91Ó°ÊÓ

Understanding the derivative of an equation is essential for analyzing the behavior of curves. Let's consider the process of finding the derivative for the equation provided in our exercise:
The equation given is \(y=-\frac{1}{2} x^{2}+k\). The derivative represents the rate at which \(y\) changes with respect to \(x\). Here, we apply the rules of differentiation: the derivative of \(x^{2}\) is \(2x\), and since it's multiplied by \(-\frac{1}{2}\), the derivative becomes \(-x\). This value, \(-x\), is also known as the slope of the tangent line to the curve at any point.

Differentiation Example

Using the power rule, which states that the derivative of \(x^n\) is \(nx^{(n-1)}\), we can easily differentiate our quadratic equation. The constant \(k\) disappears as its derivative is zero. Thus, for the equation \(y=-\frac{1}{2} x^{2}+k\), differentiation yields \(\frac{dy}{dx} = -x\). This process is a cornerstone of calculus and aids in the analysis of curve properties such as concavity and inflection points.

Two curves are orthogonal if, at their points of intersection, the tangent lines are perpendicular to each other. We can determine this condition using derivatives of the curves' equations.
If the slopes of the tangent lines at the intersection points (represented by the derivatives of the curves) are negative reciprocals of each other, the curves are orthogonal. This means if one curve has a derivative of \(f_{1}'(x)\) at the point of intersection, the other curve should have a derivative of \(-\frac{1}{f_{1}'(x)}\) for the curves to be orthogonal. The product of these derivatives should be -1, as per the orthogonality condition \(f_{1}'(x) * f_{2}'(x) = -1\).
However, as seen in the step by step solution, the family of curves defined by \(y=-\frac{1}{2} x^{2}+k\) does not satisfy this criterion. Their derivatives do not meet the aforementioned condition, and thus, the family of curves is not orthogonal because the product of their derivatives is not -1.

Misconception Alert

It's a common misconception that simply having perpendicular tangent lines at some points will make two families of curves orthogonal; however, they must satisfy the negative reciprocal relationship at all intersection points.

The slope of the tangent line to a curve at any given point is a critical concept in calculus, as it provides information about the curve's steepness and direction at that point. The slope is found by evaluating the derivative of the curve's equation at the point of interest.
For our family of curves, the derivative is a simple linear function, \(\frac{dy}{dx} = -x\), which means the slope of the tangent line varies linearly with \(x\). At any particular value of \(x\), we can calculate the tangent's slope by plugging that value into the derivative.

Visualization of Tangent Slopes

For instance, if \(x = 1\), the slope of the tangent line would be \(-1\). On a graph, this would represent a line falling to the right. Similarly, if \(x = -2\), the slope would be 2, representing a line rising to the right. These slopes help in sketching the behavior of the curve and understanding how it changes as \(x\) varies, which is particularly valuable when analyzing motion or rates of change in various scientific and engineering contexts.