1)Suppose that we have a right circular cylinder with a constant h=4 feet and th

Question

1)Suppose that we have a right circular cylinder with a constant h=4 feet and the radius is increasing at a constant rate. At what rate is the volume of this cylinder changing when the radius is 10 feet?

2)Suppose that the side length of a cube is changing. At what rate is the surface area changing when one side is 5cm?

3)Suppose that the side length of a cube is changing. At what rate is the volume changing when one side is 5cm?

4)An obect moves on a straight line so that after t seconds its distance in millimeters from its original position is given by x(t)=t^2 -2t.  a)its velocity at x=3 is? b)Find the acceleration at t=3? c)at time, t=3, the object is moving to the right?true or false.  d)at time t=3, the object is slowing down?true or false.  e)at what time does the object change direction?

Explanation / Answer

1)Volume of right circular cylinder = pi*r^2*h / 2

    d(V)/d(r) = 2 * pi * r * h / 2

                    = pi * r * h

                    = pi * 10 * 4 = 40 * 3.14 = 12.560 cubic feet per cm

2)Surface area of cube = 6*side^2

   d(s)/d(side) = 12 * side = 12 * 5 = 60 square cm

3) Volume of cube = side^3

     d(v)/d(side) = 3* side^2 = 3 * 5 * 5 =75 cubic cm per cm

4) at x =3

    3 = t^2 - 2 t

    t^2 - 2t -3 = 0

   t = 3 secs since time can not be negative

    x'(t) = 2*t - 2

    v(t) =2*t -2

       at t = 3

    v(t) = 2*3-2

    a) velocity = 4 mm/sec

   b)a(t) = 2 - 0 = 2 constant all all t

   c) at t=3 x = 3 yes therefore object is moving right

d)at t=3 object is not slowing down it is always accelerating at 2mm/sec^2

e) change of direction occurs at x=0

    i.e t^2 - 2*t =0

         t = 2 secs

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