# 数学代写|微积分代写Calculus代写|MATH141

## 数学代写|微积分代写Calculus代写|Instantaneous rate of change

Instead of asking how fast an object is moving, we can ask how fast a quantity is changing.

Example 8 The price of a gallon of ChemForm $B$ this year is given by $f(t)=0.005 t^2+0.01 t+7.924$, where $t$ is in days and $f(t)$ is in dollars. How fast is the price changing when $t=60$ ?

Before we can find the solution, we need to develop the solution method. To begin, what does it mean by “how fast is the price changing?” Think about the price of gasoline (example 8 uses ChemForm $B$ instead of gasoline, but it’s the same idea). If the price of gas goes up $\$ 0.21$this week, this is a rate of change of$\$0.03 /$ day. “How fast the price is changing” means the rate at which the price changes over time. The units are units of price (dollars) divided by units of time (days).

How do we calculate the rate of change of the price of gasoline, $\$ 0.03 /$day? We first find the change in price of$\$0.21$ for the week as price at end of week – price at beginning of week.

To find the rate of change, we then divide by the amount of time. If the week in question was noon July 3 to noon July 10, we calculate the number of days by $10-3=7$, or
day at end of week – day at beginning of week.
Using $a=$ noon July 3 and $b=$ noon July 10 , and by calling the price of gasoline $f(t)$, then $f(a)$ is the price of gasoline at noon on July 3 and $f(b)$ is the price of gasoline at noon on July 10. The average rate of change of the price of gasoline over the time period is then
$$\frac{f(b)-f(a)}{b-a} .$$
It is the same formula as for average velocity. In fact, velocity can be considered the rate of change of position over time (how fast the position of the object is changing).

## 数学代写|微积分代写Calculus代写|The definition of derivative

The derivative is defined in a similar manner as limit and continuity.
Definition 1 THE DERIVATIVE Let $f$ be a function and let $k$ be a real number in the domain of $f$. If
$$\frac{f(k+\alpha)-f(k)}{\alpha}$$
is defined and renders the same real result $L$ for every infinitesimal $\alpha$, then we write $f^{\prime}(k)=L$ and call $f^{\prime}(k)$ the derivative of $f$ at $k$. When $L$ is a real number (not $\infty$ or $-\infty$ ), then we say that $f$ is differentiable at $k$.
For the purpose of calculations we write
$$f^{\prime}(k)=\frac{f(k+\alpha)-f(k)}{\alpha},$$
with the understanding that $\alpha$ is an arbitrary infinitesimal that can be either positive or negative and that we must render the same real result for every $\alpha$.
Example 1 Let $f(x)=x^2$. Find $f^{\prime}(5)$.

Solution We use the derivative formula with $k=5$ :
\begin{aligned} f^{\prime}(5) &=\frac{f(5+\alpha)-f(5)}{\alpha}=\frac{(5+\alpha)^2-5^2}{\alpha} \ &=\frac{25+10 \alpha+\alpha^2-25}{\alpha}=\frac{10 \alpha+\alpha^2}{\alpha} \ & \approx \frac{10 \alpha}{\alpha}=10 . \end{aligned}
The value of the derivative is 10 , and we write $f^{\prime}(5)=10$.

# 微积分代考

## 数学代写|微积分代写Calculus代写|Instantaneous rate of change

$$\frac{f(b)-f(a)}{b-a} .$$

## 数学代写|微积分代写Calculus代写|The definition of derivative

$$\frac{f(k+\alpha)-f(k)}{\alpha}$$

$$f^{\prime}(k)=\frac{f(k+\alpha)-f(k)}{\alpha},$$

$$f^{\prime}(5)=\frac{f(5+\alpha)-f(5)}{\alpha}=\frac{(5+\alpha)^2-5^2}{\alpha}=\frac{25+10 \alpha+\alpha^2-25}{\alpha}=\frac{10 \alpha+\alpha^2}{\alpha} \approx \frac{10 \alpha}{\alpha}=10 \text {. }$$

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