a rate of 2 m/day, at what rate is the square's area changing when s = 10?

    area=s^2
    dA/ds=s^2=2s

    dA/dt=10+2
    change=ds/dt=2

    therefore dA/dt=dA/ds*ds/dt
    (10+2)m*2m
    =44m^2

    ReplyDelete
  25. computationApril 26, 2009 at 8:30 AM

    This comment has been removed by the author.

    ReplyDelete
  26. RamApril 27, 2009 at 2:01 PM

    computation that is realy off,
    the area diff is correct
    area=s^2
    dA/ds=s^2=2s

    bt then u knw dS/dt=2
    that means
    dA/dt = dA/ds x ds/dt
    = 4s
    Hence wen s=10
    rate of change = 4(10)
    =40

    ReplyDelete

Subscribe to: Post Comments (Atom)