Last years midterm
Q1B:
find derivative with respect to x
y = (4x+3)^1/2 / 2x
u = (4x+3)
u' = 4
T'B-TB'/B^2
1/2(4x+3)-^1/2 * 4
2(4x+3)^-1/2
(2(4x+3)-^1/2 * 2x - (4x+3)^1/2 * 2) / (4x^2)
=
(4x(4x+3)^-1/2 - 2(4x+3)^1/2) / (4x^2)
(however, this derivative is in respect to y. Can someone still tell me if the derivative is correct at least, thanks)
Q2:
Find equation of line normal to
y = 2x^3 - 3x - 1 at x = 1
y - y1 = m(x - x1)
y' = m
y' = 6x^2 - 3
y' = 6(1)^2 - 3 = 36 - 3 = 33
y = 2(1)^3 - 3(1) - 1 = 8 - 3 - 1 = 4
x = 1, y = 4
y - 4 = 33(x - 1)
y - 4 = 33x - 33
y - 37 = 33x
Q4:
A rectangular area divided into two areas, wall inside parallel to 2 opposite sides
Find max dimensions for display if total fencing available is 240 m.
(sketch: 4 sides with one partitioning wall)
2x + 3y
240 = 2x + 3y
240 - 2x = 3y
80 - 2/3x = y
A = x * y
A = x * (80 - 2/3x)
A = 80x - 2/3x^2
A' = 80 - 4/3x
x = 60 m
y = 40 m
Q6:
Evaluate integrals:
A:
(x3 - x^1/2) / x
(x^3)/x - (x^1/2)/x
x^2 - x-^1/2
(1/3x^3) - (2x^1/2)
B:
8x(2x^2-5)^4
(2x^2-5)(2x^2-5)(2x^2-5)(2x^2-5)
4x^4 - 10x^2 - 10x^2 + 25
4x^4 - 20x^2 + 25
integrate
16x^8 - 80x^6 + 100x^4 - 80x^2 + 400x^4 + 500 + 100x^4 - 500x^2 + 625
16x^8 - 80x^6 + 500x^4 - 580x^2 + 1125
simplify
16/9x^9 - 80/7x^7 + 100x^5 + 580/3x^3 + 1125x
Q7:
Find area under curve
y = 1/(x + 1)^1/2
Between x = 3 and x = 8
(x + 1)^-1/2
2(x+1)^1/2
x = 3: 2(3+1)^1/2: y = 4
x = 8: 2(8+1)^1/2: y = 6
4 - 6 = -2
Thanks for looking at it.
