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Chapter 11: Problem 78

Let \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\). a. Assume that \(\lim \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3}\right\rangle,\) which means that \(\lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0 .\) Prove that \(\lim _{t \rightarrow a} f(t)=L_{1}, \quad \lim _{t \rightarrow a} g(t)=L_{2}, \quad\) and \(\quad \lim _{t \rightarrow a} h(t)=L_{3}\). b. Assume that \(\lim _{t \rightarrow a} f(t)=L_{1}, \lim _{t \rightarrow a} g(t)=L_{2},\) and \(\lim _{t \rightarrow a} h(t)=L_{3} .\) Prove that \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3}\right\rangle\) which means that \(\lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0\).

Short Answer

Expert verified
Based on the step-by-step solution, create a short-answer question. Question: Prove that if the limit of a given vector function 饾惈(饾憽) equals another vector 饾悑, then the individual components of 饾惈(饾憽) will have the same limits as their respective components in 饾悑; and conversely, if the individual components of 饾惈(饾憽) have the same limits as the respective components of 饾悑, then the limit of 饾惈(饾憽) equals 饾悑.

Step by step solution

01

Write the given information

Given that: $$ \lim_{t \to a}\mathbf{r}(t) = \mathbf{L} = \langle L_{1}, L_{2}, L_{3} \rangle, $$ which means: $$ \lim_{t \to a}|\mathbf{r}(t) - \mathbf{L}| = 0; $$ We are asked to show that: $$ \lim_{t \to a} f(t) = L_{1}, \quad \lim_{t \to a} g(t) = L_{2}, \quad \textrm{ and } \quad \lim_{t \to a} h(t) = L_{3}. $$
02

Using the fact that 饾惈(饾憽) approaches 饾悑

As we have: $$ \lim_{t \to a}|\mathbf{r}(t) - \mathbf{L}| = 0, $$ we can use this information to find the limit of the individual components by breaking them down as follows: $$ \lim_{t \to a}| \langle f(t) - L_{1}, g(t) - L_{2}, h(t) - L_{3} \rangle | = 0; $$
03

Using the absolute vector norm

We apply the definition of the absolute value of a vector to obtain: $$ \lim_{t \to a}\sqrt{(f(t) - L_{1})^2 + (g(t) - L_{2})^2 + (h(t) - L_{3})^2} = 0; $$
04

Proving limits of individual functions

Since the square root is a continuous function, we can say that: $$ \lim_{t \to a}((f(t) - L_{1})^2 + (g(t) - L_{2})^2 + (h(t) - L_{3})^2) = 0; $$ From this, we can deduce that each individual term must approach zero as t approaches a, which means: $$ \lim_{t \to a} (f(t) - L_{1})^2 = 0, \lim_{t \to a} (g(t) - L_{2})^2 = 0, \textrm{ and } \lim_{t \to a} (h(t) - L_{3})^2 = 0; $$ Taking square roots of each of these equalities, we arrive at the required limits, as follows: $$ \lim_{t \to a} f(t) = L_{1}, \quad \lim_{t \to a} g(t) = L_{2}, \quad \textrm{ and } \quad \lim_{t \to a} h(t) = L_{3}. $$ #Part b: Proving limit of 饾惈(饾憽) when component functions have the same limits#
05

Write the given information

Given that: $$ \lim_{t \to a} f(t) = L_{1}, \quad \lim_{t \to a} g(t) = L_{2}, \quad \lim_{t \to a} h(t) = L_{3}; $$ We are asked to show that: $$ \lim_{t \to a} \mathbf{r}(t) = \mathbf{L} = \langle L_{1}, L_{2}, L_{3} \rangle, $$ which means: $$ \lim_{t \to a} |\mathbf{r}(t) - \mathbf{L}| = 0. $$
06

Express the limit using component functions

We begin by representing the limit as:| $$ \lim_{t \to a}|\langle f(t) - L_{1}, g(t) - L_{2}, h(t) - L_{3} \rangle| = 0. $$
07

Using the absolute vector norm

We can apply the definition of the absolute value of a vector to obtain: $$ \lim_{t \to a}\sqrt{(f(t) - L_{1})^2 + (g(t) - L_{2})^2 + (h(t) - L_{3})^2} = 0. $$
08

Plug in the component function limits

Now we substitute the given limits of the component functions: $$ \lim_{t \to a}(f(t) - L_{1})^2 = 0, \quad \lim_{t \to a} (g(t) - L_{2})^2 = 0, \quad \lim_{t \to a} (h(t) - L_{3})^2 = 0. $$
09

Proving the limit of 饾惈(饾憽)

Since these individual limits converge to zero, their sum also converges to zero: $$ \lim_{t \to a}\left((f(t) - L_{1})^2 + (g(t) - L_{2})^2 + (h(t) - L_{3})^2\right) = 0; $$ Thus, we can conclude that: $$ \lim_{t \to a} \mathbf{r}(t) = \mathbf{L} = \langle L_{1}, L_{2}, L_{3} \rangle, $$ which means: $$ \lim_{t \to a} |\mathbf{r}(t) - \mathbf{L}| = 0. $$

