University of Dayton



ENM 503 Differentiation and Applications

1. Find the equations of the tangent and normal to the curve y = x2 – x + 5 at (-1, 7).

y' = 2x – 1 = 2(-1) -1 = -3

Tangent: (y – 7) = -3(x + 1)

Normal: (y – 7) = (x + 1)/3

2. At what points of the curve y = x2 – 5x + 9 is the normal parallel to x + y – 1 = 0?

y' = 2x - 5 = -1 (slope) => x = 2 => y = 3

Normal: (y – 3) = -1(x – 2) or x + y – 5 = 0

3. At what point is the line 12x – y – 20 = 0 tangent to the curve y = 3x2 – 8?

x2 – 8 = 12x – 20 => 3x2 -12x + 12 = 0 or x2 – 4 x + 4 = 0 => x = 2 => y = 4

he slope is 6x = 6(2) = 12

4. Given y = 2x3 – 6x2 + 24x – 8, where is dy/dx increasing and where is it decreasing?

y' = f'(x) = 6x2 – 12x + 24 or 6(x2 – 2x + 4)

y''= f"(x) = 12x – 12 = 12(x – 1)

Observe y" < 0 if x < 1 and y" is > 0 for x > 1

5. Find two numbers whose sum is 8 and whose product is a maximum.

Given numbers x and y such that x + y = 8 and f(x, y) = xy and maximize f.

f(x) = x(8 – x) = 8x – x2

f'(x) = 8 – 2x = 0 when x = 4 and f"(x) < 0 => relative maximum

The two numbers are x = 4 and y = 4.

6. Find two numbers x and y whose sum is 16 and xy3 is a maximum.

x + y = 16 => x = 16 - y

f(y) = (16 – y)y3 = 16y3 – y4

f'(y) = 48y2 – 4y3 = 0 when y = 0, x = 16; or y = 12, x = 4

f"(y) = 96y – 12y2 = 0 when at pts of inflection at (0, 0) and (8, 0)

f"(0) = 0 Use first derivative test

f"(12) < 0 => rel max

-x4 + 48x3 -768x2 + 4096x

f'(x) = -4x3 + 144x2 – 1536x + 4096 = 0 when x = 4, 16, 16

(4, 12) is the maximum and (0, 16) is maximum and (16, 0) is minimum.

7. A wire of length L is cut into two parts one of which is bent into a square and the other an equilateral triangle. Find the sides of the square and triangle to minimize the area.

x y x/4 y/3

x + y = L

f(x, y) = x2 /16 + y2/36

f(x) = x2 /16 + (L – x)2/36

f'(x) = x/8 - (L – x)/18 = 0 => 4(L – x) = 9x => 13x = 4L or x = 4L/13, y = 9L/13

f"(x) = 1/8 + 18 > 0 => relative minimum.

8. Find extrema of y = Ln x /x

y' = [x * 1/x – Ln x * 1]/x2 = 0 when Ln x = 1 => x = e => y = Ln e / e = 1/e. (e, e-1)

Relative max or min? Max. Why?

9. Find y' for y = Ln (2 – 3x). y' = -3/(2 – 3x)

10. Find y for y = [pic]

y = [pic]= (3/5)[Ln (1 – x2) – (4/5) Ln (1 + x2)

y' = (3/5)(-2x)/(1 – 2x2) – (4/5)(2x/(1 + x2)

11. y = (Ln x)2.

Find y'. y' = 2(Ln x) /x = (2/x) Ln x = Ln X2/x = 0 when x2/x = 1 or x = 1, y = 0

y " > 0 => relative minimum.

12. Logarithmic differentiation: Find dy/dx for y = xx.

y = xx; Ln y = x Ln x ; dy/y = (1 + Ln x)dx => dy/dx = y(1 + Ln x) = xx(1 + Ln x)

13. Find dy/dx for y = x(1 – x)1/2 Ans. (2 – 3x)/2(1 – x)1/2

14. Determine the constants a and b so that the function f(x) = x3 + ax2 + bx + c has a local minimum at x = 3 and a local maximum at x = -1. Ans. a = -3, b = -9

15. You bought 2 items. You sold one for $100 making a 25% profit and the other for $100 taking a 25% loss. How did you do overall?

1.25x = 100 => x = $80; 0.75y = 100 => y = $133.33

Thus you invested $80 + $133.33 = $213.33 and got a return of $200 implying

(200 – 213.33)/200 = -6.67% loss

16. Find the area bounded by the curve y = [pic]x3 - 6x2 + 9x + 1 and the x-axis between the curve's extrema.

f(x) = x3 - 6x2 + 9x + 1

f'(x) = 3x2 – 12x + 9 = 0 when x2 – 4x + 3 = 0 = (x – 1) (x – 3) or x = 1, x = 3

f"(x) = 6x – 12 and f"(1) = -6 => (1, 5) is maximum; f"(3) = 6 => (3, 1) is minimum

Integrate the curve from a x = 1 to x = 3.

