Problem 19 - Calculus

Question

Determine \(c\) so that the straight line joining (0,3) and (5,-2) is tangent to the curve \(y=c /(x+1)\)

Step-by-Step Solution

Answer

The value of \(c\) is 4.

Step 1: Determine the Slope of the Line

The formula for the slope of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Applying this formula to the points \((0,3)\) and \((5,-2)\), we have: \[ m = \frac{-2 - 3}{5 - 0} = \frac{-5}{5} = -1. \] Thus, the slope of the line is \(-1\).

Step 2: Equation of the Line

The equation of a line using a point \((x_1, y_1)\) and a slope \(m\) can be written as \(y - y_1 = m(x - x_1)\). With point \((0,3)\) and slope \(-1\), the equation becomes: \[ y - 3 = -1(x - 0) \] Simplifying, we find: \[ y = -x + 3. \]

Step 3: Determine the Derivative of the Curve

We start by finding the derivative of the curve \(y = \frac{c}{x+1}\). Using the power rule and the chain rule for derivatives, we have: \[ \frac{dy}{dx} = -\frac{c}{(x+1)^2}. \] This represents the slope of the curve at any point \(x\).

Step 4: Find the Point of Tangency

For the line to be tangent to the curve at some point, their slopes must be equal at that point, and they must intersect. Therefore, we set \(-\frac{c}{(x+1)^2} = -1\) (the slope of the line). Simplifying gives: \[ \frac{c}{(x+1)^2} = 1 \Rightarrow c = (x+1)^2. \]

Step 5: Find Intersection Point

Set the line and the curve equal to find the point of intersection: \(-x + 3 = \frac{c}{x+1}\). Substitute \(c = (x+1)^2\) from Step 4. Thus, \(-x + 3 = x+1\). Solving gives \(x = 1\). Substituting back into the curve equation, \(y = \frac{(1+1)^2}{1+1} = 2\). So, the point of tangency is \((1,2)\).

Step 6: Solve for c

Substitute \(x=1\) back into \(c = (x+1)^2\) gives: \[ c = (1+1)^2 = 4. \] Therefore, the value of \(c\) is \(4\).

Key Concepts

slope of a line

The slope of a line represents how steep the line is. It tells us the rate at which the line rises or falls as we move from left to right. To calculate the slope (\(m\)), we use two points \((x_1, y_1)\) and \((x_2, y_2)\) on the line. The formula is:

• \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

For example, with points \((0,3)\) and \((5,-2)\), the slope is \(-1\). This means for every unit you move to the right, the line falls by one unit.
Interpreting the slope as \(-1\) is crucial when dealing with tangent lines, as the line's slope at this point will help determine specific curve behaviors.

derivative of a curve

The derivative of a curve provides the slope of the tangent line to that curve at any given point. In simpler terms, it gives us the rate of change of the curve. For the curve \(y = \frac{c}{x+1}\), the derivative is:

• \(\frac{dy}{dx} = -\frac{c}{(x+1)^2}\)

This derivative is particularly important because it tells us how the curve behaves as \(x\) changes. To find a point of tangency, the derivative must equal the slope of the tangent line. Hence, in our problem, \(\frac{dy}{dx} = -1\) is set equal to the line's slope, which is crucial for finding \(c\).

point of tangency

The point of tangency is where a line just "touches" a curve without crossing it. This point is essential because it is where the slope of the line and the curve's derivative are the same. For a tangent line, these two slopes must match.
In our exercise, to find the point of tangency, both the line equation \(-x + 3\) and the modified curve equation \(\frac{c}{x+1}\) need to intersect at the same \(x\). This intersection not only ensures a common point but also signifies where the line and curve have equal slopes. Here, determining that this occurs at \((1,2)\) helps in calculating the specific value of \(c\)\.

equation of a line

Formulating an equation of a line is key to describing its behavior in a coordinate system. When we have a point \((x_1, y_1)\) on the line and know the slope \(m\), we use the formula:

• \(y - y_1 = m(x - x_1)\)

This equation gives us a tool to predict \(y\) for any \(x\) along the line. For instance, with point \((0,3)\) and slope \(-1\), the line equation becomes \(y = -x + 3\). Finding this equation is crucial as it matches up with the curve to solve for where the tangent condition is satisfied — allowing us to navigate through complex problems, like figuring out the correct \(c\)\ value.