/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Sum and Difference Rules Prove t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Sum and Difference Rules Prove that if \(\mathbf{u}\) and \(\mathbf{v}\) are differentiable functions of \(t,\) then $$ \frac{d}{d t}(\mathbf{u}+\mathbf{v})=\frac{d \mathbf{u}}{d t}+\frac{d \mathbf{v}}{d t} $$ and $$ \frac{d}{d t}(\mathbf{u}-\mathbf{v})=\frac{d \mathbf{u}}{d t}-\frac{d \mathbf{v}}{d t} $$

Short Answer

Expert verified
The derivative rules for sums and differences are proven using the limit definition.

Step by step solution

01

Definition of Derivative

The derivative of a function with respect to a variable measures how the function changes as that variable changes. For a function \( \mathbf{u}(t) \), the derivative \( \frac{d \mathbf{u}}{d t} \) represents the rate of change of \( \mathbf{u} \) with respect to \( t \).
02

Sum of Functions

Consider the function \( \mathbf{w}(t) = \mathbf{u}(t) + \mathbf{v}(t) \). The derivative of \( \mathbf{w} \) with respect to \( t \) is given by:\[ \frac{d}{d t} (\mathbf{u} + \mathbf{v}) = \lim_{\Delta t \to 0} \frac{(\mathbf{u}(t + \Delta t) + \mathbf{v}(t + \Delta t)) - (\mathbf{u}(t) + \mathbf{v}(t))}{\Delta t} \]
03

Apply the Limit Property

Using the limit property that \( \lim_{x \to a} (f(x) + g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \), rewrite the expression:\[ \frac{d}{d t} (\mathbf{u} + \mathbf{v}) = \lim_{\Delta t \to 0} \left( \frac{\mathbf{u}(t + \Delta t) - \mathbf{u}(t)}{\Delta t} + \frac{\mathbf{v}(t + \Delta t) - \mathbf{v}(t)}{\Delta t} \right) \]
04

Separating Limits

Separate the limits of the numerator's differences over \( \Delta t \):\[ \frac{d}{d t} (\mathbf{u} + \mathbf{v}) = \left( \lim_{\Delta t \to 0} \frac{\mathbf{u}(t + \Delta t) - \mathbf{u}(t)}{\Delta t} \right) + \left( \lim_{\Delta t \to 0} \frac{\mathbf{v}(t + \Delta t) - \mathbf{v}(t)}{\Delta t} \right) \]
05

Recognizing Individual Derivatives

Recognize that the separated limits are simply the derivatives of \( \mathbf{u} \) and \( \mathbf{v} \) respectively. Thus:\[ \frac{d}{d t} (\mathbf{u} + \mathbf{v}) = \frac{d \mathbf{u}}{d t} + \frac{d \mathbf{v}}{d t} \]
06

Difference of Functions

Consider the function \( \mathbf{w}(t) = \mathbf{u}(t) - \mathbf{v}(t) \). The derivative of \( \mathbf{w} \) with respect to \( t \) follows a similar pattern as shown in the sum:\[ \frac{d}{d t} (\mathbf{u} - \mathbf{v}) = \lim_{\Delta t \to 0} \frac{(\mathbf{u}(t + \Delta t) - \mathbf{v}(t + \Delta t)) - (\mathbf{u}(t) - \mathbf{v}(t))}{\Delta t} \]
07

Apply Limit for Difference Rule

Use the same limit property focusing on subtraction:\[ \frac{d}{d t} (\mathbf{u} - \mathbf{v}) = \left( \lim_{\Delta t \to 0} \frac{\mathbf{u}(t + \Delta t) - \mathbf{u}(t)}{\Delta t} \right) - \left( \lim_{\Delta t \to 0} \frac{\mathbf{v}(t + \Delta t) - \mathbf{v}(t)}{\Delta t} \right) \]
08

Recognizing Individual Derivatives Again

The separated limits are again the derivatives of \( \mathbf{u} \) and \( \mathbf{v} \) respectively. Hence:\[ \frac{d}{d t} (\mathbf{u} - \mathbf{v}) = \frac{d \mathbf{u}}{d t} - \frac{d \mathbf{v}}{d t} \]
09

Conclusion

We have shown that the derivative of the sum of two functions is the sum of their derivatives, and the derivative of the difference is the difference of their derivatives, proving the sum and difference rules.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sum Rule
When dealing with derivatives, one of the first rules you encounter is the Sum Rule. In simple terms, this rule tells us how to differentiate the sum of two functions. Let's say you have two functions \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \), both of which are differentiable. According to the Sum Rule, if you want to find the derivative of \( \mathbf{u}(t) + \mathbf{v}(t) \), you simply take the derivative of each function individually and then add these derivatives together.

