Problem 15
Question
Given that \(\mathbf{r}(t)=\frac{\sin 2 t}{t} \mathbf{i}+(t-2)^{5} \mathbf{j}+t \ln t \mathbf{k}\), find \(\lim _{t \rightarrow 0^{+}} \mathbf{r}(t)\)
Step-by-Step Solution
Verified Answer
The limit is \( 2\mathbf{i} - 32\mathbf{j} \).
1Step 1: Understand the problem
We need to find the limit of the vector function \( \mathbf{r}(t) = \frac{\sin 2t}{t} \mathbf{i} + (t-2)^5 \mathbf{j} + t \ln t \mathbf{k} \) as \( t \) approaches 0 from the positive side.
2Step 2: Analyze the components
The function \( \mathbf{r}(t) \) is composed of three separate components: \( \frac{\sin 2t}{t} \) in the \( \mathbf{i} \)-direction, \( (t-2)^5 \) in the \( \mathbf{j} \)-direction, and \( t \ln t \) in the \( \mathbf{k} \)-direction. We need to find the limit of each component separately as \( t \to 0^+ \).
3Step 3: Compute the limit for the \( \mathbf{i} \)-component
The limit of \( \frac{\sin 2t}{t} \) as \( t \to 0^+ \) can be simplified by recognizing that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Therefore, \( \lim_{t \to 0^+} \frac{\sin 2t}{t} = 2 \).
4Step 4: Compute the limit for the \( \mathbf{j} \)-component
The expression \( (t-2)^5 \) is continuous around \( t = 0 \). So, \( \lim_{t \to 0^+} (t-2)^5 = (-2)^5 = -32 \).
5Step 5: Compute the limit for the \( \mathbf{k} \)-component
To evaluate the limit \( \lim_{t \to 0^+} t \ln t \), we make the substitution \( x = \ln t \), which implies \( t = e^x \) and \( x \to -\infty \) as \( t \to 0^+ \). Then, \( \lim_{x \to -\infty} e^x x = 0 \) because the exponential decay dominates. Thus, \( \lim_{t \to 0^+} t \ln t = 0 \).
6Step 6: Combine the limits of the components
The limit of \( \mathbf{r}(t) \) as \( t \to 0^+ \) is found by combining the limits of the three components: \( 2 \mathbf{i} - 32 \mathbf{j} + 0 \mathbf{k} \). Hence, \( \lim_{t \to 0^+} \mathbf{r}(t) = 2\mathbf{i} - 32\mathbf{j} \).
Key Concepts
Limits of Vector FunctionsComponent-wise LimitsExponential Decay DominanceContinuity in Multivariable Functions
Limits of Vector Functions
In vector calculus, understanding the limit of a vector function is crucial, as it helps us explore the behavior of vector-valued functions near a specific point. A vector function can be thought of as a collection of component functions, each representing one dimension of the vector. For instance, when you have a vector function like \( \mathbf{r}(t) = \frac{\sin 2t}{t} \mathbf{i} + (t-2)^5 \mathbf{j} + t \ln t \mathbf{k} \), it combines three separate component functions: one for \( \mathbf{i} \), one for \( \mathbf{j} \), and one for \( \mathbf{k} \).
To find the limit of such a vector function as \( t \to 0^+ \), you must calculate the limit of each component individually and then combine these results. This approach helps break down complex vector functions into manageable parts and reflects how their behavior is intertwined at a point or approaching a point in their domain.
To find the limit of such a vector function as \( t \to 0^+ \), you must calculate the limit of each component individually and then combine these results. This approach helps break down complex vector functions into manageable parts and reflects how their behavior is intertwined at a point or approaching a point in their domain.
Component-wise Limits
Component-wise limits involve assessing each part of a vector function separately. This method ensures precise calculation of the limit for vector functions like \( \mathbf{r}(t) \) mentioned in our original exercise.
- For the \( \mathbf{i} \)-component, we consider \( \frac{\sin 2t}{t} \). Using the identity \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), the limit simplifies to \( 2 \).
