14. Directional Derivatives and Gradients

The hyperboloid is a level set of the function \(f(x,y,z)=2x^2-y^2+2z^2\). The direction of the normal line is its gradient: \[ \vec\nabla f=\langle 4x,-2y,4z\rangle \] We want this to be parallel to the vector from \(P\) to \(Q\): \[ \overrightarrow{PQ}=Q-P=\langle 4,-3,-2\rangle \] To say they are parallel means one vector is a multiple of the other, i.e. \[ \vec\nabla f=\lambda\overrightarrow{PQ} \qquad \text{or} \qquad \langle 4x,-2y,4z\rangle =\lambda\langle 4,-3,-2\rangle \] where \(\lambda\) (the Greek letter lambda) is the proportionality factor. This gives \(3\) equations, \[ 4x=4\lambda \qquad -2y=-3\lambda \qquad 4z=-2\lambda, \] but there are \(4\) unknowns: \(x\), \(y\), \(z\) and \(\lambda\). We need a \(4^\text{th}\) equation. It is the equation of the hyperboloid: \[ 2x^2-y^2+2z^2=1 \] To solve these equations, we use the first three equations to express \(x\), \(y\) and \(z\) in terms of \(\lambda\) and substitute these into the equation of the hyperbola: \[ x=\lambda \qquad y=\dfrac{3}{2}\lambda \qquad z=-\,\dfrac{1}{2}\lambda \] \[ 2(\lambda)^2-\left(\dfrac{3}{2}\lambda\right)^2 +2\left(-\,\dfrac{1}{2}\lambda\right)^2=1 \] This says \((2-\,\dfrac{9}{4}+\dfrac{1}{2})\lambda^2=1\) or \(\dfrac{\lambda^2}{4}=1\) or \(\lambda=\pm2\). So the points are: \[ (x,y,z)=(2,3,-1) \qquad \text{and} \qquad (x,y,z)=(-2,-3,1) \]