To find out directional derivative $f(x.y.z)=x^2+y^2−z^2$ at $(3,4,5)$ along the curve of
intersection of the two surfaces $2x^2+2y^2−z^2=25$ and $x^2+y^2=z^2$
I am trying to parametrize above two surfaces $r(t)$ and find out vector.
But, I am not sure what value of $t$ should be and is it right way to find out directional
derivative.
If there is any other way, please explain .
On
How to do it:
In this simple example it is actually possible to give an explicit parametrization of the curve of intersection $\gamma$ in terms of elementary functions. But in general you cannot count on such luck. Fortunately it is possible to compute this directional derivative without computing $\gamma$.
We are given two surfaces $$S_1:\quad g(x,y,z)=0,\qquad S_2:\quad h(x,y,z)=0$$ that intersect transversally at a given point $P$. This means that $\nabla g(P)$ and $\nabla h(P)$ are linearly independent. The cross product $$n:=\nabla g(P)\times \nabla h(P)\ne0$$ is orthogonal to both gradients; therefore it belongs to both tangent planes and is in fact a tangent vector of the curve $\gamma=S_1\cap S_2$ at $P$, albeit not normalized.
In addition we are given a real-valued function $f$ defined in the neighborhood of $P$, and we are told to compute the directional derivative of $f$ in the direction of $n$. Since $\gamma$ has no a priori sense of direction this directional derivative is only defined up to sign, and is given by $$\pm{1\over |n|}\>\nabla f(P)\cdot n\ .$$
If you plug $z^2=x^2+y^2$ into $2x^2+2y^2-z^2=25$, you will get the intersection $x^2+y^2=25, z=5$, which is a circle. The parametric equation is then $(5\cos{t},5\sin{t},5), t\in[0,2\pi]$. And yes, you can use that to find directional derivative.