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Contour map and Intersection points

  • Thread starter Arcon
  • Start date
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1. Homework Statement

(i) Drawing a contour map for the function h(x.y) = -12-4x^2+16x-y^2-8y
(ii) (Continuing from i) at the point (1,-1,7) which direction to move to have
the maximum increase in height?

(iii) Find the point closest to the origin on the curve of intersection of the
plane 2y + 4z = 5 and the cone z^2= 4x^2 + 4y^2.

2. Homework Equations

(i) z = -12-4x^2+16x-y^2-8y (then Im stuck)

(iii) I got f = z <--(not sure if this is right),
g = 2y +4z = 5
and h = z^2= 4x^2 + 4y^2.

then gradient f = lambda * gradient g + mu * gradient h

3. The Attempt at a Solution

then from some caluculation i got two intersection point of (0,5/18,5/9) and (0, 1/10, 1/20) with (0, 1/10, 1/20) the point closest to the origin.
 

HallsofIvy

Science Advisor
Homework Helper
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1. Homework Statement

(i) Drawing a contour map for the function h(x.y) = -12-4x^2+16x-y^2-8y
(ii) (Continuing from i) at the point (1,-1,7) which direction to move to have
the maximum increase in height?

(iii) Find the point closest to the origin on the curve of intersection of the
plane 2y + 4z = 5 and the cone z^2= 4x^2 + 4y^2.

2. Homework Equations

(i) z = -12-4x^2+16x-y^2-8y (then Im stuck)
Do you understand what a "contour map" IS? No, don't set h(x,y)= z. A contour map of a function of two variables is a graph is the xy-coordinates system. Set h(x,y)= equal to a number of different constants and graph each of them. (Looks to me like you will have a number of different hyperbolas with the same asymptotes.)

(iii) I got f = z <--(not sure if this is right),
g = 2y +4z = 5
and h = z^2= 4x^2 + 4y^2.

then gradient f = lambda * gradient g + mu * gradient h
Okay, that's the "Lagrange multiplier" method.

3. The Attempt at a Solution

then from some caluculation i got two intersection point of (0,5/18,5/9) and (0, 1/10, 1/20) with (0, 1/10, 1/20) the point closest to the origin.
 

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