Rate of Change Questions? If h(t) represents the height of an object above ground level at time t and h(t) is given by h ( t ) = − 16 t 2 + 13 t + 1 find the height of the object at the time when the speed is zero.Suppose h ( t ) = t 2 + 14 t + 7. Find the instantaneous rate of change of h(t) with respect to t at t=2 .Suppose G ( x ) = 6 x 2 + x + 4. Find a number b such that G ′ ( b ) = 7Let g ( x ) = 2 x 2 + 4 x + 1. Find a value of c between 1 and 3 such that the average rate of change of g(x) from x=1 to x=3 is equal to the instantaneous rate of g(x) at x=c .Let F ( s ) = 5 s 2 + 3 s + 4. Find a value of d greater than 0 such that the average rate of change of F(s) from 0 to d equals the instantaneous rate of change of F(s) at s=1.Let f ( x ) = x 2 + x + 13. What is the value of x for which the tangent line to the graph of y = f ( x ) is parallel to the x-axis?

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Rate of Change Questions? If h(t) represents the height of an object above ground level at time t and h(t) is given by   h  (  t  )  =  −  16      t    2    +  13  t  +  1 find the height of the object at the time when the speed is zero.Suppose   h  (  t  )  =      t    2    +  14  t  +  7. Find the instantaneous rate of change of h(t) with respect to t at t=2 .Suppose   G  (  x  )  =  6      x    2    +  x  +  4. Find a number b such that       G    ′    (  b  )  =  7Let   g  (  x  )  =  2      x    2    +  4  x  +  1. Find a value of c between 1 and 3 such that the average rate of change of g(x) from x=1 to x=3 is equal to the instantaneous rate of g(x) at x=c .Let   F  (  s  )  =  5      s    2    +  3  s  +  4. Find a value of d greater than 0 such that the average rate of change of F(s) from 0 to d equals the instantaneous rate of change of F(s) at s=1.Let   f  (  x  )  =      x    2    +  x  +  13. What is the value of x for which the tangent line to the graph of   y  =  f  (  x  ) is parallel to the x-axis?

elulamami · Accepted Answer

Before starting any calculations I would first recognize the following:  h  (  t  )  =  −  16      t    2    +  13  t  +  1  v  (  t  )  =      h    ′    (  t  )  =  −  32  t  +  13  =  0We know that t=13/32 when speed is 0.  h  (  13      /    32  )  =  −  16  (  13      /    32      )    2    +  13  (  13      /    32  )  +  1  =  3.6406  h  (  t  )  =      t    2    +  14  t  +  7      h    ′    (  t  )  =  2  t  +  14      h    ′    (  2  )  =  2  (  2  )  +  14  =  18 : instantaneous rate of change of h(t) with respect to t at t=2 .      G    ′    (  b  )  =  12  b  +  1  =  0   →   b  =  −  1      /    12  g  (  x  )  =  2      x    2    +  4  x  +  1  g  (  3  )  =  2  (  3      )    2    +  4  (  3  )  +  1  =  31  g  (  1  )  =  2  (  1      )    2    +  4  (  1  )  +  1  =  7Average rate of change of g(x) from x=1 to x=3 is given by             g      (      3      )      −      g      (      1      )              (      3      −      1      )        =  (  31  −  7  )      /    2  =  12 Average rate of change is 12 Instantaneous rate of change is 4x+4 Instantaneous rate of change at x=c is 4c+44c+4=124c=8c=2Average rate of change of F(s) from 0 to   d  =            F      (      d      )      −      F      (      0      )        d    F  (  s  )  =  5      s    2    +  3  s  +  4  F  (  d  )  =  5      d    2    +  3  d  +  4F(0)=4            F      (      d      )      −      F      (      0      )        d    =            (      5              d        2            +      3      d      +      4      )      −      4        d    =  5  d  +  3Instantaneous rate of change of   F  (  s  ) at   s  =  1.       F    ′    (  s  )  =  10  s  +  3      F    ′    (  1  )  =  135d+3=135d=10d=2Lastly, When the Tangent line to   y  =  f  (  x  ) is:       f    ′    (  x  )  =  2  x  +  1If is is parallel to the x-axis, 2x+1=0 (slope is 0)2x+1=0x=−1/2It is crucial to keep track of variables when solving rate of change problems!

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New Questions in Calculus 2