Question

Find the equations of the tangent and normal lines to the graph of \(g\) at the indicated point. \(g(t)=t \sin t\) at \(\left(\frac{3 \pi}{2},-\frac{3 \pi}{2}\right)\).

Short answer

Tangent line: \(y = -x + 3\pi\); Normal line: \(y = x - 3\pi\).

Step-by-step solution

Find the Derivative of g(t)

To find the equation of the tangent line, we first need the slope of the tangent line at a given point, which is the derivative of the function at that point. Start by finding the derivative of the function \(g(t) = t \sin t\). Use the product rule where if \(u(t) = t\) and \(v(t) = \sin t\), then \(g'(t) = u'(t)v(t) + u(t)v'(t)\). By calculating, we get:\(u'(t) = 1\) and \(v'(t) = \cos t\).Thus, \(g'(t) = 1 \cdot \sin t + t \cdot \cos t = \sin t + t \cos t\).

Evaluate the Derivative at the Point

Evaluate the derivative at the specific point \(t = \frac{3\pi}{2}\).Calculate \(g'\left(\frac{3\pi}{2}\right) = \sin\left(\frac{3\pi}{2}\right) + \frac{3\pi}{2} \cos\left(\frac{3\pi}{2}\right)\).Knowing that \(\sin\left(\frac{3\pi}{2}\right) = -1\) and \(\cos\left(\frac{3\pi}{2}\right) = 0\), we find:\(g'\left(\frac{3\pi}{2}\right) = -1\).

Equation of the Tangent Line

Now that we have the slope of the tangent line \(-1\) at the point, use the point-slope form of a line \(y - y_1 = m(x - x_1)\).Here, \((x_1, y_1) = \left(\frac{3\pi}{2}, -\frac{3\pi}{2}\right)\) and the slope \(m = -1\).Substitute these values in:\(y + \frac{3\pi}{2} = -1\left(x - \frac{3\pi}{2}\right)\).Simplifying, the equation becomes \(y = -x + 3\pi\).

Equation of the Normal Line

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the slope of the tangent line. Given the slope of the tangent line is \(-1\), the slope of the normal line is \(1\).Using the point-slope form:\(y - y_1 = m(x - x_1)\), with the same point \((x_1, y_1) = \left(\frac{3\pi}{2}, -\frac{3\pi}{2}\right)\) and the slope of the normal line \(1\), we have:\(y + \frac{3\pi}{2} = 1\left(x - \frac{3\pi}{2}\right)\).Simplifying gives the equation of the normal line:\(y = x - 3\pi\).

Key concepts

Derivative

Understanding derivatives is fundamental in calculus. A derivative represents the rate of change of a function with respect to its variable. It essentially gives you the slope of the tangent line to the curve of the function at any given point.
The notation for the derivative of a function \(g(t)\) is \(g'(t)\). In this exercise, we are tasked with finding the derivative of \(g(t) = t \sin t\). This requires the use of the product rule, a rule essential when differentiating products of two functions.
To apply the product rule, if we have two functions \(u(t)\) and \(v(t)\), their product's derivative is \(u'(t)v(t) + u(t)v'(t)\). For \(g(t) = t \sin t\), let \(u(t) = t\) and \(v(t) = \sin t\). Then, \(u'(t) = 1\) and \(v'(t) = \cos t\), leading to the derivative \(g'(t) = \sin t + t \cos t\).

• Derivatives help determine the changes at specific points.
• In the context of a graph, they help calculate slopes at any desired point.
• They bridge the gap between algebraic functions and their geometric interpretations.

Understanding this principle is vital for progressing in calculus and solving a plethora of other mathematical problems.

Tangent Line

A tangent line at a specific point on a curve tells us the instantaneous rate of change of the function at that point. It is a straight line that just "touches" the curve, representing the slope given by the derivative.
In our problem, after finding the derivative \(g'(t)\), we evaluated it at \(t = \frac{3\pi}{2}\). This evaluation gave us the slope of the tangent line, which is \(-1\).
To construct the equation of the tangent line, we use the point-slope form: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point of tangency. Substituting \(m = -1\) and the point \(\left(\frac{3\pi}{2}, -\frac{3\pi}{2}\right)\), we get:

• Equation of the tangent: \(y = -x + 3\pi\).

This linear equation illustrates how the tangent connects with the graph and aids in understanding the behavior of the curve right at that point.

Normal Line

The normal line is a line perpendicular to the tangent line at the point of tangency. It represents a direction where there is no change in terms of the curve's slope.
Unlike the tangent line, the slope of the normal line is the negative reciprocal of the tangent line's slope. For our exercise, where the tangent slope is \(-1\), the normal line's slope becomes \(1\).
Using the same point-slope format, we write the equation for the normal line with slope \(1\) at the point \(\left(\frac{3\pi}{2}, -\frac{3\pi}{2}\right)\):

• Equation of the normal line: \(y = x - 3\pi\).

The normal line plays a crucial role in various applications such as physics, engineering, and computer graphics as it highlights different properties of the function. Understanding both tangent and normal lines helps in comprehending the curve's geometry better.