# 数学代写|微积分代写Calculus代写|MATH0220

#### Doug I. Jones

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## 数学代写|微积分代写Calculus代写|Differentials

So far we have denoted the derivative by the symbol $\gamma^{\prime}$ or $d y / d x$. Although either symbol stands for $\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$, the method of writing $d y / d x$ suggests that the derivative might be regarded as the ratio of two quantities, $d y$ and $d x$. This turns out to be the case. The new quantities that we now introduce are called differentials, which are defined in the next frame.

Suppose that $x$ is an independent variable, and that $y=f(x)$. Then the differential $d x$ of $x$ is defined as equal to any increment, $x_2-x_1$, where $x_1$ is the point of interest. The differential $d x$ can be positive or negative, large or small, as we please. We see that $d x$, like $x$, can be regarded as an independent variable.
The differential $d y$ is defined by the following rule:
$$d y=y^{\prime} d x$$
where $\gamma^{\prime}$ is the derivative of $y$ with respect to $x$.

Although the meaning of the derivative $\gamma^{\prime}$ is $\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$, we can see from the preceding frame that it can now be interpreted as the ratio of the differentials $d y$ and $d x$, where $d x$ is any increment of $x$ and $d y$ is defined by the rule $d y=y^{\prime} d x$.It is important not to confuse $d \gamma$ with $\Delta \gamma$. As was pointed out in frame $\mathbf{1 3 6}, \Delta \gamma$ stands for $\gamma_2-y_1=f\left(x_2\right)-f\left(x_1\right)$ where $x_2$ and $x_1$ are two given values of $x$. Both $d x$ and $\Delta x=x_2-x_1$ are arbitrary intervals, $d x$ is called a differential of $x$, and $\Delta x$ is called an increment of $x$, but their meanings are similar here.The diagram shows that $d y$ and $\Delta y$ are different quantities. Here we have set $d x=\Delta x$. The differential $d y$ is then $d y=\gamma^{\prime}\left(x_1\right) d x$, where $x_1$ indicates that the derivative has been evaluated at the point $x_1$, while the increment $\Delta \gamma$ is given by $\gamma_2-\gamma_1$. It is clear in this case that $d y$ is not the same as $\Delta \gamma$.

Although $d y$ and $\Delta y$ are different, you can see from the figure that for sufficiently small $d x$ (with $d x=\Delta x$ ), $d y$ is very close to $\Delta y$. We can write this symbolically as
$$\lim _{\Delta x \rightarrow 0} \frac{d y}{\Delta y}=1 .$$
Hence, if we intend to take the limit where $d x \rightarrow 0, d \gamma$ may be substituted for $\Delta \gamma$. Furthermore, even if we don’t take the limit, $d y$ is almost the same as $\Delta \gamma$, provided $d x$ is sufficiently small. We, therefore, often use $d y$ and $\Delta y$ interchangeably when it is understood that the limit will be taken or that the result may be an approximation.

## 数学代写|微积分代写Calculus代写|A Short Review and Some Problems

Let’s end the chapter by reviewing some of the ideas it introduced and then putting differential calculus to work.

Go to 274 .
274
Recall that the rate of change of position of a moving point with respect to time is called velocity.
(continued)

In other words, if position is related to time by a function $S(t)$, to find the velocity, we $S(t)$ with respect to
In other words, if the position and time are related by a function $S(t)$, in order to find the velocity, we differentiate $S(t)$ with respect to time (or $t$,

Go to 275 .
275
$$\frac{d}{d t} S(t)=v(t) .$$
Go to 276.
276 Try this problem.
The position of a particle along a straight line is given by the following expression,
$$S(t)=A \sin (\omega t),$$
where $A$ and $\omega$ (omega) are constants.
Find the velocity of the particle.
$$v(t)=$$
277
For the answer, go to 277.
$$v(t)=\omega A \cos (\omega t) .$$
The problem is to find the velocity, which is the rate of change of position with respect to time. In this problem, the position is $S(t)=A \sin (\omega t)$.
$$v(t)=\frac{d S}{d t}=\frac{d}{d t} A \sin (\omega t)=\omega A \cos (\omega t) .$$
(If you are not sure of the procedure here, see frame 219.)
Can you do this problem? The position of a point is given by
$$S(t)=A \sin (\omega t)+B \cos (2 \omega t) .$$

# 微积分代考

## 数学代写|微积分代写Calculus代写| differential

.微分

$$d y=y^{\prime} d x$$

$$\lim _{\Delta x \rightarrow 0} \frac{d y}{\Delta y}=1 .$$
。此外，即使我们不取极限，只要$d x$足够小，$d y$也几乎与$\Delta \gamma$相同。因此，我们经常互换使用$d y$和$\Delta y$，当我们知道将取极限或结果可能是一个近似值

## 数学代写|微积分代写微积分代写|一个简短的回顾和一些问题

274

(继续) 换句话说，如果位置和时间通过一个函数$S(t)$联系起来，为了求出速度，我们$S(t)$关于

275
$$\frac{d}{d t} S(t)=v(t) .$$

276试试这道题。粒子沿直线的位置由以下表达式给出，
$$S(t)=A \sin (\omega t),$$

$$v(t)=$$
277

$$v(t)=\omega A \cos (\omega t) .$$

$$v(t)=\frac{d S}{d t}=\frac{d}{d t} A \sin (\omega t)=\omega A \cos (\omega t) .$$
(如果你不确定这里的程序，请参阅第219帧。)

$$S(t)=A \sin (\omega t)+B \cos (2 \omega t) .$$

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