Question

Find the equations of the tangent and normal lines to the graph of \(g\) at the indicated point. \(g(\theta)=\frac{\cos \theta-8 \theta}{\theta+1}\) at (0,1).

Short answer

The tangent line is \(y = -9\theta + 1\) and the normal line is \(y = \frac{1}{9}\theta + 1\).

Step-by-step solution

Compute the derivative of g

Find the derivative of the function \(g(\theta) = \frac{\cos \theta - 8\theta}{\theta+1}\) using the quotient rule. The quotient rule is \(\left(\frac{u}{v}\right)' = \frac{vu' - uv'}{v^2}\). In this function, \(u = \cos \theta - 8\theta\) and \(v = \theta + 1\). Compute \(u'\) and \(v'\):- \(u' = -\sin\theta -8\)- \(v' = 1\)Apply the quotient rule:\[g'(\theta) = \frac{(\theta+1)(-\sin\theta -8) - (\cos\theta - 8\theta) \cdot 1}{(\theta+1)^2}\].

Simplify g'(θ) at θ = 0

Substitute \(\theta = 0\) into \(g'(\theta)\) to find the slope of the tangent line at the point (0,1).Given: \[g'(0) = \frac{(0+1)(-\sin(0) -8) - (\cos(0) - 8 \cdot 0) \cdot 1}{(0+1)^2}\]Evaluate:- \(\sin(0) = 0\)- \(\cos(0) = 1\)So we have:\[g'(0) = \frac{1(-0 -8) - (1 - 0)}{1^2} = \frac{-8 - 1}{1} = -9\].Thus, the slope of the tangent line is \(-9\).

Write the equation of the tangent line

The equation of a line is given by \(y = mx + b\). Here, \(m = g'(0) = -9\) and the point is \((0, 1)\).Using the point-slope form: \[y - 1 = -9(\theta - 0)\]Simplify to get the tangent line: \[y = -9\theta + 1\].

Write the equation of the normal line

The slope of the normal line is the negative reciprocal of the tangent line's slope. Thus, if the tangent slope is \(-9\), the normal line's slope is \(\frac{1}{9}\).Using the point-slope form with point \((0, 1)\):\[y - 1 = \frac{1}{9}(\theta - 0)\]Simplify to get the normal line:\[y = \frac{1}{9}\theta + 1\].

Key concepts

Quotient Rule

The quotient rule is a fundamental tool in calculus, especially when dealing with functions that are expressed as a division of two other functions. It's used for finding the derivative of a ratio of two differentiable functions. If you have a function defined as the quotient of two other functions, say \( u \) and \( v \), the quotient rule provides a structured way to differentiate it.

Here's the rule in essence: for a function \( \frac{u}{v} \), its derivative \( \left( \frac{u}{v} \right)' \) is given by:

• \( \left( \frac{u}{v} \right)' = \frac{vu' - uv'}{v^2} \)

This means you:

• Differentiate \( u \) to get \( u' \).
• Differentiate \( v \) to get \( v' \).
• Substitute into the formula \( \frac{vu' - uv'}{v^2} \).

In the given function \( g(\theta) = \frac{\cos \theta - 8\theta}{\theta+1} \), we identified \( u = \cos \theta - 8\theta \) and \( v = \theta + 1 \). So, applying the quotient rule made finding the derivative methodical and neat.

Derivative

A derivative measures how a function changes as its input changes. It's a way to find the slope of a line tangent to a function at any given point. This gives us vital information about the rate of change of the function.

In simple terms, the derivative tells you how fast or slow something is changing. Key use cases include:

• Finding the slope of a function at a point.
• Determining where a function is increasing or decreasing.

To compute the derivative of the function given, \( g'(\theta) \), we applied the quotient rule. Specifically, for \( \theta = 0 \), the steps let us compute:

• Substitute the value \( \theta = 0 \) into the derivative function.
• Perform arithmetic calculations: \( g'(0) = -9 \).

This resultant slope of \(-9\) is a crucial element for computing tangent and normal lines for the function.

Tangent Line

A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. It matches the slope of the curve there, and serves to approximate the curve right at that location.

Determining the tangent line involves:

• Calculating the slope of the curve at the point, which we found using derivatives.
• Using the point-slope form of a line \( y - y_1 = m(x - x_1) \).

In our solution, after finding the slope at (0,1) is \(-9\), we applied the formula:

• Using point \((0, 1)\) and slope \(-9\), the tangent line equation becomes \( y = -9\theta + 1 \).

This line reflects how the function behaves at precisely \( (0,1) \), providing a linear approximation at that point.

Normal Line

In contrast to the tangent line, the normal line at a point on a curve is perpendicular to the tangent line at that same point. The normal line gives a geometric sense of depth from the tangent line out of the plane on which the curve sits.

The slope of a normal line is the negative reciprocal of the slope of the tangent line. This means:

• If the tangent line's slope is \( m \), then the normal line slope is \( -\frac{1}{m} \).

For our function, where the tangent slope was \(-9\):

• The normal line slope becomes \( \frac{1}{9} \).
• Using the point-slope form \( y - y_1 = m(x - x_1) \), we found the equation of the normal line as \( y = \frac{1}{9}\theta + 1 \).

Thus, the normal line complements the tangent line, providing insight into the perpendicular direction relative to the curve's tangent at point (0,1).