Assignment-daixieTM为您提供莱斯特大学University of Leicester PA1710 Mathematical Physics 1.1数学物理代写代考和辅导服务！
Instructions:
Physics is a mathematical subject that relies heavily on mathematical techniques and methods to describe and understand the physical world. As a result, students studying physics typically receive training in a wide range of mathematical topics, including calculus, linear algebra, differential equations, probability theory, and more.
Integral calculus, in particular, is a core mathematical tool in physics that is used to model and analyze a wide range of physical phenomena, such as motion, forces, and energy. Series and convergence are also important concepts in physics, as they are used to represent functions and data in various forms, such as Fourier series and Taylor series.
Vector analysis is another important mathematical topic in physics that is used to describe physical quantities such as velocity, acceleration, and force. Probability distributions are also crucial in physics, as they allow us to model and understand the statistical behavior of physical systems, such as the distribution of particles in a gas or the distribution of energies in a solid.
Overall, a strong foundation in mathematics is essential for success in physics, and many of the mathematical techniques that are taught in the first year of a physics degree program provide the necessary tools for understanding and analyzing physical phenomena.
Use the definition of the derivative as a limit of difference quotients to compute the derivative of $y=x+\frac{1}{a}$ for all points $x>0$. Show all work.
Denote by $f(x)$ the function $x+\frac{1}{x}$. By definition, the derivative of $f(x)$ at $x=a$ is,
$$
f^{\prime}(a)=\lim {h \rightarrow 0} \frac{f(a+h)-f(a)}{h} . $$ The increment $f(a+h)-f(a)$ equals, $$ \left((a+h)+\frac{1}{a+h}\right)-\left(a+\frac{1}{a}\right)=h+\left(\frac{1}{a+h}-\frac{1}{a}\right) . $$ To compute the second term, clear denominators, $$ \frac{1}{a+h}-\frac{1}{a}=\frac{1}{a+h} \frac{a}{a}-\frac{1}{a} \frac{a+h}{a+h}=\frac{a-(a+h)}{a(a+h)}=\frac{-h}{a(a+h)} . $$ Thus the increment $f(a+h)-f(a)$ equals, $$ h-\frac{h}{a(a+h)} . $$ Factoring $h$ from each term, the difference quotient equals, $$ \frac{f(a+h)-f(a)}{h}=1-\frac{1}{a(a+h)} . $$ Thus the derivative of $f(x)$ at $x=a$ equals, $$ f^{\prime}(a)=\lim {h \rightarrow 0}\left(1-\frac{1}{a(a+h)}\right)=1-\frac{1}{a(a+0)}=1-\frac{1}{a^2} .
$$
Therefore the derivative function of $f(x)$ equals,
$$
f^{\prime}(x)=1-\frac{1}{x^2} .
$$
For the function $f(x)=e^{-x^2 / 2}$, compute the first, second and third derivatives of $f(x)$.
Set $u$ equals $-x^2 / 2$ and set $v$ equals $e^u$. So $v$ equals $f(x)$. By the chain rule,
$$
\frac{d v}{d x}=\frac{d v}{d u} \frac{d u}{d x} .
$$
Since $v$ equals $e^u, d v / d u$ equals $\left(e^u\right)^{\prime}=e^u$. Since $u$ equals $-x^2 / 2, d u / d x$ equals $-(2 x) / 2=-x$. Thus, back-substituting,
$$
f^{\prime}(x)=\frac{d v}{d x}=\left(e^u\right)(-x)=e^{-x^2 / 2}(-x)=-x e^{-x^2 / 2} .
$$
For the second derivative, let $u$ and $v$ be as defined above, and set $w$ equals $-x v$. So $w$ equals $f^{\prime}(x)$. By the product rule,
$$
\frac{d w}{d x}=(-x)^{\prime} v+(-x) v^{\prime}=-v-x \frac{d v}{d x} .
