Multivariable Calculus - Minimum value along a curve

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December 12th 2012, 01:09 PM
Maedbh

Multivariable Calculus - Minimum value along a curve

3. Let g(x,y)=x²+y³.
(a) Find the directional derivative of g at the point (−3, 2) in the direction (3, 5).

(b) Determine the tangent line to the curve g(x, y) = g(4, −2) at the point (4, −2).
(c) Find the tangent plane to the surface z = g(x, y) at the point (−1, 1, 2).
(d) What is the minimum of g along the curve x² + y²= 4/9?

Hey there! Looking for help on part (d) but don't know if the other information is relevant so I just posted the whole question.
Any help is appreciated!

- Maedbh
December 13th 2012, 12:28 AM
a tutor

Re: Multivariable Calculus - Minimum value along a curve

You could eliminate the x^2 in g(x,y) and differentiate with respect to y but you are probably supposed to be using Lagrange multipliers. It's a very easy example. Check your notes. You can use the other method as a check.
December 13th 2012, 06:17 AM
HallsofIvy

Re: Multivariable Calculus - Minimum value along a curve

You are asked to find the max or min of $g(x,y)= x^2+ y^3$ on the curve $f(x,y)= x^2+ y^2= 4/9$ . Look at the two gradients: $\nabla g= 2x\vec{i}+ 3y^2\vec{j}$ and $\nabla f= 2x\vec{i}+ 2y\vec{j}$ . $\nabla g$ points in the direction of greatest increase of g so you can find a maximum value by moving in the direction of $\nabla g$ until it is 0. But, here, you can't move in that direction because you must stay on the curve f(x,y)= 4/9. What you can do is move along the curve in the direction of the projection of $\nabla g$ on the curve- again until that projection is 0. That happens when $\nabla g$ is perpendicular to the curve. And since $\nabla f$ is perpendicular to the curve f(x,y)= constant, you want those two gradients parallel- that is, one is a multiple of the other. You want $\nabla g= \lambda \nabla f$ for some constant $\lambda$ . That is the "Lagrange multiplier" a tutor refers to. Surely, you seen that in class or your textbook.