- In the \( \mathbf{j} \)-component with \( (t-2)^5 \), this part of the vector function remains continuous around \( t = 0 \). Therefore, its limit directly computes to \( (-2)^5 = -32 \).
- Finally, the \( \mathbf{k} \)-component \( t \ln t \) poses a more challenging scenario. By substituting \( x = \ln t \) to change variables, we simplify the problem, recognizing that \( x \to -\infty \) as \( t \to 0^+ \), which aids in further solving this intricate component.
Exponential Decay Dominance
In calculus, the concept of exponential decay dominance often arises in problems involving limits that involve logarithms or exponential functions. It notably impacts the solutions involving terms like \( t \ln t \) in vector functions.
As seen in our exercise, we transformed \( t \ln t \) into an expression \( e^x x \) with the substitution \( x = \ln t \), simplifying it for analysis. As \( t \to 0^+ \), or equivalently \( x \to -\infty \), the exponential term \( e^x \) diminishes more rapidly than \( x \) increases in magnitude.
This rapid decrease to zero means that the limit of \( t \ln t \) results in \( 0 \). The decay of \( e^x \) emerges so strongly that it controls the behavior of the product, rendering the entire product infinitesimal as \( x \to -\infty \). This dominance of decay is why the \( \mathbf{k} \)-component in the limit of the vector function ultimately vanishes.
As seen in our exercise, we transformed \( t \ln t \) into an expression \( e^x x \) with the substitution \( x = \ln t \), simplifying it for analysis. As \( t \to 0^+ \), or equivalently \( x \to -\infty \), the exponential term \( e^x \) diminishes more rapidly than \( x \) increases in magnitude.
This rapid decrease to zero means that the limit of \( t \ln t \) results in \( 0 \). The decay of \( e^x \) emerges so strongly that it controls the behavior of the product, rendering the entire product infinitesimal as \( x \to -\infty \). This dominance of decay is why the \( \mathbf{k} \)-component in the limit of the vector function ultimately vanishes.
Continuity in Multivariable Functions
Understanding continuity in multivariable functions is essential as vector and scalar functions in several dimensions exhibit more complex behaviors than their single-variable counterparts. A function is continuous at a point if it can be defined at that point and the limit approaching the point equals the function's value at that point.
For our specific vector function, \((t-2)^5\) demonstrates continuity around \(t = 0\). In this case, the component function is a polynomial expression that remains well-defined and continuous, ensuring straightforward computation of its limit.
This concept of continuity assists in verifying when particular limits you calculate are reliable. If each component function is continuous at the point of interest and the entire function is defined there, then continuity implies that the vector function's limit can be determined solely through its components:
For our specific vector function, \((t-2)^5\) demonstrates continuity around \(t = 0\). In this case, the component function is a polynomial expression that remains well-defined and continuous, ensuring straightforward computation of its limit.
This concept of continuity assists in verifying when particular limits you calculate are reliable. If each component function is continuous at the point of interest and the entire function is defined there, then continuity implies that the vector function's limit can be determined solely through its components:
- This ensures a smoother transition from evaluating individual limits to concluding the behavior of the full vector function.
- Through component continuity, it becomes easier for students to trust the direct computation of limits from each function.
Other exercises in this chapter
Problem 15
Find the first partial derivatives of the given function. $$ z=5 x^{4} y^{3}-x^{2} y^{6}+6 x^{5}-4 y $$
View solution Problem 15
In Problems, find the directional derivative of the given function at the given point in the indicated direction. $$ f(x, y)=(x y+1)^{2} ;(3,2) \text { , in the
View solution Problem 15
Fill in the blank or answer true/false. Where appropriate, assume continuity of \(P, O\), and their first partial derivatives. Work done by a force \(F\) along
View solution Problem 15
Sketch the region \(D\) whose volume \(V\) is given by the iterated integral. $$ \int_{0}^{4} \int_{0}^{3} \int_{0}^{2-2 / 3} d x d z d y $$
View solution