$$
By the last paragraph,
$$
\frac{d v}{d x}=-x e^{-x^2 / 2}
$$
Substituting in,
$$
f^{\prime \prime}(x)=\frac{d w}{d x}=-e^{-x^2 / 2}-x\left(-x e^{-x^2 / 2}\right)=-e^{-x^2 / 2}+x^2 e^{-x^2 / 2}=\left(x^2-1\right) e^{-x^2 / 2} .
$$
For the third derivative, take $u$ and $v$ as above, and set $z$ equals $\left(x^2-1\right) v$. So $z$ equals $f^{\prime \prime}(x)$. By the product rule,
$$
\frac{d z}{d x}=\left(x^2-1\right)^{\prime} v+\left(x^2-1\right) v^{\prime}=2 x v+\left(x^2-1\right) \frac{d v}{d x} .
$$
By the first paragraph,
$$
\frac{d v}{d x}=-x e^{-x^2 / 2}
$$
Substituting in,
$$
f^{\prime \prime \prime}(x)=\frac{d z}{d x}=2 x e^{-x^2 / 2}+\left(x^2-1\right)\left(-x e^{-x^2 / 2}\right)=2 x e^{-x^2 / 2}+\left(-x^3+x\right) e^{-x^2 / 2}=\left(-x^3+3 x\right) e^{-x^2 / 2} .
$$
A function $y=f(x)$ satisfies the implicit equation,
$$
2 x^3-9 x y+2 y^3=0 .
$$
The graph contains the point $(1,2)$. Find the equation of the tangent line to the graph of $y=f(x)$ at $(1,2)$.
Differentiating both sides of the equation gives,
$$
\frac{d}{d x}\left(2 x^3-9 x y+2 y^3\right)=\frac{d}{d x}(0)=0
$$
Because the derivative is linear,
$$
\frac{d}{d x}\left(2 x^3-9 x y+2 y^3\right)=2 \frac{d\left(x^3\right)}{d x}-9 \frac{d(x y)}{d x}+2 \frac{d\left(y^3\right)}{d x} .
$$
Of course $d\left(x^3\right) / d x$ equals $3 x^2$. By the product rule,
$$
\frac{d(x y)}{d x}=\frac{d(x)}{d x} y+x \frac{d y}{d x}=y+x \frac{d y}{d x}
$$
For the last term, the chain rule gives,
$$
\frac{d\left(y^3\right)}{d x}=\frac{d\left(y^3\right)}{d y} \frac{d y}{d x}=3 y^2 \frac{d y}{d x}
$$
Substituting in gives,
$$
\frac{d}{d x}\left(2 x^3-9 x y+2 y^3\right)=2\left(3 x^2\right)-9\left(y+x \frac{d y}{d x}\right)+2\left(3 y^2\right) \frac{d y}{d x}=\left(6 x^2-9 y\right)+\left(6 y^2-9 x\right) \frac{d y}{d x}
$$
By the first paragraph, $d / d x\left(2 x^3-9 x y+2 y^3\right)$ equals 0 . Substituting in gives the equation,
$$
\left(6 x^2-9 y\right)+\left(6 y^2-9 x\right) \frac{d y}{d x}=0
$$
Subtracting the first term from each side gives,
$$
\left(6 y^2-9 x\right) \frac{d y}{d x}=\left(9 y-6 x^2\right) .
$$
Dividing both sides by $\left(6 y^2-9 x\right)$ gives,
$$
\frac{d y}{d x}=\frac{9 y-6 x^2}{6 y^2-9 x}=\frac{3 y-2 x^2}{2 y^2-3 x}
$$
Finally, plugging in $x$ equals 1 and $y$ equals 2 gives,
$$
\frac{d y}{d x}=\frac{3(2)-2(1)^2}{2(2)^2-3(1)}=\frac{6-2}{8-3}=\frac{4}{5} .
$$
Therefore, the equation of the tangent line is,
$$
y=\frac{4}{5}(x-1)+2,
$$
which simplifies to,
$$
y=(4 / 5) x+6 / 5
$$
