# 微积分应用 Calculus and its Applications MATH08058

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

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}

## 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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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,
$$\phi(2)=\langle 4,12\rangle$$

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

## 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 $z=x^{2}+y^{3}$