[pic]= 6.

17. Write the equation of the tangent to the curve y = x3 - 6x2 + 9x + 1 where the curve’s slope is a minimum.

Slope function is y' = 3x2 – 12x + 9 which we seek a minimum.

y" = 6x – 12 = 0 at x = 2 and the slope is -3

y'" = 6 > 0 => that x = 2; y = 3 is the minimum

(y – 3) = -3(x – 2)

18. With L units of string to make a square and a circle, where should the string be cut to ensure maximum area? minimum area?

x/4 L – x = ( D

f(x) = [pic] 0 [pic] x [pic] L

f ‘(x) = [pic]

[pic]

f(0) = [pic] => all for the circle; absolute maximum

f(L) = [pic] => all for the square

f ‘(x) = [pic]

f([pic]) = [pic]; absolute minimum

19. An apple orchard now has 30 trees per acre, and the average yield is 400 apples per tree. For each additional tree planted per acre, the average yield per tree is reduced by 10 apples. How many trees per acre will yield the largest crop of apples?

Let x = number of additional trees planted [0, 40]

f(x) = (400 – 10x)(30 + x) = 1200 + 400x – 10x2 on [0, 40].

f(0) = 12000; f(40) = 0

f ‘(x) = 400 – 20x = 0 when x = 5 trees

f ‘’(x) = -20 => relative maximum

f(5) = 350 * 35 = 12,250 apples

20. A person standing on a bridge throws a stone upward. Exactly 3 seconds later the stone passes the person on the way down and 2 seconds later lands in the water below. Find the initial velocity of the stone and the height of the bridge above the water.

a(t) = -g => v(t) = - gt + v0 => s(t) = -gt2/2 + v0t + s0

v(1.5) = 0 => -1.5g + v0 = 0 => v0 = + 1.5g

v(3) = -v0 = -3g + v0 => v0 = 1.5g

s(5) = 0 = -12.5g + 7.5g + s0 => s0 = 5g.

Note that max s occurs when v = 0 (i.e., changing directions from + to -)

21. Solve for x: [pic]

-3 < [pic]

Case 1: x – 6 > 0 => x > 6 => -3x + 18 < 2x – 5 < 3x – 18

23 < 5x or x > 23/5 and x > 13 and x > 6 => x > 13

Case 2: x – 6 < 0 => x < 6 => -3x + 18 > 2x - 5 > 3x – 18

=> 5x < 23 and x < 6 and x < 13 => x < 23/5

Answer (- (, 23/5) ( (13, ()

22. Solve the equation e2x - 7ex + 10 = 0.

Let y = ex, Then y2 – 7 y + 10 = 0 => (y – 2)(y – 5) = 0 => y = 2 and y = 5

ex = 2 => x = Ln 2 and x = Ln 5.

24. Current I = E/(R + r) where E and r are constants. Power P = I2R. Show that P is a maximum when R = r.

dI/dR = -E/(R + r)2; dP/dR = I2*1 + R*2I * dI/dR = 0

= I2 + 2RI * (-E/(R + r)2 = 0

=> I2 = 2ERI/(R + r)2

1 = 2R/(R + r)

R + r = 2R

R = r

25. An object has equation of motion s = t2 – t. Find the position, velocity, and acceleration at t = 1.

s = t3 – t => s(1) = 0

v = 3t2 – 1 => v(1) = 2

a = 6t => a(1) = 6

26. An arrow is shot straight up from ground level with an initial velocity of 128 ft/sec. How high will it go and when will it hit the ground? Neglect air resistance and assume that g = 32 ft/sec2.

a(t) = -32

v(t) = -32t + v0 where v0 = 128

s(t) = -16t2 +128t + s0 where s0 = 0

s'(t) = -32t + 128 = 0 when t = 4 sec => max height at t = 4

s(4) = -16(42) + 128(4) = 256 ft maximum height and the arrow reaches the ground when

s = 0 => -16t2 + 128t = 0 when t(-16t + 128) = 0 or when t = 8 seconds; thus t = 8 secs.

27. An arrow is shot straight up from the ground level with velocity v0. What is the velocity when it hits the ground?

A = -32 => v = -32t + v0 => s = -16t2 + v0t + s0

s = 0 when -16t2 + v0t = 0 => t = v0/16

v(v0/16) = -32(v02 /256) + v0 * v0/16 = -v0 ft/sec => don't shoot arrows straight up.

28. A car accelerates from 0 to 60 mph (88ft/sec) in 30 seconds. Find the assumed constant acceleration. Average velocity is 44 ft per second in 30 seconds or a = 44/30.

Assuming a = kt. Find k and determine the speed after one minute.

A = 44/225 t v = kt2/2 = 88 at t = 30 seconds => k = 176/900 = 44/225

V = 44/225(602)/2 = 352 ft/sec or (240 mph)

29. Related Rates A man 6 ft tall is walking away from a 15 foot lamppost at 6ft/sec.

At what rate is the end of his shadow moving away from the lamppost?

dx/dt = 6 ft/sec and dy/dt = ?

