232. 

$$
\text { 1) }\left\{\begin{array}{l}
x=k \sin t+\sin k t \\
y=k \cos t+\cos k t ;\left(\frac{d y}{d x}\right)_{t=0} ?
\end{array}\right.
$$

What is the geometric meaning of the result?
2) $\left\{\begin{array}{l}x=\alpha^{2}+2 \alpha \\ y=\ln (\alpha+1) ; \frac{d^{2} y}{d x^{2}} ?\end{array}\right.$
3) $\left\{\begin{array}{l}x=1+e^{a \varphi} \\ y=a \varphi+e^{-a_{\varphi} \varphi} ; \frac{d^{3} y}{d x^{3}} ?\end{array}\right.$
Solution. 1) We find the derivatives of $x$ and $y$ with respect to the parameter $t$:

$$
\frac{d x}{d t}=k \cos t+k \cos k t ; \quad \frac{d y}{d t}=-k \sin t-k \sin k t
$$

The desired derivative of $y$ with respect to $x$ is found as the ratio of the derivatives of $y$ and $x$ with respect to $t$:

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=-\frac{k(\sin t+\sin k t)}{k(\cos t+\cos k t)}=-\frac{2 \sin \frac{t+k t}{2} \cos \frac{t-k t}{2}}{2 \cos \frac{t+k t}{2} \cos \frac{t-k t}{2}}=-\operatorname{tg} \frac{k+1}{2} t$.

At $t=0$, we get $\frac{d y}{d x}=0$. According to the geometric meaning of the derivative (§ 1), at the point $(0 ; k+1)$, where $t=0$, the tangent to the graph of the given function is parallel to the $O x$ axis.

2) We find the derivatives of $x$ and $y$ with respect to the parameter $\alpha$:

$$
\frac{d x}{d a}=2 \alpha+2 ; \frac{d y}{d \alpha}=\frac{1}{\alpha+1}
$$

and the desired derivative of $y$ with respect to $x$:

$$
y^{\prime}=\frac{d y}{d x}=\frac{d y}{d \alpha}: \frac{d x}{d \alpha}=\frac{1}{2(\alpha+1)^{2}}=\frac{1}{2}(\alpha+1)^{-2}
$$

Next, we find the derivative of $y^{\prime}$ with respect to $\alpha$, and then the desired second derivative of $y$ with respect to $x$ as the ratio of the derivatives of $y^{\prime}$ and $x$ with respect to $\alpha$:

$$
\frac{d y^{\prime}}{d \alpha}=-(\alpha+1)^{-3} ; \quad y^{\prime \prime}=\frac{d y^{\prime}}{d x}=\frac{d y^{\prime}}{d \alpha}: \frac{d x}{d \alpha}=\frac{-(\alpha+1)^{-3}}{2(\alpha+1)}=-\frac{1}{2(\alpha+1)^{4}}
$$

3) Using the general formulas (A) for the derivatives of a function given parametrically, we obtain

$$
\begin{aligned}
& y^{\prime}=\frac{d y}{d x}=\frac{d y}{d \varphi}: \frac{d x}{d \varphi}=\frac{a-a e^{-a \varphi}}{a e^{a \varphi}}=e^{-a \varphi}-e^{-2 \alpha \varphi} ; \\
& y^{\prime \prime}=\frac{d y^{\prime}}{d x}=\frac{d y^{\prime}}{d \varphi}: \frac{d x}{d \varphi}=\frac{2 a e^{-2 a \varphi}-a e^{-\alpha \varphi}}{a e^{a \varphi}}=2 e^{-3 a \varphi}-e^{-2 a \varphi} \\
& y^{\prime \prime \prime}=\frac{d y^{\prime \prime}}{d x}=\frac{d y^{\prime \prime}}{d \varphi}: \frac{d x}{d \varphi}=\frac{2 a e^{-2 a \varphi}-6 a e^{-3 a^{\prime} \varphi}}{a e^{a \varphi}}=2 e^{-3 a \varphi}-6 e^{-4 a \varphi} \\
& \text { 233. }\left\{\begin{array}{l}
x=t^{2} \\
y=t^{3} ; \frac{d y}{d x}
\end{array}\right. \\
& \text { 234. }\left\{\begin{array}{l}
x=\frac{3 a t}{1+t^{3}} \\
y=\frac{3 a t^{2}}{1+t^{3}} ; \frac{d y}{d x} ?
\end{array}\right. \\
& \text { 235. }\left\{\begin{array}{l}
x=a \cos t \\
y=a \sin t ; \frac{d^{2} y}{d x^{2}} ?
\end{array}\right. \\
& \text { 236. }\left\{\begin{array}{l}
p=\cos \alpha+\alpha \sin \alpha \\
q=\sin \alpha-\alpha \cos \alpha ; \frac{d^{2} q}{d p^{2}} ?
\end{array}\right. \\
& \text { 237. }\left\{\begin{array}{l}
x=z^{2} \\
y=z^{3}+z ;\left(\frac{d^{2} y}{d x^{2}}\right)_{z=1}
\end{array}\right\} \\
& \text { 238. }\left\{\begin{array}{l}
x=a \cos ^{3} t \\
y=a \sin ^{3} t
\end{array}\left(\frac{d^{2} y}{d x^{2}}\right)_{t=\frac{\pi}{6}} ?\right.
\end{aligned}
$$

## § 11. Tangent and Normal to a Plane Curve. Angle Between Two Curves

If a plane curve is referred to a rectangular coordinate system (Fig. 33), then the equations of the tangent and normal to it at the point $M\left(x_{0}, y_{0}\right)$ are:

$$
y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right) ; \quad y-y_{0}=-\frac{1}{y_{0}^{\prime}}\left(x-x_{0}\right)
$$

where $y_{0}^{\prime}$ is the value at the point $x_{0}$ of the derivative $\frac{d y}{d x}$ from the equation of the curve.

The direction of the curve at each of its points is determined by the direction of the tangent to it at that point. The angle between two intersecting curves is defined as the angle between two lines tangent to the curves at their point of intersection (Fig. 34) by the formula

$$
\operatorname{tg} \varphi=\frac{k_{1}-k_{2}}{1+k_{1} k_{2}}
$$

where $k_{1}$ and $k_{2}$ are the slopes of the tangents to the curves at their point of intersection $P\left(x_{0}, y_{0}\right)$, i.e., the particular values at the point $x_{0}$ of the derivatives of $y$ with respect to $x$ from the equations of these curves:

$$
k_{1}=\operatorname{tg} \alpha_{1}=\left(\frac{d y_{1}}{d x}\right)_{x=x_{0}} ; k_{2}=\operatorname{tg} \alpha_{2}=\left(\frac{d y_{2}}{d x}\right)_{x=x_{0}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-080.jpg?height=223&width=347&top_left_y=215&top_left_x=192)

Fig. : 3

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-080.jpg?height=210&width=315&top_left_y=225&top_left_x=763)

Fig. 34