qr code generator visual basic 2010 FIGURE 316 in .NET framework

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FIGURE 316
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3 A P P L I C A T I O NS O F T H E D E R I V A T I V E
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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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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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34 Falling Bodies
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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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FIGURE 317
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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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EXAMPLE
Suppose that a falling body hits the ground with velocity 100 ft/sec What was the initial height of the body
3 A P P L I C A T I O NS O F T H E D E R I V A T I V E
SOLUTION
With notation as developed above, we know that velocity is given by h ( t) = 32t + 0 We have taken v0 to be 0 because the body is a falling body; it had no initial push If T is the time at which the body hits the ground, then we know that 100 = h ( T ) = 32 T As a result, T = 25/8 sec When the body hits the ground, its height is 0 Thus we know that 0 = h( T ) = h( 25/8) = 16 ( 25/8) 2 + h0 We may solve for h0 to obtain h0 = 625 4
Thus all the information about our falling body is given by h( t) = 16t2 + At time t = 0 we have h( 0) = 625 4 625 4
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