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We know that volume and radius of a sphere are related by the formula V = 4 3 r 3 ( )

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The free variable in this problem is time, so we differentiate equation ( ) with respect to time t It is important that we keep the Chain Rule in mind as we do so1 The result is dV 4 dr = 3r 2 dt 3 dt ( )

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Now we are given that dV /dt = 20 Our question is posed at the moment that V = 64 But, according to ( ) , this means that r = 3 48/ Substituting these values into equation ( ) yields 20 = Solving for dr/dt yields dr 5 = 2/3 dt 48 1/3 Thus the radius is increasing at the specified rate 4 3 3

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A 13-foot ladder leans against a wall (Figure 315) The foot of the ladder begins to slide away from the wall at the rate of 1 foot per minute When the foot is 5 feet from the wall, at what rate is the top of the ladder falling

The h / d t rate ( 2 ft/s ) is given in the problem as is the height ( 120ft ) and the distance out (250 ft ) The x value for these conditions can be calculated from the Pythagorean theorem:

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Let h( t) be the height of the ladder at time t and b( t) be the distance of the base of the ladder to the wall at time t Then the Pythagorean theorem tells us that h( t) 2 + b( t) 2 = 132 We may differentiate both sides of this equation with respect to the variable t (which is time in minutes) to obtain 2 h( t) h ( t) + 2 b( t) b ( t) = 0 Solving for h ( t) yields h ( t) = b( t) b ( t) h( t)

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At the instant the problem is posed, b( t) = 5, h( t) = 12 (by the Pythagorean theorem), and b ( t) = 1 Substituting these values into the equation yields h ( t) = 5 5 1 = ft/min 12 12

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x 2 =s2-1202 or x=J2502 -1202 =219

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It begins to expand very slowly so that its diagonal is increasing at the rate of 1 millimeter per minute At the moment that the diagonal is 100 millimeters, at what rate is the area increasing

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