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Solutions to Applications of Derivatives Practice Problems for AP Calculus

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By — McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these solutions can be found at: Applications of Derivatives Practice Problems for AP Calculus

Part A The use of a calculator is not allowed.

  1. Volume: V =
  2. Surface Area: S=

  3. Pythagorean Theorem yields
  4. x2 + y2 =(13)2

    At x =5, (5)2 + y2 =132 y = ±12,

    since y > 0, y =12.

    Therefore,

  5. See Figure 8.6-1. Using similar triangles, with y the length of the shadow you have:
  6. Differentiate:

     Solutions To Practice Problems

  7. See Figure 8.6-2. Volume of a cone
  8. Using similar triangles, you have

    reducing the equation to

     Solutions To Practice Problems

  9. See Figure 8.6-3.
  10. Step 1. Using the Pythagorean Theorem, you have x2 + y2 =z2. You also have

     Solutions To Practice Problems

    Step 2. Differentiate:

     

    At x =120, both cars have traveled 3 hours and thus, y =3(30)=90. By the Pythagorean Theorem, (120)2 +(90)2 =z2 z =150.

    Step 3. Substitute all known values into the equation:

    Step 4. The distance between the two cars is increasing at 50 mph at x =120.

  11. See Figure 8.6-4.
  12.  Solutions To Practice Problems

    Step 1. Applying the Pythagorean Theorem, you have

    Step 2.

    Step 3.

    Step 4. First Derivative Test:

    Thus at x =6, the area A is a relative maximum.

    Step 5. Check endpoints. The domain of A is [9/2, 9]A(9/2)=0; and A(9)=0. Therefore, the maximum area of an isosceles triangle with the perimeter of 18 cm is cm2. (Note that at x =6, the triangle is an equilateral triangle.)

  13. See Figure 8.6-5.
  14. Step 1. Let x be the number and be its reciprocal.

    Step 2. s = with 0 < x < 2.

    Step 3.

     Solutions To Practice Problems

    Step 4.

    Step 5. Thus at x =1, s is a relative minimum. Since it is the only relative extremum, thus, at x =1, it is the absolute minimum.

  15. See Figure 8.6-5.
  16. Step 1. Volume: V =x (8–2x )(15–2x) with 0 ≤ x ≤ 4.

    Step 2. Differentiate: Rewrite as V =4x3 – 46x2 +120x

    Step 3. Set V =0 12x2 – 92x + 120=0 3x2 – 23x +30=0. Using the quadratic formula, you have defined for all real numbers.

    Step 4. Second Derivative Test:

    Step 5. Check endpoints. At x =0, V =0 and at x =4, V =0. Therefore, at x = V is the absolute maximum.

  17. See Figure 8.6-6.
  18.  Solutions To Practice Problems

    Step 1. Distance Formula:

    Step 2. Let S = Z2, since S and Z have the same maximums and minimums.

    Step 3.

    Step 4. Second Derivative Test:

    Thus at x =1, Z has a minimum, and since it is the only relative extremum, it is the absolute minimum.

    Step 5. At x =1,

  19. Step 1. Average Cost:
  20. Step 2.

    Step 3. at x =0 which is not in the domain.

    Step 4. Second Derivative Test:

    Thus at x =2, the average cost is a minimum.

  21. See Figure 8.6-7.
  22.  Solutions To Practice Problems

    Step 1. Area:

      A =x(200 – 2x )=200x – 2x2 with 0 ≤ x ≤ 100

    Step 2. A '(x )=200 – 4x

    Step 3. Set A '(x )=0 200 – 4x =0; x =50.

    Step 4. Second Derivative Test:

      A '(x )= – 4; thus at x =50, the area is a relative maximum.
      A(50)=5000 m2.

    Step 5. Check endpoints.

      A(0)=0 and A(100)=0; therefore at x =50, the area is the absolute maximum and 5000 m2 is the maximum area.

    Part B Calculators are allowed.

  23. Step 1. Let h be the height of the trough and 4 be a side of one of the two equilateral triangles. Thus, in a 30–60 right triangle,
  24. Step 2. Volume:

    Step 3. Differentiate with respect to t.

    Step 4. Substitute known values:

  25. See Figure 8.6-8.
  26.  Solutions To Practice Problems

    Step 1. tan θ = S/3000

    Step 2. Differentiate with respect to t.

