Question

Find the point on the surface curve \(y=x^{2}\) nearest the line \(y=x-1 .\) Identify the point on the line.

Short answer

Answer: The nearest point on the surface curve y=x^2 to the line y=x-1 is (q+1, q), where q is a parameter that represents x-1.

Step-by-step solution

Find the distance between a point on the curve and a point on the line

To find the distance between a point \((x_1, y_1)\) on the curve \(y=x^2\) and a point \((x_2, y_2)\) on the line \(y=x-1\), we can use the distance formula: \(D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\). Since \(y_1 = x_1^2\) and \(y_2 = x_2 - 1\), we can rewrite the distance formula as: \(D = \sqrt{(x_2-x_1)^2 + [(x_2-1) - x_1^2]^2}\).

Minimize the distance

In order to minimize the distance, we'll compute the derivative of the distance function with respect to \(x_2\). However, to simplify the problem, let's minimize the distance squared, \(D^2\). This is because \(D^2\) will have a minimum value at the same point \(D\) has a minimum value. So, our function to minimize is \(D^2 = (x_2-x_1)^2 + [(x_2-1) - x_1^2]^2\).

Compute the derivative with respect to \(x_2\)

To find the minimum distance, we'll take the derivative of \(D^2\) with respect to \(x_2\) and set it equal to zero. Since \(x_1^2\) is equal to \(y_1\), we can rewrite the function as \(D^2 = (x_2-x_1)^2 + (x_2-y_1-1)^2\). Then, compute the derivative: \(\frac{d(D^2)}{dx_2} = 2(x_2-x_1) + 2(x_2-y_1-1)\). Now we set \(\frac{d(D^2)}{dx_2}=0\) and solve for \(x_2\). \(0=2(x_2 - x_1) + 2(x_2 - y_1 - 1)\) \(0 = 2(x_2 - x_1) + 2(x_2 - x_1^2 - 1)\) #=> \(x_2=\frac{x_1+x_1^2+1}{2}\).

Find the point on the curve

Now we need to eliminate one of the variables, either \(x_1\) or \(x_2\). Since \(x_1^2 = y_1\), we can substitute this into our equation for \(x_2\). \(x_2 = \frac{x_1+x_1^2+1}{2}\) Plug \(y_1 = x_1^2\) into the equation: \(x_2 = \frac{x_1+y_1+1}{2}\) Now, plug this value of \(x_2\) into the line equation \(y_2 = x_2 - 1\) to find the corresponding point \((x_1, y_1)\) on the curve. \(y_1 + 1 = \frac{x_1+y_1+1}{2}\) Multiplying both sides by 2: \(2y_1+2= x_1+y_1+1 => x_1 = y_1 + 1\). Finally we have \((x_1, x_1^2)\) on the curve and \((\frac{x_1+x_1^2+1}{2},x_1^2)\) on the straight line. Since \(x_1 = y_1+1 \Rightarrow x_1-1=y_1=q\): So we get the point as \((\frac{q+2}{2},q)\) on the line and \((q+1, q)\) on the curve.

Key concepts

Distance Formula

The distance formula is a key tool used to find the distance between two points in a plane. The formula is derived from the Pythagorean theorem. It applies to points \( (x_1, y_1) \) and \( (x_2, y_2) \), and is represented by: \[D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \,\]where:

• \(x_1, y_1\) are the coordinates of the first point, and
• \(x_2, y_2\) are the coordinates of the second point.

To tackle problems involving optimal distances, this formula helps us express the separation between geometric entities, such as curves and lines, in terms of the variables at play. By focusing on minimizing the distance squared (which simplifies differentiation), we can uncover the optimal relationship between different elements under consideration.

Derivative

Derivatives are fundamental in calculus, and they help us understand how functions change. In optimization problems, derivatives are used to find critical points where functions reach their minimum or maximum values. For a function \( f(x) \), its derivative is denoted as \( f'(x) \) and is expressed as:\[f'(x) = \frac{d}{dx}f(x) \,\]which gives the rate of change of the function with respect to \(x\). In the context of minimizing the distance between a point on a curve and a line, \( D^2 \) represents the squared distance function. By computing the derivative \( \frac{d(D^2)}{dx_2} \), we can locate where this function achieves its minimum. Finding this critical point involves setting the derivative to zero and solving for \(x_2\). This procedure hinges on differentiating effectively to determine where the smallest distance occurs.

Curve and Line Intersection

Curve and line intersections relate to finding points in the plane where a curve meets or gets closest to a straight line. In our exercise, the curve \( y = x^2 \) represents a parabolic path, while \( y = x - 1 \) defines a straight line.To find where they intersect or approach each other:

• The parabolic equation \( y = x^2 \) represents a set of points where each \(x\) gives a corresponding \(y\).
• The linear equation \( y = x - 1 \) expresses a line where each \(x\) yields a \(y\) that satisfies the line.

The exercise seeks the nearest point on the curve to this line, not merely where they intersect. Thus, the task becomes aligning their differences by adjusting the \(x_2\) such that the distance \( D \) is minimized, affirming the closest possible proximity point between curve and line geometrically.

Minimization Problem

A minimization problem in calculus involves finding the input value that causes a function to achieve its smallest conceivable output value. Here, the goal is to find the point on the parabola \( y = x^2 \) nearest to the line \( y = x - 1 \).Key steps in solving a minimization problem:

• Express the quantity you want to minimize—in this case, the distance squared \( D^2 \) between the curve and the line.
• Use calculus, primarily derivatives, to determine critical points by setting the derivative of the distance function to zero.
• Substitute variables to simplify and ultimately solve for one of them to find specific coordinates.

With these calculated coordinates, you reveal the point on the curve closest to the given line, illustrating how abstract calculus tools find practical geometric solutions.