Tag Archives: MATH08058代考

微积分应用 Calculus and its Applications MATH08058

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这是一份ed.ac.uk爱丁堡大学MATH08058作业代写的成功案例

微积分应用 Calculus and its Applications MATH08058
问题 1.

\text { Evaluate } \int \frac{x}{\sqrt{\left(x^{4}+4\right)}} d x


证明 .

Put $\mathrm{x}^{2}=\mathrm{t} ; \therefore 2 \mathrm{x} d \mathrm{~d}=\mathrm{dt}$.
$$
\begin{aligned}
&\therefore \text { the given integral }=\frac{1}{2} \int \frac{1}{\sqrt{\left(t^{2}+4\right)}} d t \
&=\frac{1}{2} \sinh ^{-1}\left(\frac{t}{2}\right)=\frac{1}{2} \sinh ^{-1}\left(\frac{x^{2}}{2}\right) .
\end{aligned}
$$

英国论文代写Viking Essay为您提供作业代写代考服务

MATH08058  COURSE NOTES ：

Put $x^{3}=t, \therefore 3 x^{2} d x=d t$.
$$
\begin{aligned}
&\therefore \text { the given integral }=\frac{1}{3} \int \frac{d t}{\sqrt{\left(t^{2}-9\right)}} \
&=\frac{1}{3} \cosh ^{-1}\left(\frac{t}{3}\right)=\frac{1}{3} \cosh ^{-1}\left(\frac{x^{3}}{3}\right) .
\end{aligned}
$$








微积分及其运用 Calculus and its Applications MATH08058 

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这是一份ed.ac.uk爱丁堡大学MATH08058作业代写的成功案例

微积分及其运用 Calculus and its Applications MATH08058 
问题 1.

At time $t$ the coordinates of a car are $\phi(t)=\left(t^{2}, t^{3}\right)$. We wish to determine how fast, and in what direction, the car is heading at time $t=2$. First, we differentiate the parameterization:
$$
\phi^{\prime}(t)=\left\langle 2 t, 3 t^{2}\right\rangle
$$
Thus,
$$
\phi(2)=\langle 4,12\rangle
$$

证明 .

The direction of the car is thus the direction that this vector is pointing. Its speed is given by the magnitude of this vector:
$$
|\langle 4,12\rangle|=\sqrt{4^{2}+12^{2}}=4 \sqrt{10}
$$

英国论文代写Viking Essay为您提供作业代写代考服务

MATH08051  COURSE NOTES ：

We find two vectors tangent to the graph of $z=x^{2}+y^{3}$ This surface is parameterized by
$$
\Psi(x, y)=\left(x, y, x^{2}+y^{3}\right)
$$
The desired point is at $\Psi(2,1)$. To find two tangent vectors we simply take the partial derivatives of $\Psi$ and evaluate at $(2,1)$.
$$
\frac{\partial \Psi}{\partial x}=\langle 1,0,2 x\rangle
$$
and so,
$$
\frac{\partial \Psi}{\partial x}(2,1)=\langle 1,0,4\rangle
$$
Similarly,
$$
\frac{\partial \Psi}{\partial y}=\left\langle 0,1,3 y^{2}\right\rangle
$$
and so,
$$
\frac{\partial \Psi}{\partial y}(2,1)=\langle 0,1,3\rangle
$$
We conclude $\langle 1,0,4\rangle$ and $\langle 0,1,3\rangle$ are two vectors tangent to the graph of $z=x^{2}+y^{3}$