Problem 9 - Calculus Early Transcendentals

Question

Find an equation of the tangent(s) to the curve at the given point. Then graph the curve and the tangent(s). $$x=6 \sin t, \quad y=t^{2}+t ; \quad(0,0)$$

Step-by-Step Solution

Answer

The tangent line at (0,0) is \(y = \frac{1}{6}x\).

Step 1: Parametric Equations and Point Check

First, confirm the point \((0,0)\) is on the curve defined by the parametric equations \(x=6\sin t\) and \(y=t^2+t\). To satisfy \(x=0\), we need \(6\sin t=0\), implying \(t=0,\pi\), etc. Applying \(t=0\) in \(y=t^2+t\) gives \(y=0\). Thus, \((0,0)\) corresponds to \(t=0\).

Step 2: Find Derivatives

To find the slope of the tangent at \(t=0\), determine the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). Compute:\[ \frac{dx}{dt} = 6\cos t \] and \[ \frac{dy}{dt} = 2t + 1 \].

Step 3: Slope of the Tangent Line

Use the chain rule to find \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t+1}{6\cos t} \]. At \(t=0\), \(\frac{dy}{dt}=1\), \(\frac{dx}{dt}=6\) and therefore, \( \frac{dy}{dx} = \frac{1}{6} \).

Step 4: Equation of the Tangent Line

Using the point-slope form for tangents, the equation is: \(y-y_1 = m(x-x_1)\). Substituting \(m = \frac{1}{6}\), \((x_1, y_1) = (0,0)\), gives:\[ y = \frac{1}{6}x \].

Step 5: Graphing the Curve and Tangent

Plot the graph of the curve parameterized by \((6\sin t, t^2 + t)\). Then draw the tangent line given by \(y = \frac{1}{6}x\). At \(t=0, \pi,\ldots\), the curve intersects the tangent line at \((0,0)\), confirming tangency.

Key Concepts

Parametric Equations

Parametric equations describe a set of related mathematical values where one or more variables are expressed as functions of a separate parameter. For instance, in the exercise dealing with the parametric equations \(x=6\sin t\) and \(y=t^2+t\), \(t\) is the parameter. It controls the movement along the curve represented by \(x\) and \(y\). Parametric equations offer an efficient way to represent complex curves using one variable, commonly modeled in physics or engineering problems.

In our problem, we start by verifying whether the point \((0,0)\) lies on the curve. By setting the equations equal to zero, \(6\sin t = 0\) implies \(t=0,\pi,\ldots\). Substituting \(t=0\) into \(y=t^2+t\), we see \(y=0\), confirming that \((0,0)\) is indeed a point on the curve.

Derivatives of Parametric Functions

Derivatives help us understand how one variable changes in relation to another. In parametric functions, we calculate derivatives with respect to a common parameter like \(t\). These derivatives provide crucial information about the curve's slope at any given point.

To find these derivatives, we compute \(\frac{dx}{dt}=6\cos t\) and \(\frac{dy}{dt}=2t+1\). These derivatives indicate how \(x\) and \(y\) change as \(t\) varies. For determining the slope or the tangent's steepness on the curve, these derivatives become foundational in later steps.

Slope of Tangent Line

The slope of the tangent line represents how steep a line is touching the curve at a specific point. To calculate this slope in parametric equations, we use the chain rule to determine \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\). This is derived from the ratio \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).

In this exercise, we have \(\frac{dy}{dt} = 2t + 1\) and \(\frac{dx}{dt} = 6\cos t\). Evaluating at \(t=0\), \(\frac{dy}{dx} = \frac{1}{6}\), giving the slope of the tangent line at point \((0,0)\). The slope signifies how the curve rises or falls as the curve runs along the x-axis.

Point-Slope Form

The point-slope form of a line's equation helps us quickly find a line equation when given a point on the line and the line's slope. The formula is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line.

In our exercise, the slope \(m\) is \(\frac{1}{6}\) and the point is \((0,0)\). Plugging these into the point-slope form gives \(y = \frac{1}{6}x\). This formula describes the tangent line intersecting the parametric curve at the given point.

Graphing Parametric Curves

Graphing parametric curves provides a visual understanding of the curve's behavior and geometry. When plotting, each value of \(t\) generates a point \((x, y)\), creating a smooth, continuous path.

In this task, we graph the parametric curve defined by \((6\sin t, t^2 + t)\) and its tangent line \(y = \frac{1}{6}x\). By observing the plot, we see where the tangent line just touches the curve at \((0,0)\), illustrating the concept of tangency. Graphically verifying solutions helps solidify understanding and validate theoretical findings.