[related rates][section 12]

(1.)  xy = 6 and dx/dt = 5            given

      Find dy/dt when x = 3.      here is the problem

       x(dy/dt) + y(dx/dt) = 0   take the derivative implicitly

         3y = 6            replace x with 3
         __  ___
         3    3          divide each side by 3

           y = 2           divide and cancel

          3(dy/dt) + 2(5) = 0    make  substitutions

          3(dy/dt) + 10 = 0       multiply


                    -10  -10     subtract 10 from each  side
       _______________________
           3(dy/dt)     = -10      subtract
           _________      ____
              3             3         divide each side by 3

             dy/dt = -10/3                  cancel

(2.)   Given x/y = 2 and dx/dt = 4;   Find dy/dt when

         x = 2.

        2/y = 2             replace x with 2

      2 = 2y             multiply each side by y, cancel
    ___  ____
     2     2              divide each side by 2

        1 = y               divide and cancel



        y(dx/dt) - x(dy/dt)
      ______________________ = 0    take the derivative
               y2                      implicitly

       1(4) - 2(dy/dt)
     ____________________ = 0       make substitutions
            12   

     4 - 2(dy/dt) = 0           multiply and divide

      -4 + 2(dy/dt) = 0        multiply thru by -1

  +    4            +  4       add 4 to each side
   _______________________
           2(dy/dt) = 4          add
           ________  ___
              2       2    divide each side by 2

              dy/dt = 2      divide and cancel

(3.)  A 10-foot ladder is leaning against the side of a

     house.   As the foot of the ladder is pulled away from

     the house, the top of the ladder slides down along the

     side of the house.   Suppose the foot of the ladder is

     pulled away at a rate of 2 ft/sec.   How fast is the top

     of the ladder sliding down when the foot is 8 ft from

     the house?

     [use a 6 - 8 - 10 pythagorean triangle]

      a2 + b2 = c2     use the pythagorean theorem

      2a(da/dt) + 2b(db/dt) = 2c(dc/dt)   take the derivative

                                         implicitly

        a(da/dt) + b(db/dt) = c(dc/dt)  divide thru by 2, cancel

        6(da/dt) + 8(2) = 10(0)    make substitutions

         6(da/dt) + 16 = 0             multiply

              -     16  -16     subtract 16 from each side
    _____________________________
            6(da/dt)  =  -16            subtract
            ________    _____
                6        6         divide each side by 6

               da/dt = -8/3         reduce and cancel

(4.)  Two roads intersect at right angles.  A car travelling

      80 km/hr reaches the intersection half an hour before a

     bus that is travelling on the other road at 60 km/hr.  How

    fast is the distance between the car and the bus increasing

    1 hour after the bus reaches the intersection?

     a2 + b2 = c2      use the pythagorean theorem

     1202 + 602 = c2    replace a with 120 and b with 60

     14,400 + 3600 = c2           square

        18,000 = c2             add like terms

        (3600)(5) = c2       factor like this
                  _
           c = 60
5    take the square root of each side

      2a(da/dt) + 2b(db/dt) = 2c(dc/dt)

 [take the derivative implicitly]

        a(da/dt) + b(db/dt) = c(dc/dt)  divide thru by 2, cancel
                                 _
      (120)(80) + (60)(60) = (60
5)(dc/dt)  make substitutions
                         _
       7200 + 3600 = (60
5)(dc/dt)      multiply
                         _
            10,800 = (60
5)(dc/dt)    add like terms
           ________  _____________
               _          _
            60
5       605          divide each side by this
               _
          180/
5 = dc/dt         reduce and cancel
                  _                                        _
      dc/dt = 180
5/5          multiply top and bottom by 5
                 _
      dc/dt = 36
5                divide
             
(5.)  A sailor standing on the edge of a dock 15 ft above the

     lake surface is pulling in his boat by means of a line

     attached to the boat's bow.  He pulls in the line at

     a rate of 5 ft/min.  How fast is the boat approaching the

     foot of the dock when the boat is 20 ft away?

