Submitted by: Submitted by kichi90
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Words: 1100
Pages: 5
Category: Business and Industry
Date Submitted: 07/29/2012 09:39 PM
Q1. Derive the equation of a straight line that passes through (-2, 3) and (1, -5).
Answer:
(x₁, y₁) = (−2, 3) and (x₂, y₂) = (1, -5), so
a₁ = y₂ - y₁x₂ - x₁
= -5-31-(-2)
= - 83
a₀ = y₁ − a₁ x₁
= 3 – ( - 83 ) (-2)
= 3 – 163
= - 73
y = - 73 - 83 x
3y = -7 – 8x or 3y = -8x – 7
Q2. Where are the axis of symmetry, x-intercept and y-intercept for y = 4x² + 2x – 1.
Answer:
y = 4x² + 2x – 1
a₀ = -1
a₁ = 2
a₂ = 4, so for the axis of symmetry:
x = -a₁2a₂
= -22(4)
= - 28
d = a₁² - 4 a₂ a₀
= (2)² - 4 (4) (-1)
= 20, there are 2 intercepts.
The x-intercepts are the solution to 0 = 4x² + 2x – 1, which can be solved by the quadratic equation:
x = -a₁ ± a₁² - 4 a₂ a₀2a₂
= -2 ± 2² - 4 (4) (-1)2(4)
= -2 ± 208
= - 0.809 or 0.309
The y-intercept are the solution to y = a₂ x² + a₁ x + a₀,
x = 0
y = a₀
y = 2(0)² - 3(0) + 1
= 1
y-intercept is 1.
Q3. Find the first and second derivatives for
a. y = 3x + 1.
b. y = log 3.
c. y = log x² - x² + 3 exp (4x²).
Answer:
a. y = 3x + 1
yʹ = 3
yʺ = 0
b. y = log 3
yʹ = 0
yʺ = 0
c. y = log x² - x² + 3 exp (4x²)
yʹ = 2xx² - 2x + 24x exp 4x²
= 2x - 2x + 24x exp 4x²
yʺ = - 2x-2 - 2 + 192x² exp 4x²
Q4. Find the equation for the tangent line at x = 1 for y = 1/x.
Answer:
In this case, f(x) = 1/x, x₀ = 1 and y₀ =1/ x₀ = 1. Moreover, f ʹ(x) = - 1x² which equal to -1 when x = 1.
Moreover, the tangent must pass through the point of interest which is (1, 1). So, the tangent line is
y - y₀ = f ʹ(x₀)(x - x₀)
y – 1 = - 1 (x – 1)
y = - x + 1 + 1
y = - x + 2
Q5. Find all the first and second order partial derivatives for
y = Ax1α x2β, A > 0, 0 < α + β ≤ 1.
Demonstrate that the Young’s theorem holds in this case. Using total differential, show how much would y change if x₁ changes from 1 to 1.01 and x₂ changes from 1 to 0.99 simultaneously?...