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Chapter 17: Problem 19

Scalar line integrals Evaluate the following line integrals along the curve \(C\). \(\int_{C}(2 x+y) d s ; C\) is the line segment \(\mathbf{r}(t)=\langle 3 t, 4 t\rangle,\) for \(0 \leq t \leq 2\)

Short Answer

Expert verified
Based on the step by step solution provided above, answer the following question: Q: Find the scalar line integral of the function f(x, y) = 2x + y along the curve given by the parametric equations r(t) = <3t, 4t> for 0 鈮� t 鈮� 2. A: The scalar line integral of the given function along the curve is equal to 200.

Step by step solution

01

Parameterize the curve

The curve \(C\) is given by the parametric equations: $$\mathbf{r}(t) = \langle 3t, 4t \rangle $$ for \(0 \leq t \leq 2\). So we have \(x(t) = 3t\) and \(y(t) = 4t\).
02

Compute the differential element \(ds\)

Now we need to find the differential element \(ds\) which is given by: $$ds = \sqrt{(dx/dt)^2 + (dy/dt)^2} dt$$ To find \(ds\), first compute \(dx/dt\) and \(dy/dt\). We have: $$dx/dt = \frac{d}{dt}(3t) = 3$$ $$dy/dt = \frac{d}{dt}(4t)= 4$$ Now we can compute \(ds\) as follows: $$ds = \sqrt{(3)^2 + (4)^2} dt = 5 dt$$
03

Replace \(x\) and \(y\) by their parametric expressions and rewrite \(ds\) in terms of \(dt\)

Now we need to substitute \(x(t)\) and \(y(t)\) into the scalar function \(f(x, y)\) as well as replace \(ds\) by its corresponding expression in terms of \(dt\): $$f(x(t), y(t)) = 2(3t) + 4t = 10t$$ Thus, the scalar line integral becomes: $$\int_C (2 x+y) ds = \int_{0}^{2} 10t \cdot 5 dt$$
04

Evaluate the integral

Now we can evaluate the definite integral: $$\int_{0}^{2} 10t \cdot 5 dt = 50 \int_{0}^{2} t dt = 50 \left[\frac{1}{2}t^2\right]_0^2 = 50(2^2) - 50(0) = 200$$ Therefore, the scalar line integral of the given function along the given curve is equal to 200.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Equations
Parametric equations are a powerful tool for representing curves in mathematics. They describe a path or a curve in space by expressing the coordinates as functions of a single variable, typically denoted by t. This t can be envisioned as time, guiding a point along the curve as it changes.

For instance, the given parametric equations of the curve C are
\[ \mathbf{r}(t) = \langle 3t, 4t \rangle \]
Here, x(t) = 3t and y(t) = 4t define the position of a point on the line segment for each value of t from 0 to 2. By representing curves parametrically, we gain the flexibility to work with more complex shapes and paths compared to traditional Cartesian equations.
Differential Element ds
The differential element ds represents an infinitesimally small segment of the curve's length. It's a crucial part of computing line integrals as it measures how much 'distance' we are moving along the curve with each tiny step in the parameter t.

Mathematically, for a curve described by parametric equations x(t) and y(t), ds is given by:
\[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt \]
By finding the derivatives dx/dt and dy/dt, and then applying the Pythagorean theorem to these rates of change, we obtain the formula for ds. In our specific case, it simplifies to 5 dt, indicating a proportional relationship between ds and dt on a straight line.
Definite Integral Evaluation
Evaluating definite integrals is a fundamental process in calculus that allows us to calculate the accumulation of quantities, like area under a curve. For line integrals, it helps in summing up values along a curve.

It involves finding the antiderivative of the function being integrated and then applying the Fundamental Theorem of Calculus. Specifically:
\[ \int_{a}^{b} f(t) dt = F(b) - F(a) \]
where F is the antiderivative of f. In the context of our problem, we evaluated
\[ \int_{0}^{2} 10t \cdot 5 dt \]
by finding the antiderivative of 10t, multiplying by 5, and then applying the evaluated limits from 0 to 2, leading to the result of 200.
Line Integral Computation
Line integral computation involves integrating a function over a curve, which in our case, is the scalar line integral. It's a tool for summing values of a function along a path in a vector field. A scalar line integral, specifically, calculates the integral of a scalar field over a curve, often representing physical concepts like work done by a force field.

In our example, once we have the parametric representation and ds, we substitute these into the integral, rewriting it with respect to dt as a single-variable integral. Then we perform the definite integral evaluation to find the total accumulated value along the curve, which gave us the final answer of 200 for the given problem.

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Most popular questions from this chapter

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