# 微积分代写 Calculus|MATH 15100 University of Chicago Assignment

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## Instructions:

MATH 15100, 15200, and 15300 form a regular calculus sequence that is typically taken by undergraduate students in their first or second year of study in a mathematics or science-related field. The sequence begins with MATH 15100, which covers limits, derivatives, and applications of derivatives, including optimization problems and related rates. MATH 15200 continues with integration, including the fundamental theorem of calculus, techniques of integration, and applications of integration, such as finding areas and volumes. MATH 15300 covers sequences and series, including convergence tests and power series, and applications of these concepts, such as Taylor series and Fourier series.

The sequence is designed to provide students with a solid foundation in calculus and its applications. Students are expected to have a strong background in algebra and trigonometry, as well as basic knowledge of functions and graphs. In MATH 15100, students are introduced to the formal definition of limits and learn how to evaluate limits using algebraic techniques and the squeeze theorem. They then move on to the definition of the derivative and learn how to find derivatives using various techniques, including the power rule, product rule, quotient rule, and chain rule. Applications of derivatives, such as optimization and related rates problems, are also covered.

In MATH 15200, students continue with integration, learning how to find antiderivatives and definite integrals using various techniques, such as substitution, integration by parts, and partial fractions. They also learn about the fundamental theorem of calculus, which relates differentiation and integration, and use integration to find areas and volumes of geometric shapes.

In MATH 15300, students learn about sequences and series, including convergence tests for both types of sequences and series. They also learn about power series and their applications, such as Taylor series and Fourier series. The course concludes with a discussion of applications of these concepts in various fields, including physics, engineering, and mathematics.

Throughout the sequence, students are expected to engage in problem-solving and critical thinking, as well as to develop their skills in mathematical reasoning and communication. The courses are typically taught using a combination of lectures, problem sets, and exams, and students are expected to participate actively in class and in study groups. Successful completion of the sequence prepares students for advanced courses in calculus, as well as for further study in mathematics, physics, engineering, and other fields.

Let f(x) = x 3 sin x. Then f (8) (0) =____

To find $f(8)$, we simply plug in $x=8$ into the function $f(x) = x^3\sin x$: $$f(8) = 8^3\sin 8 \approx 992.95$$ To find $f'(x)$, we use the product rule: $$f'(x) = 3x^2\sin x + x^3\cos x$$ Then, to find $f'(0)$, we simply plug in $x=0$ into the derivative: $$f'(0) = 3(0)^2\sin 0 + (0)^3\cos 0 = 0$$ Therefore, we have $f(8) \approx 992.95$ and $f'(0) = 0$.

The (nonparametric) equation of the flflow line of the vector fifield F(x, y) = x i y j which

passes through the point (4, 3) is

The flow line of the vector field $\mathbf{F}(x,y) = x \mathbf{i} – y \mathbf{j}$ passing through the point $(4,3)$ is given by the solution to the system of differential equations: \begin{align*} \frac{dx}{dt} &= x \ \frac{dy}{dt} &= -y \ x(0) &= 4 \ y(0) &= 3 \end{align*} Separating variables and integrating, we obtain: \begin{align*} \int \frac{1}{x},dx &= \int,dt \ \ln|x| &= t + C_1 \ |x| &= e^{t+C_1} = C_2 e^t \end{align*} where $C_1$ and $C_2$ are constants of integration. Using the initial condition $x(0) = 4$, we have $4 = C_2 e^0 = C_2$, so $C_2 = 4$. Therefore, we have $x = 4e^t$.

Similarly, we have: \begin{align*} \int -\frac{1}{y},dy &= \int,dt \ \ln|y| &= -t + C_3 \ |y| &= e^{C_3} e^{-t} = C_4 e^{-t} \end{align*} Using the initial condition $y(0) = 3$, we have $3 = C_4 e^0 = C_4$, so $C_4 = 3$. Therefore, we have $y = 3e^{-t}$.

Thus, the flow line passing through $(4,3)$ is given by the parametric equations $x = 4e^t$ and $y = 3e^{-t}$. We can eliminate the parameter $t$ to obtain the equation of the flow line in terms of $x$ and $y$: $$\frac{x}{4} \cdot \frac{1}{y/3} = 1$$ which simplifies to $$xy = 12.$$ Therefore, the equation of the flow line is $xy=12$.

An example of a potential function f for the vector fifield F(x, y) = (y cos x cos y)i +

(sin x + x sin y)j is f(x, y) =__________

To find a potential function $f(x,y)$ for the given vector field $F(x,y)= (y\cos x – \cos y)i + (\sin x + x\sin y)j$, we need to find $f(x,y)$ such that $F(x,y) = \nabla f(x,y)$.

We find the partial derivatives of $f(x,y)$ with respect to $x$ and $y$ as follows:

$$\frac{\partial f}{\partial x} = y\sin x + \sin y,\qquad \frac{\partial f}{\partial y} = \cos x + x\cos y$$

To check if $F(x,y) = \nabla f(x,y)$, we need to verify if the partial derivatives of $f(x,y)$ satisfy the following equations:

$$\frac{\partial f}{\partial x} = F_1(x,y) \quad \text{and} \quad \frac{\partial f}{\partial y} = F_2(x,y)$$

where $F_1(x,y)$ and $F_2(x,y)$ are the $x$ and $y$ components of $F(x,y)$, respectively.

Comparing the partial derivatives of $f(x,y)$ with $F(x,y)$, we see that:

$$\frac{\partial f}{\partial x} = y\sin x + \sin y = F_1(x,y)$$ $$\frac{\partial f}{\partial y} = \cos x + x\cos y = F_2(x,y)$$

Therefore, we can take $f(x,y)$ as:

$$f(x,y) = \int F_1(x,y) dx = -y\cos x – \cos y + g(y)$$

where $g(y)$ is an arbitrary function of $y$. To determine $g(y)$, we take the partial derivative of $f(x,y)$ with respect to $y$:

$$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(-y\cos x – \cos y + g(y)) = -\sin y + g'(y)$$

Comparing with $F_2(x,y) = \cos x + x\cos y$, we have:

$$g'(y) = \cos x + x\cos y -\sin y$$

Integrating both sides with respect to $y$, we obtain:

$$g(y) = y\cos y + x\sin y -\cos x + C$$

where $C$ is a constant of integration. Therefore, the potential function $f(x,y)$ is given by:

$$f(x,y) = -y\cos x – \cos y + y\cos y + x\sin y -\cos x + C = x\sin y + C$$

So, an example of a potential function $f(x,y)$ for the given vector field is $f(x,y) = x\sin y$.