Question

Sketch the graph of the given parametric equation and find its length. \(x=3 t^{2}+2, y=2 t^{3}-1 / 2 ; 1 \leq t \leq 4\)

Short answer

The curve length is approximately 136.324 units.

Step-by-step solution

Understand the Parametric Equations

Given are parametric equations: \(x = 3t^2 + 2\) and \(y = 2t^3 - \frac{1}{2}\) with the parameter \(t\) ranging from 1 to 4. The task involves sketching the graph described by these equations and finding the arc length of the curve.

Compute Derivatives

To find the arc length of the curve, calculate the derivatives of \(x\) and \(y\) with respect to \(t\): \(\frac{dx}{dt} = \frac{d}{dt}(3t^2 + 2) = 6t\), and\(\frac{dy}{dt} = \frac{d}{dt}(2t^3 - \frac{1}{2}) = 6t^2\).

Set Up Arc Length Integral

The formula for the arc length \(L\) of a parametric curve \((x(t), y(t))\) from \(t = a\) to \(t = b\) is given by:\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \].Substitute the derivatives:\[ L = \int_{1}^{4} \sqrt{(6t)^2 + (6t^2)^2} \, dt \].

Simplify the Integral

Simplify the expression inside the square root:\( (6t)^2 + (6t^2)^2 = 36t^2 + 36t^4 = 36(t^2 + t^4) = 36t^2(1 + t^2) \).Hence, \[ L = \int_{1}^{4} 6t \sqrt{1 + t^2} \, dt \].

Solve the Integral

To solve the integral \(\int 6t \sqrt{1 + t^2} \, dt\), use the substitution \(u = 1 + t^2\) which leads to \(du = 2t \, dt\) or \(t \, dt = \frac{du}{2}\).Transform the integral:\[ L = \int \sqrt{u} \cdot 3 \, du \] which becomes \( \int 3u^{1/2} \, du\).Integrating gives \( L = 3 \cdot \frac{2}{3} u^{3/2} = 2[u^{3/2}] \), back-substitute \( u = 1 + t^2 \).

Evaluate the Integrated Expression

Substitute back and evaluate the expression from \(t = 1\) to \(t = 4\):\[ L = 2 [(1 + (4)^2)^{3/2} - (1 + (1)^2)^{3/2}] \].This simplifies to:\[ L = 2 [(17)^{3/2} - (2)^{3/2}] \].\((17)^{3/2} \approx 70.99 \) and \((2)^{3/2} \approx 2.828\).Therefore, \[ L = 2 [70.99 - 2.828] = 2 \times 68.162 \ \approx 136.324 \].

Sketch the Graph

Plot points for selected values of \(t\) (e.g., \(t = 1, 2, 3, 4\)) using the equations \(x = 3t^2 + 2\) and \(y = 2t^3 - \frac{1}{2}\). Connect them smoothly to form the curve. Identify the behavior and shape of the path based on these points.

Key concepts

Calculus

Calculus is a branch of mathematics that deals with the concepts of change and motion. It provides tools to model real-world situations and to understand how things change over time. The two main areas of calculus are differentiation and integration.

• **Differentiation** involves finding the derivative of a function, which represents the rate of change. It answers questions like "how fast is something changing?"
• **Integration**, on the other hand, deals with finding the area under a curve. It can be used to find total quantities, such as area or volume.

In problems involving parametric equations, calculus plays a crucial role in determining properties of the curve, like its arc length. By differentiating the parametric equations, we can understand how the curve behaves and further solve complex problems by integrating the results.

Arc Length

Finding the arc length of a curve is a common problem in calculus. The arc length is the distance along the curve between two points. For parametric equations, this involves integrating over a given interval.To calculate the arc length for the parametric curve defined by:- \ \( x = f(t) \) and \( y = g(t) \), the formula used is: \ \[ L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \].This formula combines the changes in both the \(x\) and \(y\) directions to find the total length of the curve. To solve it:- Compute the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). - Substitute these into the arc length formula. - Simplify if necessary, and integrate over the specified interval. Once integrated, it provides the length of the path traced by the parametric equations between the given parameter values.

Derivatives

Derivatives are fundamental to calculus and describe how a function changes as its input changes. In the context of parametric equations, derivatives help us understand the instantaneous rate of change of the curve's position in the \(x\) and \(y\) directions.To find these derivatives:- Differentiate the parametric equations with respect to the parameter \(t\).- For example, given \(x = 3t^2 + 2\), the derivative \( \frac{dx}{dt} = 6t \).- Likewise, for \(y = 2t^3 - \frac{1}{2}\), the derivative \( \frac{dy}{dt} = 6t^2 \).These derivatives reveal how fast the curve is moving horizontally and vertically as \(t\) changes. They are also crucial in computing the arc length. This is because the arc length formula uses the Pythagorean theorem, combining horizontal and vertical changes to measure the distance along the curve itself.

Parametric Curves

Parametric curves provide a flexible way to describe a path or trajectory in the plane by using equations in terms of a third variable, often labeled as \(t\). This is different from regular functions where \(y\) is directly a function of \(x\).

• Each value of \(t\) gives a point \((x(t), y(t))\), mapping out the curve as \(t\) varies.
• Parametric equations offer an advantage in representing complex curves, like loops or spirals, where a single function \(y=f(x)\) wouldn't suffice.

In our exercise, the equations \(x = 3t^2 + 2\) and \(y = 2t^3 - \frac{1}{2}\) trace a curve from \(t = 1\) to \(t = 4\). This introduces a method for sketching by evaluating specific points on the curve:1. Select various values of \(t\) within the given interval.2. Compute the corresponding \((x, y)\) values.3. Plot these points to visualize the trajectory of the parametric curve.