Question

True or False The slope of the normal line to the curve \(x=3 \cos t, y=3 \sin t\) at \(t=\pi / 4\) is \(-1 .\) Justify your answer.

Short answer

True. The slope of the normal line to the curve at \(t=\pi / 4\) is -1.

Step-by-step solution

Find the Parametric Equations and their Derivatives

We are provided with the parametric equations: \(x=3 \cos t\) and \(y=3 \sin t\). To get the slope of the tangent line to the curve at any point, we need to find the derivative of y with respect to x, which is \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). First, we need to determine the derivatives of x and y with respect to t. The derivative of \(x=3 \cos t\) with respect to t, \(\frac{dx}{dt}\) is \(-3 \sin t\). Also, \(\frac{dy}{dt}\) which is the derivative of \(y=3 \sin t\) with respect to t is \(3 \cos t\).

Find the Slope of the Tangent Line

Next, substitute the derivatives of x and y into the formula \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). Therefore, \(\frac{dy}{dx} = \frac{3 \cos t}{-3 \sin t} = - \cot t\). To find the slope at \(t=\pi / 4\), substitute this value into the equation for the slope. This gives \(\frac{dy}{dx}|_{t=\pi/4} = - \cot(\pi / 4) = -1\). This is the slope of the tangent line.

Determine the Slope of the Normal Line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is -1, the negative reciprocal of -1 is also -1. Therefore, the slope of the normal line to the curve at \(t=\pi / 4\) is indeed -1.

Key concepts

Parametric Equations Derivatives

Understanding the derivatives of parametric equations is essential for analyzing the behavior of curves. A parametric curve is defined by two equations, where both x and y are expressed in terms of a third variable, typically t. For the curve given by the parametric equations \(x=3 \cos t\) and \(y=3 \sin t\), we find the derivatives with respect to t to determine the rate of change of each coordinate as a function of t.

To find the slope of a curve at a specific point when dealing with parametric equations, we need to compute the derivative of y with respect to x, denoted as \(\frac{dy}{dx}\). This can be done by dividing the derivative of y with respect to t, \(\frac{dy}{dt}\), by the derivative of x with respect to t, \(\frac{dx}{dt}\). A critical step involves finding these individual derivatives accurately to ensure the slope of the tangent line is correctly determined.

Tangent and Normal Lines Calculus

In calculus, the concepts of tangent and normal lines to curves are closely connected. The tangent line to a curve at a given point is the line that 'just touches' the curve at that point. It has the same direction as the curve's direction of travel at that point, characterized by its slope. This slope can often give us significant information about the behavior of the curve, such as direction or rate of change.

The normal line, on the other hand, is perpendicular to the tangent line. In two dimensions, the slope of the normal line is the negative reciprocal of the slope of the tangent line. To illustrate, if a tangent line has a slope of \(m\), the slope of the normal line will be \(\frac{-1}{m}\). The ability to switch between these two lines is a powerful tool in different applications of calculus like optimization and curve sketching.

Slope Calculation

Slope calculation is fundamental in understanding how steep a line is. In a Cartesian coordinate system, the slope is the measure of the steepness or the incline of a line and is quantified by the vertical change divided by the horizontal change between two distinct points on the line. For the tangent line, this is often simplified in calculus to finding the derivative of the function that describes the curve at a certain point.

When working with parametric equations, the process for calculating slope differs slightly because we work with derivatives with respect to a third variable, t, rather than directly between y and x. As seen in the exercise, the slope of the tangent line is found by using \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\), which reflects the rate of change of y in relation to x along the curve. In the given problem, the calculation revealed that the tangent at \(t=\frac{\pi}{4}\) has a slope of -1, which also means the normal line at this point will have the same slope since the negative reciprocal of -1 is -1.