    Step 3. At t =5; S =100(5)2 =2500; Thus, Z2 =(3000)2 +(2500)2 = 15,250,000. Therefore, Z = ±500 , since Z > 0, Z =500 . Substitute known values into the question:

    =0.197 radian/sec. The angle of elevation is changing at 0.197 radian/sec, 5 seconds after lift off.

  27. See Figure 8.6-9.
  28.  Solutions To Practice Problems

  29. The water level in the cylinder is rising at

  30. Step 1. Let x be the distance of the foot of the ladder from the higher wall. Let y be the height of the point where the ladder touches the higher wall. The slope of the ladder is
  31. Step 2. Phythagorean Theorem:

    Step 3. Enter The graph of y1 is continuous on the interval x > 8. Use the [Minimum] function of the calculator and obtain x =14.604; y =19.731. Thus the minimum value of l is 19.731 or the shortest ladder is approximately 19.731 feet.

  32. Step 1.
  33. Step 2. Enter:

    Step 3. Use the [Minimum] function in the calculator and obtain x =790.6.

    Step 4. Verify the result with First Derivative Test. Enter y2= d(2500/x +.02+004x, x); Use the [Zero] function and

    Apply the First Derivative Test:

    Thus the minimum average cost per unit occurs at x =790.6. (The graph of the average cost function is shown in Figure 8.6-10.)

     Solutions To Practice Problems

  34. See Figure 8.6-11.
  35.  Solutions To Practice Problems

    Step 1. Area A =(2x )(2y ); 0 ≤ x ≤ 5 and 0 ≤ y ≤ 2.

    Step 2. 4x2 +25y2 =100; 25y2 =100 – 4x2.

    Since y≥ 0

    Step 3.

    Step 4.

    Step 5. Verify the result with the First Derivative Test. Enter

    Use the [Zero] function and obtain x =3.536. Note that:

    The function f has only one relative extremum. Thus it is the absolute extremum. Therefore, at x =3.536, the area is 20 and the area is the absolute maxima.

  36. See Figure 8.6-12.
  37.  Solutions To Practice Problems

    Step 1. Distance formula:

      l2 =x2 + y2; x > 0.5 and y > 4

    Step 2. The slope of the hypotenuse:

    Step 3.

    Step 4. Enter y 1= and use the [Minimum] function of the calculator and obtain x =2.5.

    Step 5. Apply the First Derivative Test. Enter y2=d(y1(x ), x ) and use the [Zero] function and obtain x =2.5. Note that:

    Since f has only one relative extremum, it is the absolute extrenum.

    Step 6. Thus at x =2.5, the length of the hypotenuse is the shortest. At The vertices of the triangle are (0, 0), (2.5, 0) and (0, 5).

  1. Rewrite: y = [sin(cos(6x – 1))]2
  2. Thus, =2 [sin (cos (6x – 1))]

            × [cos(cos(6x – 1))]

            × [–sin(6x – 1)] (6)

          = – 12 sin(6x – 1)

            × [sin(cos(6x – 1))]

            × [cos(cos(6x – 1))].

  3. As x → ∞, the numerator approaches 0 and the denominator increases without bound (i.e., ∞).
  4. Thus, the

  5.  
    1. Summarize the information of f ' on a number line.
    2. Solutions to Cumulative Review Problems

      Since f has only one relative extremum, it is the absolute extremum. Thus, at x = 3, it is an absolute minimum.

    3. The function f is decreasing on the interval (–∞, 3) and increasing on (3, ∞).
    4. Solutions to Cumulative Review Problems
    5. No change of concavity Solutions to Cumulative Review Problems No point of inflection.

    6. The function f is concave upward for the entire domain (–∞,∞).
    7. Possible sketch of the graph for f (x). See Figure 8.7-1.
    8. Solutions to Cumulative Review Problems

  6. (Calculator) See Figure 8.7-2.
  7. Solutions to Cumulative Review Problems

    Step 1.   Differentiate:

    Step 2.   Set

    Step 3.   Solve for

    Step 4.   Thus,

  8. (Calculator) See Figure 8.7-3.
  9. Step 1. Distance formula:

    Solutions to Cumulative Review Problems

    Step 2. Enter:

    Use the [Minimum] function of the calculator and obtain x = .65052 and y1 = .44488. Verify the result with the First Deriative Test. Enter y2 = d(y1 (x), x ) and use the [Zero] function and obtain x = .65052.

    Solutions to Cumulative Review Problems

    Thus the shortest distance is approximately 0.445.

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