     Use a 15 - 20 - 25   right triangle.

     a2 + b2 = c2    use the pythagorean theorem

    2a(da/dt) + 2b(db/dt) = 2c(dc/dt)    take the derivative

                                           implicitly

     a(da/dt) + b(db/dt) = c(dc/dt)   divide thru by 2, cancel

     15(0) + 20(db/dt) = 25(5)    make substitutions

         20(db/dt) = 125             multiply add
         _________  _____
            20       20            divide each side by 20

               db/dt = 6.25     divide and cancel

(6.)  An airplane at a height of 1000 m is flying horizontally

      at a velocity of 500 km/hr and passes directly over an

     observer.   How fast is the plane receding from the

     observer when it is 1500 m away, (along a direct line
 
    of sight), from the observer?

      a2 + b2 = c2     use the pythagorean theorem

     10002 + b2 = 15002    replace a with 1000 & c with 1500

      1,000,000 + b2 = 2,250,000    square 1000 and 1500

     -1,000,000      -1,000,000  subtract this fr ea side
   __________________________________
                  b2 = 1,250,000       subtract

                  b2 = (5)(250,000)    factor like this
                            _
                    b = 500
5    take square roots

    a2 + b2 = c2    use the pythagorean theorem again

  2a(da/dt) + 2b(db/dt) = 2c(dc/dt)   take the derivative

                                         implicitly

    a(da/dt) + b(db/dt) = c(dc/dt)   divide thru by 2

    1000(0) + (500
5)(500,000) = 1500(dc/dt)

    [make substitutions]
              _
         (500
5)(500,000) = 1500(dc/dt)   multiply, add
               _
            (5
5)(5000) = 15(dc/dt)   divide thru by 100
           _
    25,000
5 = 15(dc/dt)     multiply
   __________  _________
       15        15          divide each side by 15
            _
   (5000/3)
5 = dc/dt       reduce and cancel

(7.)  Sand is dropped onto a conical pile at a rate of 15 m3/min.

     Suppose the height of the pile is always equal to its

    diameter.  How fast is the height increasing when the

    pile is 7 m high?

     V = (1/3)(pi)r2h          use this formula

     V = (1/3)(pi)(h/2)2h    replace r with h/2

     V = (h3/12)(pi)            multiply

    dV/dt = (h2/4)(pi)(dh/dt)    take the derivative implicitly

    15 = (72/4)(pi)(dh/dt)   replace dV/dt with 15 & h with 7

    15 = (49/4)(pi)(dh/dt)      square 7

    60 = 49(pi)(dh/dt)        multiply each side by 4, cancel
   ___  ______________
49pi    49pi               divide each side by 49pi

    60
 _________ = dh/dt               cancel
   49pi

(8.)  A rock is thrown into a pool of water.  A circular wave

     leaves the point of impact and travels so that its radius

     increases at a rate of 25 cm/sec.  How fast is the

    circumference of the wave increasing when the radius is

     1 m?

     C = 2
r    use the circumference formula

    dC/dt = 2
∏(dr/dt)    take the derivative implicitly

    dC/dt = 2∏(25)
    make substitutions

    dC/dt = 50
             multiply

(9.)  Bacteria grow in circular colonies.  The radius of one

      colony is increasing at the rate of 4 mm/day.  On

      Wednesday, the radius of the colony is 1 mm.  How fast

     is the area of the colony changing one week later?

      A = ∏r2     use the area formula for circle

     dA/dt = 2∏r(dr/dt)   take the derivative implicitly

     dA/dt = 2∏(7)(4)    replace r with 7, replace dr/dt with 4

     dA/dt = 56∏                multiply


(10.)  A spherical mothball is dissolving at a rate of 8pi

       cm3/hr.  How fast is the radius of the mothball de-

      creasing when the radius is 3 cm.

       V = (4/3)
∏r3          use the volume formula

     dV/dt = 4
∏r2(dr/dt)   take the derivative implicitly

      8∏ = 4∏(3)2(dr/dt)   replace dV/dt with 8∏

                             replace r with 3

     8∏ = 36∏(dr/dt)         multiply
    ____  __________
     36∏    36∏             divide each side by 36∏

    2/9 = dr/dt           reduce, cancel