]> 14.4 Extrema on a Curve in Three Dimensions

## 14.4 Extrema on a Curve in Three Dimensions

A curve $C$ in three dimensions can be defined by two equations (that is as the intersection of two surfaces) or by use of a single parameter as in two dimensions.

If $q$ is an extreme values of $F$ on $C$ we cannot have $∇ ⟶ F · t ⟶$ non-zero at argument $q$ , by our general principle; otherwise $F$ will be larger on one side of $q$ and smaller on the other than its value at $q$ on $C$ .

The implications of this condition are different here however. We can no longer say that $∇ ⟶ F$ points in some particular direction at an extremal point. Rather it must be normal to some particular direction, that of the tangent vector to $C$ at such points.

When $C$ is described by two equations, $G = 0$ and $H = 0 , t ⟶$ is in the direction of $∇ ⟶ G × ∇ ⟶ H$ , and the statement that $∇ ⟶ F$ has no component in that direction is the statement that $∇ ⟶ F$ lies in the plane of $∇ ⟶ G$ and $∇ ⟶ H$  and so the volume of their parallelepiped is 0 and the determinant whose columns are all these grads must be 0.

This condition and $G = 0$ and $H = 0$ determine $x , y$ and $z$ at critical points.

Another way to state the same condition is to use two Lagrange Multipliers, say $c$ and $d$ and write $∇ ⟶ F = c ∇ ⟶ G + d ∇ ⟶ H$ . We can solve the three equations obtained by writing all three components of this vector equation and use them and $G = 0$ and $H = 0$ , to solve for $c , d , x , y$ , and $z$ .

Exercises:

14.6 Given a curve defined as the intersection of the surfaces defined by equations $x y z = 1$ , and $x 2 + 2 y 2 + 3 z 2 = 7$ , find equations determining the critical points of $2 x 3 − y 3$ by the determinantal approach.

14.7 Write the equations for the critical points obtained using the Lagrange Multipliers approach for the same problem.

14.8 We seek the critical points for $F$ on the curve $x = 5 sin ⁡ t , y = 3 cos ⁡ 3 t , z = sin ⁡ 2 t$ , for $t = 0$ to $2 π$ , with $F = x 2 + y 2 + z 2$ . Write equations for them.