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Key Concepts

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

Vector Calculus
Vector calculus is an essential part of mathematics, especially in fields like physics and engineering. It helps us understand and work with functions that have multiple variables and components. In vector calculus, vectors are used to represent quantities that have both magnitude and direction, making them quite different from regular scalar values. These vectors can consist of several "components"; in the given exercise, we have a vector \(\mathbf{r}(t)\) with components \(f(t), g(t), h(t).\)
This subject aims to extend traditional calculus concepts, such as derivatives and integrals, to vector fields. It allows for the analysis of complex systems, such as fluid dynamics or electromagnetism, where multiple dimensions are involved.
  • Gradients: These describe the rate and direction of change in scalar fields.
  • Divergence: It measures the tendency of a vector field to originate from or converge into a point.
  • Curl: This indicates the rotation or twisting of a vector field around a point.
By understanding these concepts, vector calculus provides tools to solve real-world problems that involve multivariable functions, making it indispensable in scientific research.
Componentwise Limits
Componentwise limits are an essential concept when dealing with vector-valued functions. They allow us to break down complex vectors into individual components and analyze them separately. Suppose we have a vector function \(\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle\). The componentwise limit involves determining the limit of each component function as the variable \(t\) approaches a specific value \(a\).
In our exercise, we're given that the entire vector converges to a point \(\mathbf{L} = \langle L_1, L_2, L_3 \rangle\). However, to analyze it componentwise means checking if:
  • \(\lim_{{t \to a}} f(t) = L_1\)
  • \(\lim_{{t \to a}} g(t) = L_2\)
  • \(\lim_{{t \to a}} h(t) = L_3\)
This approach simplifies the problem by ensuring that each component of the vector function approaches a specific limit value. It effectively transforms the multi-dimensional problem into smaller, easier to solve, scalar problems, allowing you to simultaneously analyze the behavior of all components.
Continuity of Vector Functions
In mathematical terms, the continuity of a function guarantees that small changes to the input result in small changes to the output. This concept is extended to vector-valued functions in a very intuitive way. A vector function \(\mathbf{r}(t)\) is continuous at a point \(a\) if the vector approaches its limit \(\mathbf{L}\) as \(t\) approaches \(a\). It means that the entire expression \(|\mathbf{r}(t) - \mathbf{L}|\) goes to zero.
This principle applies to each component of the vector. If \(f(t), g(t)\), and \(h(t)\) are continuous, their combined function \(\mathbf{r}(t)\) is also continuous. Each individual function must be continuous over its domain for this property to hold.
When proving continuity in vector functions, remember:
  • The limit of the vector function \(\mathbf{r}(t)\) coincides with the vector \(\mathbf{L}\).
  • All component functions \(f(t), g(t), h(t)\) must be approached separately and must independently meet their limits \(L_1, L_2, L_3\) respectively.
Thus, confirming componentwise continuity affirms the overall continuity of a vector-valued function. This ensures that the function behaves in a predictable manner as its variable approaches a given point.

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Most popular questions from this chapter

Compute the following derivatives. $$\frac{d}{d t}\left(\left(3 t^{2} \mathbf{i}+\sqrt{t} \mathbf{j}-2 t^{-1} \mathbf{k}\right) \cdot(\cos t \mathbf{i}+\sin 2 t \mathbf{j}-3 t \mathbf{k})\right)$$

Derivative rules Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are differentiable functions at \(t=0\) with \(\mathbf{u}(0)=\langle 0,1,1\rangle, \mathbf{u}^{\prime}(0)=\langle 0,7,1\rangle\) \(\mathbf{v}(0)=\langle 0,1,1\rangle,\) and \(\mathbf{v}^{\prime}(0)=\langle 1,1,2\rangle .\) Evaluate the following expressions. a. \(\left.\frac{d}{d t}(\mathbf{u} \cdot \mathbf{v})\right|_{t=0}\) b. \(\left.\frac{d}{d t}(\mathbf{u} \times \mathbf{v})\right|_{t=0}\) c. \(\left.\frac{d}{d t}(\cos t \mathbf{u}(t))\right|_{t=0}\)

Suppose the vector-valued function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is smooth on an interval containing the point \(t_{0} .\) The line tangent to \(\mathbf{r}(t)\) at \(t=t_{0}\) is the line parallel to the tangent vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\) that passes through \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right) .\) For each of the following functions, find an equation of the line tangent to the curve at \(t=t_{0} .\) Choose an orientation for the line that is the same as the direction of \(\mathbf{r}^{\prime}\). $$\mathbf{r}(t)=\left\langle e^{t}, e^{2 t}, e^{3 t}\right\rangle ; t_{0}=0$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The vectors \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\) are parallel for all values of \(t\) in the domain. b. The curve described by the function \(\mathbf{r}(t)=\left\langle t, t^{2}-2 t, \cos \pi t\right\rangle\) is smooth, for \(-\infty

Suppose the vector-valued function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is smooth on an interval containing the point \(t_{0} .\) The line tangent to \(\mathbf{r}(t)\) at \(t=t_{0}\) is the line parallel to the tangent vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\) that passes through \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right) .\) For each of the following functions, find an equation of the line tangent to the curve at \(t=t_{0} .\) Choose an orientation for the line that is the same as the direction of \(\mathbf{r}^{\prime}\). $$\mathbf{r}(t)=\langle 2+\cos t, 3+\sin 2 t, t\rangle ; t_{0}=\pi / 2$$
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