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Chapter 3: Problem 32

Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$x^{2}+y^{2}=25, \quad(3,-4)$$

Short Answer

Expert verified
The point is on the curve. Tangent: \(y = \frac{3}{4}x - \frac{25}{4}\); Normal: \(y = -\frac{4}{3}x\).

Step by step solution

01

Verify the Point on the Curve

Substitute the point (3, -4) into the equation of the curve, which is given as \(x^2 + y^2 = 25\). We will check if the equation holds true.\[x^2 + y^2 = 3^2 + (-4)^2 = 9 + 16 = 25\]Since both sides of the equation are equal, the point (3, -4) lies on the curve.
02

Find the Derivative

To find the tangent line, first we need to calculate the derivative of \(x^2 + y^2 = 25\) implicitly with respect to \(x\).\[2x + 2y\frac{dy}{dx} = 0\]Solving for \(\frac{dy}{dx}\), we get:\[\frac{dy}{dx} = -\frac{x}{y}\]
03

Calculate the Slope of the Tangent Line

Using the derivative, find the slope of the tangent line at the point (3, -4).\[\frac{dy}{dx} = -\frac{3}{-4} = \frac{3}{4}\]The slope of the tangent line is \(\frac{3}{4}\).
04

Equation of the Tangent Line

Using the point-slope form, \(y - y_1 = m(x - x_1)\) where \((x_1, y_1)\) is the point (3,-4) and \(m\) is the slope \(\frac{3}{4}\):\[y + 4 = \frac{3}{4}(x - 3)\]Simplifying, the equation of the tangent line is:\[y = \frac{3}{4}x - \frac{25}{4}\]
05

Find the Slope of the Normal Line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Thus, the slope is \(-\frac{4}{3}\).
06

Equation of the Normal Line

Again using the point-slope form for the equation of the normal line,\[y + 4 = -\frac{4}{3}(x - 3)\]Simplifying, the equation of the normal line is:\[y = -\frac{4}{3}x \ \ \ + 0\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Line
A tangent line is a straight line that touches a curve at a single point without crossing it. This line represents the curve's slope at that specific location. In our exercise, the curve is described by the equation \( x^2 + y^2 = 25 \). We need to find the tangent line at the point \( (3, -4) \). Here鈥檚 how we tackle it:
  • The point \( (3, -4) \) has been verified to be on the curve.
  • To find the tangent line, we calculate the derivative \( \frac{dy}{dx} = -\frac{x}{y} \). This formula comes from implicit differentiation applied to the curve's equation.
  • At the point \( (3, -4) \), the slope of the tangent line is \( \frac{3}{4} \).
  • Using point-slope form \( y - y_1 = m(x - x_1) \) where \( m \) is the slope and \( (x_1, y_1) \) is the point, we get the tangent line's equation as \( y = \frac{3}{4}x - \frac{25}{4} \).
Understanding tangent lines is crucial because they provide an approximation of the curve at a given point. This approximation becomes exact at the point of tangency.
Normal Line
The normal line is another line related to the curve, but unlike the tangent line, the normal line is perpendicular to the tangent line at the point of tangency. This perpendicular characteristic makes it useful to understand how the curve behaves around a specific point. For our curve \( x^2 + y^2 = 25 \) at the point \( (3, -4) \), here's what happens:
  • Since the tangential slope at \( (3, -4) \) is \( \frac{3}{4} \), the slope of the normal line can be calculated as the negative reciprocal of the tangent slope, which is \(-\frac{4}{3}\).
  • Using the point-slope formula again, the equation for the normal line is found to be \( y = -\frac{4}{3}x + 0 \).
The significance of the normal line is that it offers insight into the symmetry of the problem and is used in various applications, such as in calculating boundaries and stress analysis in physics.
Derivative Calculation
Before diving into the equations of the tangent and normal lines, we need to understand how to find the derivative using implicit differentiation. In problems like \( x^2 + y^2 = 25 \), directly solving for \( y \) is not straightforward, so we use implicit differentiation. This method helps differentiate expressions where \( y \) cannot be isolated easily.Here's how it works:
  • Implicit differentiation involves differentiating both sides of the equation with respect to \( x \).
  • For the equation \( x^2 + y^2 = 25 \), differentiate to get \( 2x + 2y\frac{dy}{dx} = 0 \).
  • Solve for \( \frac{dy}{dx} \) to find the derivative: \( \frac{dy}{dx} = -\frac{x}{y} \).
  • This equation tells us the slope of the tangent line wherever it is evaluated on the curve.
Understanding derivative calculation through implicit differentiation is critical as it allows us to deal with more complex relationships where variables are interdependent.

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Most popular questions from this chapter

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I .\) Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\). c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying $$|x-a|<\delta \Rightarrow|f(x)-L(x)|<\epsilon$$ for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=x 2^{x}, \quad[0,2], \quad a=1$$

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