Question

For the following exercises, find the arc length of the curve on the indicated interval of the parameter.$x=4t+3,y=3t-2,0\le t\le 2$

Short answer

The arc length of the curve is 10 units.

Step-by-step solution

Understand the Arc Length Formula

The arc length of a curve defined by parametric equations $x=f(t)$ and $y=g(t)$ over an interval $a\le t\le b$ is given by the formula: $$L={\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$$

Differentiate the Parametric Equations

Differentiate $x=4t+3$ with respect to $t$: $\frac{dx}{dt}=4$. Differentiate $y=3t-2$ with respect to $t$: $\frac{dy}{dt}=3$.

Substitute into the Arc Length Formula

Substitute $\frac{dx}{dt}=4$ and $\frac{dy}{dt}=3$ into the arc length formula: $$L={\int }_0^2\sqrt{4^2+3^2}dt$$ This simplifies to $$L={\int }_0^2\sqrt{16+9}dt={\int }_0^2\sqrt{25}dt={\int }_0^{2}5dt$$

Evaluate the Integral

Evaluate the integral ${\int }_0^{2}5dt$: $$L=5\times [t{]}_0^2=5\times (2-0)=10$$

Conclusion

The arc length of the curve for the given parametric equations over the interval $0\le t\le 2$ is $10$.

Key concepts

Parametric Equations

Parametric equations are a powerful way to describe a curve in the plane, involving expressions for the coordinates as functions of a third variable, typically denoted as $t$ (the parameter). In our exercise, the functions $x=4t+3$ and $y=3t-2$ describe the motion along a curve, where $t$ varies from 0 to 2.

These equations allow us to express both $x$ and $y$ in terms of $t$ rather than finding $y$ solely as a function of $x$. This is very helpful when the curve may not easily be described by a single function of $x$. Parametric equations prove advantageous for complex shapes and facilitate the calculation of various calculus problems, including computing arc lengths.

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes. In the context of parametric equations, differentiation helps us ascertain how the $x$ and $y$ coordinates change with respect to the parameter $t$.

For our parametric equations, we differentiate $x=4t+3$ and $y=3t-2$ to find:

• The derivative of $x$ with respect to $t$, noted as $\frac{dx}{dt}=4$.
• The derivative of $y$ with respect to $t$, denoted as $\frac{dy}{dt}=3$.

These derivatives indicate constant rates of change, suggesting a linear relationship storming through a straight path, making it simpler to handle in terms of calculations.

Integrals

Integral calculus relates to finding the total accumulation of certain quantities, such as area under the curve or net displacement. In the context of this exercise, we use integrals to sum up small pieces of the curve's length into a total arc length.

By substituting our derivatives from the differentiation step into the arc length formula, we set up the integral: $${\int }_0^2\sqrt{(4{)}^2+(3{)}^2}dt$$This evaluates into a simpler form due to constant derivatives where integration becomes straightforward.

Curve Length Calculation

Once the stage is set with parameters, derivatives, and the integral expression, we're ready to calculate the arc length of the curve. The integral simplifies as: $${\int }_0^{2}5dt$$which is easy to solve, resulting in the expression $5\times t$ evaluated from 0 to 2.

So, we compute: $$L=5\times (2-0)=10$$

The final answer $L=10$ signifies that the total length of the curve described by the given parametric equations over the interval is 10 units. This calculation is often crucial not only in theoretical mathematics but also in areas like physics, engineering, and computer graphics to ascertain distances traced along a path.