# 微积分代写|CALCULUS I MATH101 University of Liverpool Assignment

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To add more details, calculus is a branch of mathematics that deals with the study of rates of change and accumulation. It is divided into two main branches: differential calculus and integral calculus.

Differential calculus deals with the study of rates of change, such as slopes of tangent lines, velocity, and acceleration. It involves concepts such as derivatives, limits, and continuity. Derivatives are used to calculate the instantaneous rate of change of a function at a specific point.

Integral calculus deals with the study of accumulation, such as finding the area under a curve or the volume of a solid. It involves concepts such as integrals, limits, and the fundamental theorem of calculus. Integrals are used to calculate the total amount of a quantity that accumulates over an interval of time or a region of space.

Calculus has numerous practical applications in various fields such as physics, engineering, economics, and biology. It provides a powerful tool for solving real-world problems by modeling them mathematically and then applying calculus techniques to obtain solutions.

$\mathrm{Q}=$ top of the ladder: $\overrightarrow{\mathrm{OQ}}=\langle 0, L \sin \theta\rangle ; \quad \mathrm{R}=$ bottom of the ladder: $\overrightarrow{\mathrm{OR}}=\langle-L \cos \theta, 0\rangle$.
Midpoint: $\overrightarrow{\mathrm{OP}}=\frac{1}{2}(\overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}})=\left\langle-\frac{L}{2} \cos \theta, \frac{L}{2} \sin \theta\right\rangle$.
Parametric equations: $x=-\frac{L}{2} \cos \theta, y=\frac{L}{2} \sin \theta$.

It looks like you have provided the vector representations and parametric equations for three points in a coordinate plane: point Q at the top of a ladder, point R at the bottom of the ladder, and the midpoint P between points Q and R.

The vector representation of point Q is $\overrightarrow{\mathrm{OQ}}=\langle 0, L \sin \theta\rangle$, which means that point Q is located at the origin of the coordinate plane (since the x-coordinate is 0) and has a y-coordinate of $L \sin \theta$.

The vector representation of point R is $\overrightarrow{\mathrm{OR}}=\langle-L \cos \theta, 0\rangle$, which means that point R is located on the x-axis (since the y-coordinate is 0) and has an x-coordinate of $-L \cos \theta$.

The vector representation of point P is found by taking the average of the vectors for points Q and R: $\overrightarrow{\mathrm{OP}}=\frac{1}{2}(\overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}})=\left\langle-\frac{L}{2} \cos \theta, \frac{L}{2} \sin \theta\right\rangle$. This means that point P is located at $x=-\frac{L}{2} \cos \theta$ and $y=\frac{L}{2} \sin \theta$.

The parametric equations for these points are simply the x and y coordinates expressed as functions of the parameter $\theta$: $x=-\frac{L}{2} \cos \theta, y=\frac{L}{2} \sin \theta$. These equations allow you to plot the points on the coordinate plane as $\theta$ varies.

a) $\frac{d}{d t}(\vec{R} \cdot \vec{R})=\vec{V} \cdot \vec{R}+\vec{R} \cdot \vec{V}=2 \vec{R} \cdot \vec{V}$.

We can use the product rule of differentiation and the fact that the derivative of a constant vector is zero to find the derivative of $\vec{R} \cdot \vec{R}$ with respect to time:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = \frac{d}{dt}(\sum_{i=1}^{n} R_i^2) = \sum_{i=1}^{n} \frac{d}{dt}(R_i^2)$$

Using the chain rule, we have:

$$\frac{d}{dt}(R_i^2) = 2R_i \frac{dR_i}{dt}$$

Therefore:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = \sum_{i=1}^{n} 2R_i \frac{dR_i}{dt} = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt}$$

Using the product rule of differentiation, we have:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} + 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} + 2\sum_{i=1}^{n} \frac{dR_i}{dt} R_i$$

Using the dot product rule, we have:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} + 2\sum_{i=1}^{n} \frac{dR_i}{dt} R_i = 2\vec{R} \cdot \vec{V}$$

Therefore, we have:

$$\frac{d}{d t}(\vec{R} \cdot \vec{R})=\vec{V} \cdot \vec{R}+\vec{R} \cdot \vec{V}=2 \vec{R} \cdot \vec{V}.$$

b) Assume $|\vec{R}|$ is constant: then $\frac{d}{d t}(\vec{R} \cdot \vec{R})=2 \vec{R} \cdot \vec{V}=0$, i.e. $\vec{R} \perp \vec{V}$.

Starting with $\vec{R}\cdot\vec{R}=|\vec{R}|^2$, we can take the derivative with respect to time $t$ using the chain rule:

$$\frac{d}{d t}(\vec{R} \cdot \vec{R})=\frac{d}{d t}(|\vec{R}|^2)=2|\vec{R}|\frac{d|\vec{R}|}{dt}.$$

Since we are assuming $|\vec{R}|$ is constant, we have $\frac{d|\vec{R}|}{dt}=0$, so

$$\frac{d}{d t}(\vec{R} \cdot \vec{R})=2|\vec{R}|\frac{d|\vec{R}|}{dt}=0.$$

Therefore, $\vec{R}\cdot\vec{V}=0$ because the dot product of any two vectors is zero if and only if the vectors are perpendicular. Thus, we have $\vec{R}\perp\vec{V}$ as required.