Friday, August 28, 2009

Wording Clothfunny Baby




Task Unit 1: Functions VV

* In what follows, n is the number of their group.
* Let M be the remainder after dividing n by 2. Exercise your group is assigned to the M +1.

1. Given the role

h(x,y)=10x^2 y-nx^2-(n-1) y^2-x^4-2y^4

and the point P = ( a, b ) of your domain a = n / 3 and b = (-1) n n / 4. T is the trace of the graph z = h ( x, y ) with the plane x = a (if n is even) or the plane y = b (if n is odd .) Determine an equation of the tangent to the trace above. Illustrate graphically.

Solution:

Given the role


h (x, y) = 10x ^ 2 and-nx ^ 2 - (n-1) y ^ 2-x ^ 4-2y ^ 4

with n = 5 , replacing n we have:

h (x, y) = 10x ^ 2-5x ^ 2-4y ^ 2-x ^ 4-2y ^ 4

Point = (a, b)

point = (5 / 3, -5 / 4)

now evaluate our function h (x, y) in point (a, b) , ie point (5 / 3, -5 / 4)

h (5 / 3, -5 / 4) = 10 × (5 / 3) ^ 2 × (-5 / 4) -5 × (5 / 3) ^ 2-4 × (-5 / 4) ^ 2 - (5 / 3) ^ 4-2 × (-5 / 4) ^ 4

h (5 / 3, -5 / 4) =- 67.46 = Z_0

Then the point evaluate is

(x, y, z) = (5 / 3, -5 / 4, -67.46)


Then we derive with respect to x with what we get:

h_x (x, y) = 20xy-10x-4x ^ 3

h_x (5 / 3, -5 / 4) = 20 × (5 / 3) × (-5 / 4) -10 × (5 / 3) -4 × (5 / 3) 〗 ^ 3

h_x (5 / 3, -5 / 4) =- 2075/27 = m

The plan as presented in (x, z) to observe the line in R ^ 2 , where we find the equation that represents our chart.

The plane (yellow) is represented by y = -5 / 4 (That is our curve is centered on this plane)

red line represents the line we're looking for.

Then

Equation of Straight in R ^ 2 , then:

Z-Z_0 = m (x-x_0) Sea t = x-x_0

Then

Equation of Straight in R ^ 3 (Parametric Equation)


Now Equation vector of the tangent line to which the plane y =- 5 / 4 y la funcion

h(5/3,-5/4)=10×(5/3)^2×(-5/4)-5×(5/3)^2-4×(-5/4)^2-(5/3)^4-2×(-5/4)^4

(x,y,z)=t(1,0,-2075/27)+(5/3,-5/4,-67,46)










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