So, mathematically speaking, if \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \) are functions of \( t \), the rule is expressed as:
  • \( \frac{d}{dt} (\mathbf{u} + \mathbf{v}) = \frac{d \mathbf{u}}{dt} + \frac{d \mathbf{v}}{dt} \)
This can be visualized as breaking down a complex task into simpler, more manageable tasks. You differentiate each component function and then combine your results, making it a straightforward way to handle the sum of functions.
Difference Rule
Closely related to the Sum Rule is the Difference Rule. This rule addresses how to differentiate the difference between two functions. It's almost like a mirror image of the Sum Rule, but instead of adding the derivatives, we subtract them.

Imagine you have two functions \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \) that are differentiable. To find the derivative of their difference \( \mathbf{u}(t) - \mathbf{v}(t) \), the Difference Rule tells us to take the derivative of each function separately and subtract one from the other.

Here’s how it looks mathematically:
  • \( \frac{d}{dt} (\mathbf{u} - \mathbf{v}) = \frac{d \mathbf{u}}{dt} - \frac{d \mathbf{v}}{dt} \)
By implementing the Difference Rule, you handle subtractions much like additions, only instead of summing up changes, you're calculating the difference in changes between the two functions.
Derivative of a Function
The derivative of a function is a core concept in calculus and serves as a measure of how a function changes as its input changes. In essence, it tells you the rate at which the function's value is changing at any given point.

Suppose you have a function \( \mathbf{u}(t) \); its derivative, denoted by \( \frac{d \mathbf{u}}{dt} \), indicates how \( \mathbf{u} \) changes with respect to \( t \). This conceptual tool is crucial when analyzing any dynamic system where variables are in continual flux.

The act of finding a derivative is known as differentiation, and it uses specific rules like the Sum Rule and Difference Rule to systematically determine these rates of change.
Limit Definition of the Derivative
The fundamental way to define a derivative is through the limit process. This definition gives the derivative its formal mathematical meaning and illustrates how small changes in a function's input lead to changes in output.

For a function \( \mathbf{u}(t) \), the derivative \( \frac{d \mathbf{u}}{dt} \) is defined as:
  • \( \lim_{\Delta t \to 0} \frac{\mathbf{u}(t + \Delta t) - \mathbf{u}(t)}{\Delta t} \)
This means you take the difference between the values of the function at two points very close together, divide by the change in the input \( \Delta t \), and see what value this quotient approaches as \( \Delta t \) becomes infinitely small.

Limit definition is not just a formula but a concept that underpins the derivative. It provides an essential understanding that enhances the comprehension of more advanced calculus topics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At time \(t=0,\) a particle is located at the point \((1,2,3) .\) It travels in a straight line to the point \((4,1,4),\) has speed 2 at \((1,2,3)\) and constant acceleration \(3 \mathbf{i}-\mathbf{j}+\mathbf{k}\) . Find an equation for the position vector \(\mathbf{r}(t)\) of the particle at time \(t .\)

Firing from \(\left(x_{0}, y_{0}\right)\) Derive the equations $$ \begin{aligned} x &=x_{0}+\left(v_{0} \cos \alpha\right) t \\ y &=y_{0}+\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2} \end{aligned} $$ (see Equation ( 5\()\) in the text) by solving the following initial value problem for a vector \(\mathbf{r}\) in the plane. $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}} \\ {\text { Initial conditions: }} & {\mathbf{r}(0)=x_{0} \mathbf{i}+y_{0} \mathbf{j}} \\ {} & {\frac{d \mathbf{r}}{d t}(0)=\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(v_{0} \sin \alpha\right) \mathbf{j}}\end{array} $$

Ellipse a. Show that the curve \(\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(1-\cos t) \mathbf{k}\) \(0 \leq t \leq 2 \pi,\) is an ellipse by showing that it is the intersection of a right circular cylinder and a plane. Find equations for the cylinder and plane. b. Sketch the ellipse on the cylinder. Add to your sketch the unit tangent vectors at \(t=0, \pi / 2, \pi,\) and \(3 \pi / 2 .\) c. Show that the acceleration vector always lies parallel to the plane (orthogonal to a vector normal to the plane). Thus, if you draw the acceleration as a vector attached to the ellipse, it will lie in the plane of the ellipse. Add the acceleration vectors for \(t=0, \pi / 2, \pi,\) and 3\(\pi / 2\) to your sketch. d. Write an integral for the length of the ellipse. Do not try to evaluate the integral; it is nonelementary. e. Numerical integrator Estimate the length of the ellipse to two decimal places.

As mentioned in the text, the tangent line to a smooth curve \(\mathbf{r}(t)=f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}\) at \(t=t_{0}\) is the line that passes through the point \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right)\) parallel to \(\mathbf{v}\left(t_{0}\right),\) the curve's velocity vector at \(t_{0} .\) In Exercises \(33-36,\) find parametric equations for the line that is tangent to the given curve at the given parameter value \(t=t_{0}\) . $$ \mathbf{r}(t)=(a \sin t) \mathbf{i}+(a \cos t) \mathbf{j}+b t \mathbf{k}, \quad t_{0}=2 \pi $$

Evaluate the integrals in Exercises \(21-26\) $$ \int_{0}^{1}\left[t^{3} \mathbf{i}+7 \mathbf{j}+(t+1) \mathbf{k}\right] d t $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.