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I tried it to relate with concept of maxima and minima but not able to get the idea behind this method

ViHdzP
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MUKESH KUMAR
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  • What do you mean by the "partial derivative method"? You'll need to add some more context to this question if you want it to be answered. – ViHdzP Dec 04 '19 at 15:58
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    This is pretty much a duplicate of https://math.stackexchange.com/q/1941210/265466. A pair of intersecting lines is a degenerate hyperbola, and the intersection point is its center. – amd Dec 04 '19 at 17:36
  • To add to my previous comment, the intersection point corresponds to the saddle point of a particular hyperboloid—the pair of intersecting lines are generated by taking a horizontal slice through the saddle point. Details are in the linked question. – amd Dec 04 '19 at 19:38

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Without more detail, I can hazard a guess. There are multiple ways to use the partial derivative to solve this problem, though I think linear algebra would be the easiest.

We set it up as a problem in two independent variables.

$$x_1=a_1t+b_1$$ $$y_1=c_1t+d_1$$ $$x_2=a_2\lambda+b_2$$ $$y_2=c_2\lambda+d_2$$

So given $\lambda$ and $t$, we have a distance function:

$$d(t,\lambda)^2=S(t,\lambda)=[(a_1t+b_1)-(a_2t+b_2)]^2+[(c_1\lambda+d_1)-(c_2\lambda+d_2)]^2$$

The total derivative of S is $dS=\frac{\partial S}{\partial t}dt+\frac{\partial S}{\partial \lambda}d\lambda$.

Intuitively if imprecisely: In the one dimensional case, $dy=f'(x)dx.$ Why do we need $f'(x)=0$? Because $\Delta y\approx f'(x)\Delta x$. Whether we have a positive or negative value of $\Delta x$, function values are greater are less than their values at x. So we have a y value that's greater than or less than $f(x)$ "close" to $(x,f(x))$. Consequently, $f(x)$ is not a local minimum or a local maximum. If we want local extrema, we need $f'(x)=0$ at our extrema candidate. We also need to be mindful of possible extrema occurring at extreme allowable values of the independent variables, in this case, $\lambda$ and $t$.

By similar reasoning, we need $\partial S/\partial t=0$ and $\partial S/\partial \lambda=0$.

So:

$$\frac{\partial S}{\partial t}=0=2(a_1-a_2)[(a_1-a_2)t+(b_1-b_2)]$$ $$\frac{\partial S}{\partial \lambda}=0=2(c_1-c_2)[(c_1-c_2)\lambda+(d_1-d_2)]$$

We solve for $t$ and $\lambda$ to get are local extrema candidates. Our extreme t and lambda are positive and negative $\infty$ for each. We know the lines are either parallel, skew, or they intersect, so we have trivial maxima at these extremes and we can prove that there are no minima.

We plug our $\lambda$ and $t$ into $S(t,\lambda)$ to find the square of our minimum distance. If this is zero, we have an intersection, otherwise, our lines are skew.

TurlocTheRed
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