Question

For the following exercises, find the area of the region. $$x=t^2,y=\ln (t),0\le t\le e$$

Short answer

The area is $\frac{e^2}{2}$.

Step-by-step solution

Understand the Parametric Equations

The problem provides the parametric equations: $x=t^2$ and $y=\ln (t)$, where $t$ ranges from 0 to $e$. We are asked to find the area under the curve described by these parametric equations.

Set up the Area Integral for Parametric Equations

For parametric curves, the area under the curve from $t=a$ to $t=b$ is given by the integral $A={\int }_a^{b}y\cdot \frac{dx}{dt}dt$. In our problem, $a=0$ and $b=e$. Thus, we have: $$A={\int }_0^e\ln (t)\cdot \frac{d}{dt}(t^2)dt.$$

Compute the Derivative $\frac{dx}{dt}$

Differentiate $x=t^2$ with respect to $t$ to find $\frac{dx}{dt}$. This gives us $\frac{dx}{dt}=2t$.

Substitute in the Integral

Substitute $y=\ln (t)$ and $\frac{dx}{dt}=2t$ into the integral: $$A={\int }_0^e\ln (t)\cdot 2tdt=2{\int }_0^{e}t\ln (t)dt.$$

Use Integration by Parts

To evaluate $\int t\ln (t)dt$, use integration by parts. Let $u=\ln (t)$ and $dv=tdt$. Then $du=\frac{1}{t}dt$ and $v=\frac{t^2}{2}$. Apply the integration by parts formula $\int udv=uv-\int vdu$.

Apply Integration by Parts

Calculate: $$\int t\ln (t)dt=\frac{t^2}{2}\ln (t)-\int \frac{t^2}{2}\cdot \frac{1}{t}dt.$$ Simplify the second integral: $\int \frac{t^2}{2t}dt=\int \frac{t}{2}dt=\frac{t^2}{4}.$

Evaluate the Antiderivative

The antiderivative becomes $\frac{t^2}{2}\ln (t)-\frac{t^2}{4}$. Multiply by 2 for the original area problem: $$2(\frac{t^2}{2}\ln (t)-\frac{t^2}{4})=t^2\ln (t)-\frac{t^2}{2}.$$

Compute the Definite Integral

Evaluate from $t=0$ to $t=e$: $${[t^2\ln (t)-\frac{t^2}{2}]}_0^e.$$ Calculate at $t=e$: $e^2\ln (e)-\frac{e^2}{2}=e^2-\frac{e^2}{2}=\frac{e^2}{2}.$ The term at $t=0$ is $0$.

Conclude the Computation

The computation results in $\frac{e^2}{2}$. Hence, the area of the region enclosed by this parametric curve is $\frac{e^2}{2}$.

Key concepts

Area Under Curve

Finding the area under a curve is a fundamental skill in calculus. It's all about measuring the size of the region that lies between the curve and the x-axis over a certain interval. In this case, we're dealing with parametric equations, which can describe more complex curves that depend on a parameter such as time.
The area enclosed by a parametric curve can be calculated using an integral. Usually, we use definite integrals that compute the area between two points on the parameter. Here's a step-by-step:

• Identify the parametric equations. These are given by the functions for x and y.
• Set up the integral based on the parametric equations. You integrate the product of y and the derivative of x with respect to the parameter over the respective limits.
• Calculate the integral to find the exact area under the curve.

Understanding how to find the area under a parametric curve helps in solving a wider range of mathematical problems beyond traditional Cartesian coordinates.

Integration by Parts

Integration by parts is a technique that helps solve integrals involving the product of two functions, and it's a bit like the product rule for differentiation but in reverse.
In calculus, integration by parts uses the formula:$$\int udv=uv-\int vdu$$Here's how you apply it:

• Choose which function to differentiate (u) and which to integrate (dv).
• Differentiate u to get du, and integrate dv to get v.
• Substitute into the formula to simplify the integral.
• Calculate the resulting integrals to solve the problem.

The trick is selecting u and dv wisely—often functions that simplify through differentiation or that you know how to integrate efficiently. In our exercise, the choice of $u=\ln (t)$ and $dv=tdt$ makes the products and integrals manageable. Practicing this technique allows you to manipulate more complex integrals in everyday calculus problem solving.

Calculus Problem Solving

When approaching calculus problems, especially those involving integrals and derivatives, a systematic approach is crucial.
This systematic method involves:

• Understanding the problem: Identify what is being asked and translate it into mathematical terms using equations and derivatives.
• Setting up the problem: Formulate the correct integral or derivative equations. Pay close attention to given limits or specific conditions.
• Solve the equations: Employ different calculus methods like substitution, integration by parts, or partial fractions.
• Validate the solution: By checking units or evaluating plausibility, confirm the solution aligns with the problem's requirements.

Problem solving in calculus is much like piecing together a puzzle. Each technique you master enriches your toolbox, helping tackle novel challenges beyond textbook exercises. Gaining comfort with these strategies is key to becoming adept at calculus.

Parametric Integral

In parametric equations, both x and y depend on an independent parameter, often t. Calculating an integral in this form requires understanding how each variable changes with t.
Here's how to compute a parametric integral step-by-step:

• Identify the derivatives necessary for substitution. Here, it involves $\frac{dx}{dt}$, which you find by differentiating the parametric equation for x with respect to the parameter.
• Setup the area integral: Combine the y function and the derivative you found into the integral, covering the specified range of the parameter.
• Solve the integral: Simplify and evaluate it using appropriate techniques, such as integration by parts if needed.

A parametric integral is particularly useful for finding areas under curves that are not efficient to represent using conventional Cartesian equations. By mastering this, you are better prepared for dynamic mathematical problems that involve motion or time-dependent scenarios.