# 数学代写|微积分代写Calculus代写|MATH1111

## 数学代写|微积分代写Calculus代写|Motion examples

Example 1 A particle moves along a number line according to the equation $s(t)=\frac{t^3}{3}-t$, starting at time $t=0$. Units are centimeters and seconds. (a) In what direction is the particle moving at time $t=0$ ? (b) What is the average velocity of the particle during the interval $0<t<2$ ? (c.) Does the particle ever change directions? (d) What is the speed of the particle at time $t=0$ ? (e) What is the acceleration of the particle at time $t=2$ ?
Solution (a) Direction is part of velocity (velocity is speed and direction), so we need to calculate the velocity at time $t=0$ :
\begin{aligned} v(t)=s^{\prime}(t) &=\frac{1}{3} \cdot 3 t^2-1=t^2-1 \ v(0) &=0^2-1=-1 \mathrm{~cm} / \mathrm{s} . \end{aligned}
Because the velocity at time $t=0$ is negative, the particle is moving in the negative direction (left).

(b) Average velocity was discussed in section 1.8. The formula for average velocity over the time interval $[a, b]$ is
$$\frac{s(b)-s(a)}{b-a} .$$
Using $a=0$ and $b=2$ gives
$$\frac{s(2)-s(0)}{2-0}=\frac{\frac{8}{3}-2-\left(\frac{0}{3}-0\right)}{2}=\frac{\frac{2}{3}}{2}=\frac{1}{3} .$$
The average velocity during the interval $0 \leq t \leq 2$ is $\frac{1}{3} \mathrm{~cm} / \mathrm{s}$.
(c) When might a particle change directions? Consider a car on a narrow driveway. In order for the car to change directions (for instance, change from moving forward into the driveway to moving backward out of the driveway), the car must stop. Stopping means that the speed is zero (the same as velocity is zero). Therefore, we can determine where the particle might change directions by solving the equation $v(t)=0$ :
\begin{aligned} v(t)=t^2-1 &=0 \ t^2 &=1 \ t &=\pm 1 . \end{aligned}

## 数学代写|微积分代写Calculus代写|Rates of change in economics

A business that sells goods or services receives money, called revenue, from its sales. The amount of revenue received depends on the amount of goods or services sold. The revenue function describes the relationship between revenue and sales; $R(x)$ represents the revenue from selling $x$ units of goods or services. Likewise, the cost function is represented by $C(x)$ and gives the cost of procuring, manufacturing, or providing $x$ units of goods and services. Profit is revenue minus cost, so the profit function $P(x)=R(x)-C(x)$ represents the profit from producing and selling $x$ units.

The adjective marginal means the effect of “one more.” Marginal revenue is the additional revenue from selling one more unit, marginal cost is the additional cost of producing one more unit, and marginal profit is the additional profit from producing and selling one more unit.

It turns out that under most circumstances, “marginal” can be considered a synonym for “derivative.” To see why, let’s consider the local linearity diagram in figure 1 . The slope of the tangent line is $R^{\prime}(k)=\frac{d y}{d x}$. Note that $d x$ represents additional units sold whereas $\Delta y$ represents additional revenue. If $d x=1$ and the gap is small enough, then $R^{\prime}(k)=\frac{d y}{1}=d y$ is close to $\Delta y$, which is the additional revenue from one additional unit sold (i.e., the marginal revenue). Marginal revenue is therefore usually equated with $R^{\prime}(x)$, marginal cost with $C^{\prime}(x)$, and marginal profit with $P^{\prime}(x)$.

# 微积分代考

## 数学代写|微积分代写Calculus代写|Motion examples

(d) 粒 子在时间的速度是多少 $t=0$ ? (e) 粒子在时间的加速度是多少 $t=2$ ?

$$v(t)=s^{\prime}(t)=\frac{1}{3} \cdot 3 t^2-1=t^2-1 v(0) \quad=0^2-1=-1 \mathrm{~cm} / \mathrm{s} .$$

(b) $1.8$ 节讨论了平均速度。时间间隔内平均速度的公式 $[a, b]$ 是
$$\frac{s(b)-s(a)}{b-a} .$$

$$\frac{s(2)-s(0)}{2-0}=\frac{\frac{8}{3}-2-\left(\frac{0}{3}-0\right)}{2}=\frac{\frac{2}{3}}{2}=\frac{1}{3} .$$

(c) 粒子何时可以改变方向? 考虑在狭窎车道上行驶的汽车。为了让汽车改变方向 (例如，从向前驶入车 道变为向后驶出车道)，汽车必须停下来。停止意味着速度为零（与速度为零相同)。因此，我们可以 通过求解方程来确定粒子可能改变方向的位置 $v(t)=0$ :
$$v(t)=t^2-1=0 t^2 \quad=1 t=\pm 1 .$$

## 数学代写|微积分代写Calculus代写|Rates of change in economics

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