15 (y – x)/6 = y/15 => 3y = 5x

6 3 dy/dt = 5 dx/dt = 30 => dy/dt = 10 ft/sec.

x y - x

31. Express in sigma notation 1 + 2 + 3 + … + 10

10 + 17 + 26 + … + (n2 + 1)

1 + 3 + 5 + … + (2n – 1)

32. Find the area under the curve y = x2 from x = 0 to x = 1.

The region bounded by y = x2 and y = 0 and between x = 1 and x = 2.

33. Evaluate the integrals: [pic]

34. A man has three sons. He says to a woman that the product of their ages is 36 and that the sum of their ages is on the building across the street. The woman asks for another clue and the man replies that the oldest has red hair. What are the ages of the three sons?

35. Test Z = x2 + y2 for relative maxima and minima.

Zx = 2x; Zy = 2y => critical point is (0, 0)

Zxx = 2; Zyy = 2; D = 2*2 - 02 = 4 > 0 => (0, 0) is relative minimum.

37. Test Z = x2 – y2 for relative maxima and minima.

Zx = 2x; Zy = -2y; critical pointa t (0, 0)

D = -4 < 0 => saddle point at (0, 0)

38. Test Z = x3 + y3 -3xy for relative maxima and minima.

Zx = 3x2 - 3y; Zy = 3y2 – 3x

y = x2; x = y2 => x = x4 or x(x3 – 1) = 0; x = 0, y= 0; x = 1, y = 1

Zxx = 6x; Zyy = 6y

Zxy = -3 Zyx = -3

D(0, 0) = -9 < 0 => a saddle point

D(1, 1) = 6 * 6 – 9 = 27 with Zxx and Zyy > 0 => relative minimum.

39. Find 3 positive numbers x, y, and z such that x + y + z = 12 and x2yz is a maximum.

W = x2yz = x2y(12 – x – y) = 12x2y – x3y – x2y2

Wx = 24xy – 3x2y – 2xy2 = xy(24 – 3x – 2y)

Wy = 12x2 – x3 -2x2y = x2(12 – x – 2y)

Clearly x = 0 or y = 0 will not yield a maximum.

(solve '((3 2 24)(1 2 12))) ( (x = 6, y = 3) => z = 3 => (6, 3, 3)

Did you catch the equal power of y and z?

40. Constrained Extrema; Lagrange Multipliers

Find the minimum value of W = x2 + y2 + z2 subject to x + y + z = 1,

L = x2 + y2 + z2 + (( x + y + z – 1)

Lx = 2x + (; Ly = 2y + (; Lz = 2z + (; L( = x + y + z – 1; all set to zero imply

x = y = z = 1/3 = W

41. Find max of W = xy + z subject to x2 + y2 + z2 = 1

L = xy + z + ((x2 + y2 + z2 - 1)

Lx = y + 2(x; Ly = x + 2(y; Lz = 1 + 2(z; L( = x2 + y2 + z2 – 1; all set to zero imply

y2z + 2(xyz = 0; x2z + 2(yz = 0; xy + 2(xyz = 0;

y2z = x2z = xy From symmetry, x = y and x = y = 0 => z = ( 1

Max is (0, 0, 1) or W = 1

42. Find 2 numbers whose sum is 11 and whose product is a maximum. 5½

f(x) = x (11 – x) => f'(x) = 11 -2x = 0 when x = y = 11/2

43. Find the slope of the curve x3 = (y – x2)2 at (1, 2). Use implicit differentiation.

3x2 = 2(y – x2)1(y' – 2x); now solve for y'; y' = 3x2/2(y – x2) + 2x and evaluate at (1, 2).

y' = 3/2 + 2 = 7/2.

44. Determine where y = f(x) = (x – 1)3 + 1 is concave up and where concave down.

Seek y"; y ' = 3(x – 1)2; y" = 6(x – 1); when positive concave up; when negative concave down.

y" > 0 or positive when x > 1 and y" < 0 when x < 1. The point (1, 1) is a point of inflection.

45. Determine concavity for y = x2.

y" = 2 which is always positive and thus always concave up.

46 Show that the line 6x - y - 9 = 0 is tangent to y = x2.

6x – x2 – 9 = 0 => (x – 3)2 = 0 and y' = 2x = 6 at x = 3

At the point (3, 9) the slope is 6 => (y – 9) = 6 (x – 3)

or y = 6x - 9

47. Integrate [pic]

To solve for A, cove r up the x-4 factor in the denominator and substitute 4 for x in the other terms to get 7(4 – 16)/2*7= -6 for A. Repeat for B to get

7(4 – 4)/-2*5 = 0, and for C = 7(4 – 9)/-7*-5 = 1

Thus the integral is -6 Ln (x – 4) + Ln (x + 3). Note that the numerator can be factored into 7(2 – x)(2 + x) and the (x – 2) terms may be canceled.

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