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A sponge is in the shape of a right circular cone (Figure 316) As it soaks up water, it grows in size At a certain moment, the height equals 6 inches, and is increasing at the rate of 03 inches per second At that same moment, the radius is 4 inches, and is increasing at the rate of 02 inches per second How is the volume changing at that time

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We know that the volume V of a right circular cone is related to the height h and the radius r by the formula 1 2 r h 3

Now the numbers can be put into the formula for dx/dt :

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Differentiating both sides with respect to the variable t (for time in seconds) yields 1 dr dh dV = 2r h + r 2 dt 3 dt dt

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Substituting the values for r, dr/dt, h, and dh/dt into the right-hand side yields 1 24 1 dV = 2 4 ( 02) 6 + 42 ( 03) = [96 + 48] = dt 3 3 5

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When the kite is 250 ft away from the girl, at a height of 120ft , and the string is going out at 2 ft/s ,the kite is moving 23 fi/s horizontally

equilateral triangle is expanding so that its side length increases by 1 millimeter per hour When the side length is 100 millimeters, how is the area increasing

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It is known that, near the surface of the earth, a body falls with acceleration 2 (due to gravity) of about 32 ft/sec If we let h(t) be the height of the object at time t (measured in seconds), then our information is that h (t) = 32 Observe the minus sign to indicate that height is decreasing Now we will do some organized guessing What could h be It is some function whose derivative is the constant 32 Our experience indicates that polynomials decrease in degree when we differentiate them That is, the degree goes from 5 to 4, or from 3 to 2 Since, h is a polynomial of degree 0, we therefore determine that h will be a polynomial of degree 1 A moment s thought then suggests that h (t) = 32t This works! If h (t) = 32t then h (t) = [h (t)] = 32 In fact we can do a bit better Since constants differentiate to zero, our best guess of what the velocity should be is h (t) = 32t + v0 , where v0 is an undetermined constant Now let us guess what form h(t) should have We can learn from our experience in the last paragraph The antiderivative of 32t (a polynomial of degree 1) should be a polynomial of degree 2 After a little fiddling, we guess 16t 2 And this works The antiderivative of v0 (a polynomial of degree 0) should be a polynomial of degree 1 After a little fiddling, we guess v0 t And this works Taking all this information together, we find that the antiderivative of h (t) = 32t + v0 is h(t) = 16t 2 + v0 t + h0 ( )

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Notice that we have once again thrown in an additive constant h0 This does no harm: h (t) = [ 16t 2 ] + [v0 t] + [h0 ] = 32t + v0 , just as we wish And, to repeat what we have already confirmed, h (t) = [h (t)] = [ 32t] + [v0 ] = 32 We now have a general formula (namely ( )) for the position of a falling body at time t (Recall that we were first introduced to this formula in Section 26) See Figure 317 Before doing some examples, we observe that a falling body will have initial velocity 0 Thus 0 = h (0) = 32 0 + v0 Hence, for a falling body, v0 = 0 In some problems we may give the body an initial push, and then v0 will not be zero

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These first two problems have utilized the Pythagorean theorem as their defining, or "getting started" equation Related rate problems use a variety of defining statements to

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Suppose that a falling body hits the ground with velocity 100 ft/sec What was the initial height of the body

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( 7, Geometry of Planar Cam Pro les)

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