Question

Parametric equations for a curve are given. (a) Find \(\frac{d y}{d x}\). (b) Find the equations of the tangent and normal line(s) at the point(s) given. (c) Sketch the graph of the parametric functions along with the found tangent and normal lines. \(x=t^{2}-t, y=t^{2}+t ; \quad t=1\)

Short answer

(a) \( \frac{dy}{dx} = \frac{2t + 1}{2t - 1} \); (b) Tangent: \( y = 3x + 2 \), Normal: \( y = -\frac{1}{3}x + 2 \); (c) Sketch includes curve and lines.

Step-by-step solution

Differentiate x with respect to t

First, find \( \frac{dx}{dt} \). The given equation is \( x = t^2 - t \). Differentiating with respect to \( t \), we get: \[ \frac{dx}{dt} = 2t - 1. \]

Differentiate y with respect to t

Next, find \( \frac{dy}{dt} \). The given equation is \( y = t^2 + t \). Differentiating with respect to \( t \), we get: \[ \frac{dy}{dt} = 2t + 1. \]

Calculate dy/dx using chain rule

Use the chain rule \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). Thus, \[ \frac{dy}{dx} = \frac{2t + 1}{2t - 1}. \]

Find dy/dx at t = 1

Substitute \( t = 1 \) into \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{2(1) + 1}{2(1) - 1} = \frac{3}{1} = 3. \]

Find the point on the curve at t = 1

Substitute \( t = 1 \) into the parametric equations \( x = t^2 - t \) and \( y = t^2 + t \): \[ x = 1^2 - 1 = 0 \] \[ y = 1^2 + 1 = 2 \] The point is \((0, 2)\).

Find the equation of the tangent line

The slope at the point is 3. Using the point-slope form \( y - y_1 = m(x - x_1) \), where \((x_1, y_1) = (0, 2)\): \[ y - 2 = 3(x - 0) \] Simplifies to \( y = 3x + 2 \).

Find the equation of the normal line

The slope of the normal line is the negative reciprocal of the tangent slope, \(-\frac{1}{3}\). Using the same point: \[ y - 2 = -\frac{1}{3}(x - 0) \] Simplifies to \( y = -\frac{1}{3}x + 2 \).

Sketch graph with lines

Sketch the curve by substituting several values of \( t \) into the parametric equations. Along with the curve \((x,y) = (t^2-t, t^2+t)\), draw the tangent line \( y = 3x + 2 \) and the normal line \( y = -\frac{1}{3}x + 2 \).

Key concepts

Tangent Line

The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point without crossing it. It's essentially the best linear approximation of the curve at that particular location. To find the tangent line of a curve given by parametric equations, such as - \( x = t^2 - t \) - \( y = t^2 + t \), you first need to determine the slope of the tangent line at your point of interest. To do this, calculate \(\frac{d y}{d x}\), which represents the rate of change of \( y \) with respect to \( x \). This involves finding how quickly \( y \) changes as \( x \) changes, which gives us the slope of the tangent line. Let’s take the specific example from the problem, which involves finding the tangent line at the point where \( t = 1 \). First, evaluate \( \frac{d y}{d x} \) by substituting \( t = 1 \) into the derived expression \( \frac{2t + 1}{2t - 1} \), and obtain the slope \( 3 \). Next, using the point-slope form of a line, - Point: \((0, 2)\) - Slope: \(3\). The equation becomes \( y = 3x + 2 \). This line is the tangent to the curve at the given point.

Normal Line

A normal line to a curve at a particular point is an important geometric concept. It's a line that is perpendicular to the tangent line at that point. Understanding the normal line can help in visualizing how a curve behaves locally around that point. To find the equation of the normal line, first calculate the slope of the tangent line as explained before. For our specific case, the slope of the tangent line at \( t = 1 \) was determined to be \( 3 \). The slope of the normal line is the negative reciprocal of the tangent slope. Here, that means turning the tangent slope \( 3 \) into \( -\frac{1}{3} \). Once you have the slope, use the point-slope form of a line with the slope of \( -\frac{1}{3} \) and point \((0, 2)\): \[ y - 2 = -\frac{1}{3}(x - 0) \] simplifies to \( y = -\frac{1}{3}x + 2 \). This line is perpendicular to the tangent line and passes through the point \((0, 2)\).

Differentiation

Differentiation is a fundamental concept in calculus that represents how a function changes as its input changes. It's a way to find the rate at which something is happening, which is crucial when dealing with curves described by parametric equations. In our exercise, we use differentiation to find derivatives that tell us how \( x \) and \( y \) change with respect to the parameter \( t \). For the curve given by: - \( x = t^2 - t \) - \( y = t^2 + t \) We first differentiate these with respect to \( t \): \[ \frac{dx}{dt} = 2t - 1 \] \[ \frac{dy}{dt} = 2t + 1 \] These derivatives are essential in using the chain rule to find \( \frac{d y}{d x} \), which represents the slope of the tangent line to the curve at any point.

Chain Rule

The chain rule is a powerful tool in calculus used for differentiating composite functions. It tells us how to differentiate the outer function based on the differentiation of the inner function. In terms of parametric equations, it provides a method to find \(\frac{d y}{d x}\) when both \(x\) and \(y\) are given in terms of another variable, typically \(t\). In our problem, \(x = t^2 - t\) and \(y = t^2 + t\) are both functions of \(t\). To find \(\frac{dy}{dx}\), we use: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] The chain rule effectively combines partial derivatives \(\frac{dy}{dt} = 2t + 1\) and \(\frac{dx}{dt} = 2t - 1\) to express how \(y\) changes relative to \(x\). This step is crucial to determining the slope of the tangent line at any point along the curve, especially for \(t = 1\). The result, \(\frac{3}{1} = 3\), is used to write the equation for the tangent line as we explored in earlier sections.