Animated Solution for Mathematics - Differentiation: The shortest distance between the line $x-y=1$ and the curve $x^2=2y$ is:

To find the shortest distance between the parabola $x^2=2y$ and the line $x-y=1$, we utilize the geometric principle that the shortest distance occurs at the point where the tangent to the parabola is parallel to the given line.

First, we determine the slope of the line $x-y=1$. Rearranging this into slope-intercept form, we get $y=x-1$, which indicates that the slope $m$ is $1$.

For the parabola $x^2=2y$, we differentiate with respect to $x$ to find the slope of the tangent at any arbitrary point. Differentiating both sides yields:

This simplifies to the slope function:

Since the tangent must be parallel to the line, we set the slope of the tangent equal to the slope of the line:

This value represents the $x$-coordinate of the point on the parabola closest to the line. Substituting $x=1$ back into the parabola equation $x^2=2y$, we find:

Thus, the point of closest approach is $P(1,\frac{1}{2})$.

We now calculate the perpendicular distance from point $P(1,\frac{1}{2})$ to the line $x-y-1=0$ using the distance formula:

Substituting the values $a=1$, $b=-1$, $c=-1$, $x_1=1$, and $y_1=\frac{1}{2}$:

Simplifying the numerator:

The denominator is $\sqrt{1+1}=\sqrt{2}$. Therefore, the shortest distance is: