Textbook Question
79–82. {Use of Tech} Visualizing tangent and normal lines
b. Graph the tangent and normal lines on the given graph.
(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)
2. Intro to Derivatives
Tangent Lines and Derivatives
3:23 minutesProblem 3.81
Textbook Question
For what value of c is the curve y = c/ (x + 1) tangent to the line through the points (0, 3) and (5, -2)?
Verified step by step guidance1
First, find the equation of the line passing through the points (0, 3) and (5, -2). Use the slope formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Substitute the given points to find the slope.
With the slope \( m \) calculated, use the point-slope form of the line equation: \( y - y_1 = m(x - x_1) \). Substitute one of the points, say (0, 3), and the slope to find the equation of the line.
Next, find the derivative of the curve \( y = \frac{c}{x + 1} \) to determine its slope at any point \( x \). Use the quotient rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u = c \) and \( v = x + 1 \).
Set the derivative of the curve equal to the slope of the line found in step 2. This will give you an equation involving \( c \) and \( x \). Solve for \( c \) in terms of \( x \).
Finally, find the point of tangency by equating the curve \( y = \frac{c}{x + 1} \) to the line equation found in step 2. Solve for \( x \) and substitute back to find the value of \( c \) that makes the curve tangent to the line.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Tangent Line
A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line at that point is equal to the derivative of the curve's function at that point. Understanding how to find the slope of the tangent line is crucial for determining the conditions under which the curve and the line are tangent.
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Slopes of Tangent Lines
Slope of a Line
The slope of a line is a measure of its steepness, calculated as the change in the y-coordinates divided by the change in the x-coordinates between two points on the line. For the line through the points (0, 3) and (5, -2), the slope can be calculated using the formula (y2 - y1) / (x2 - x1). This slope will be essential for comparing it to the slope of the tangent line to the curve.
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Derivative
The derivative of a function at a point represents the rate of change of the function's value with respect to changes in its input. In this context, finding the derivative of the curve y = c/(x + 1) will allow us to determine the slope of the tangent line at any point on the curve. Setting this derivative equal to the slope of the line will help us find the value of c for which the curve is tangent to the line.
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