Problem 20 - Calculus

Question

Choose \(b, c, d\) so that the two parabolas \(y=x^{2}+b x+c\) and \(y=d x-x^{2}\) are tangent to each other at \(x=1, y=0\).

Step-by-Step Solution

Answer

The values are \(b = -3\), \(c = 2\), and \(d = 1\).

Step 1: Determine the point of tangency

The problem states that the two parabolas are tangent at the point \((x, y) = (1, 0)\). So, \(x = 1\) and \(y = 0\) is a point on both parabolas.

Step 2: Plug the point into both parabolas

First, plug \(x = 1\) and \(y = 0\) into each parabola equation:For the first parabola:\[0 = 1^2 + b \cdot 1 + c \Rightarrow 0 = 1 + b + c\]For the second parabola:\[0 = d \cdot 1 - 1^2 \Rightarrow 0 = d - 1\]

Step 3: Solve for d

From the equation \(d - 1 = 0\), we find:\[d = 1\]

Step 4: Equate the derivatives at the tangency point

Since the curves are tangent, their derivatives at \(x = 1\) must be equal. First calculate the derivatives of the parabolas.For the first parabola \(y = x^2 + bx + c\):\[y' = 2x + b\]For the second parabola \(y = dx - x^2\):\[y' = d - 2x\]For \(x = 1\), equate the derivatives:\[2 \cdot 1 + b = d - 2 \cdot 1\]

Step 5: Solve for b

Substitute \(d = 1\) from the earlier solution:\[2 + b = 1 - 2\]This simplifies to: \[b = -3\]

Step 6: Solve for c using the point substitution equation

Using \(1 + b + c = 0\), substitute \(b = -3\):\[1 - 3 + c = 0\]Solve for \(c\):\[c = 2\]

Key Concepts

Parabolas

Understanding parabolas is key to mastering many concepts in mathematics, especially in calculus problem solving. A parabola is a symmetric curve on a plane. The standard equation for a parabola is usually given as \(y = ax^2 + bx + c\). This equation is quadratic, meaning it has an exponent of 2 on the \(x\) term.

• The shape of the parabola is determined by the coefficients \(a\), \(b\), and \(c\).
• The parabola opens upwards if \(a > 0\) and downwards if \(a < 0\).
• The vertex of the parabola gives us the highest or lowest point on the curve, depending on the direction of opening.

In problems where you need to determine values like \(b\) and \(c\), you're essentially setting constraints on the shape and position of the parabola. This involves ensuring certain points lie on the curve or that it touches (is tangent to) another curve, as seen in the given exercise.

Tangent

The notion of tangency is essential when dealing with curves like parabolas. A tangent line is a straight line that touches a curve at precisely one point, never crossing it. This concept is essential in calculus, where it represents the slope of the curve at that point. For two curves to be tangent to each other:

• They must share a common point.
• Their slopes must be equal at that point, meaning their derivatives are equal.

In the exercise, the two parabolas are required to be tangent at \(x = 1, y = 0\). This implies not only that the parabolas pass through this point but also that their derivatives at \(x = 1\) must be the same. This equality of slopes leads us directly into our next core concept, derivatives.

Derivatives

Derivatives are a fundamental principle in calculus and are used to understand the rate at which a function is changing at any given point. When it comes to parabolas or any other curves:- The derivative at a point is the slope of the tangent at the curve at that point.- For the parabola \(y = x^2 + bx + c\), the derivative \(y' = 2x + b\) gives the slope at any point \(x\).- For the parabola \(y = dx - x^2\), the derivative \(y' = d - 2x\) serves a similar purpose.To find if two curves are tangent at a point, equate their derivatives at that point. In this problem, derivatives help us find \(b\) and \(d\) such that the parabolas remain tangent at \((1, 0)\). You calculate the derivatives and set them equal at \(x = 1\) to determine the unknown coefficients.

Algebraic Equations

Algebraic equations are essential tools for solving problems involving parabolas, tangents, and derivatives. They help us set conditions and constraints that must be fulfilled for a problem's requirements to be met.In the exercise:- We started by substituting the tangency point \((1,0)\) into the equations of the parabolas to form equations involving \(b\) and \(c\), earning us \(b + c + 1 = 0\) for the first parabola and \(d = 1\) for the second after simplification.- Then, by equating the derivatives \(2 + b = 1 - 2\) we used algebra to solve for \(b\) and found \(b = -3\).- Finally, by substituting back, we found \(c\) by solving the remaining algebraic equation.These steps show how algebraic equations allow for the determination of unknown variables, making them indispensable in calculus problem solving.