Example. Find the derivative $y_{x}^{\prime}$, if

\[
\left\{\begin{array}{l}
x=\ln \left(t+\sqrt{1+t^{2}}\right) \\
y=\sqrt{1+t^{2}}-\ln \frac{1+\sqrt{1+t^{2}}}{t}
\end{array}\right.
\]
Solution. We calculate:

$$
\begin{aligned}
& \frac{d x}{d t}=\frac{1}{t+\sqrt{1+t^{2}}}\left(1+\frac{t}{\sqrt{1+t^{2}}}\right)=\frac{1}{\sqrt{1+t^{2}}} \\
& \frac{d y}{d t}=\frac{t}{\sqrt{1+t^{2}}}-\frac{t}{1+\sqrt{1+t^{2}}} \frac{\frac{t^{2}}{\sqrt{1+t^{2}}}-\left(1+\sqrt{1+t^{2}}\right)}{t^{2}}=\frac{\sqrt{t^{2}+1}}{t}
\end{aligned}
$$

Substituting the obtained results into formula (1), we get

$$
\left\{\begin{array}{l}
x=\ln \left(t+\sqrt{1+t^{2}}\right) \\
y^{\prime}=\frac{1+t^{2}}{t}
\end{array}\right.
$$

Answer. $\left\{\begin{array}{l}x=\ln \left(t+\sqrt{1+t^{2}}\right), \\ y^{\prime}=\frac{1+t^{2}}{t} .\end{array}\right.$

Conditions of the Problems. Find the derivatives of the functions given parametrically.

1. $\left\{\begin{array}{l}x=t^{2}+1 / t^{3}, \\ y=\sin \left(t^{3} / 3+3 t\right) .\end{array}\right.$
2. $\left\{\begin{array}{l}x=\sqrt{t^{2}-2 t} \\ y=\sqrt[3]{t-1}\end{array}\right.$
3. $\left\{\begin{array}{l}x=\ln \left(t^{2}+1\right), \\ y=\sqrt{t^{2}+1} .\end{array}\right.$
4. $\left\{\begin{array}{l}x=\sqrt{1-t}, \\ y=\operatorname{tg} t .\end{array}\right.$
5. $\left\{\begin{array}{l}x=\ln \left(1-t^{2}\right), \\ y=\arcsin t .\end{array}\right.$
6. $\left\{\begin{array}{l}x=\ln \operatorname{tg} t, \\ y=1 / \sin t .\end{array}\right.$
7. $\left\{\begin{array}{l}x=\cos ^{3} t, \\ y=\sin ^{3} t .\end{array}\right.$
8. $\left\{\begin{array}{l}x=t-\sin t, \\ y=1-\cos t .\end{array}\right.$
9. $\left\{\begin{array}{l}x=\ln \sin t, \\ y=\sqrt{1-t^{2}} .\end{array}\right.$
10. $\left\{\begin{array}{l}x=\sqrt{1-t^{2}}, \\ y=\arcsin t .\end{array}\right.$

Answers.

1. $\left\{\begin{array}{l}x=\left(t^{2}+1\right) / t^{3} \\ y^{\prime}=-t^{4} \cos \left(t^{3} / 3+3 t\right) \text {. }\end{array}\right.$
2. $\left\{\begin{array}{l}x=\sqrt{t^{2}-2 t}, \\ y^{\prime}=\sqrt{t^{2}-2 t} /\left(3 \sqrt[3]{\left.(t-1)^{5}\right)} .\right.\end{array}\right.$
3. $\left\{\begin{array}{l}x=\ln \left(t^{2}+1\right) \\ y^{\prime}=\frac{1}{2} \sqrt{t^{2}+1}\end{array}\right.$
4. $\left\{\begin{array}{l}x=\sqrt{1-t}, \\ y^{\prime}=-2 \sqrt{1-t} / \cos ^{2} t .\end{array}\right.$
5. $\left\{\begin{array}{l}x=\ln \left(1-t^{2}\right), \\ y^{\prime}=-\sqrt{1-t^{2}}(2 t)\end{array}\right.$
6. $\left\{\begin{array}{l}x=\ln \operatorname{tg} t \\ y^{\prime}=-\cos ^{2} t / \sin t .\end{array}\right.$
7. $\left\{\begin{array}{l}x=\cos ^{3} t \\ y^{\prime}=-\operatorname{tg} t .\end{array}\right.$
8. $\left\{\begin{array}{l}x=t-\sin t \\ y^{\prime}=\operatorname{ctg}(t / 2)\end{array}\right.$
9. $\left\{\begin{array}{l}x=\ln \sin t \\ y^{\prime}=-t \operatorname{tg} t / \sqrt{1-t^{2}} .\end{array}\right.$
10. $\left\{\begin{array}{l}x=\sqrt{1-t^{2}}, \\ y^{\prime}=-1 / t .\end{array}\right.$

## 4.7. Tangent and Normal to a Curve Given Parametrically

Problem Statement. Derive the equations of the tangent and normal to the curve

$$
\left\{\begin{array}{l}
x=f(t) \\
y=g(t)
\end{array}\right.
$$

at the point $A$, corresponding to the parameter value $t=t_{0}$.

Solution Plan. If the function $y(x)$ has a finite derivative at the point $a$, then the equation of the tangent is

$$
y=y(a)+y^{\prime}(a)(x-a)
$$

If $y^{\prime}(a)=\infty$, then the equation of the tangent is $x=a$.

If $y^{\prime}(a) \neq 0$, then the equation of the normal is

$$
y=y(a)-\frac{1}{y^{\prime}(a)}(x-a)
$$

If $y^{\prime}(a)=0$, then the equation of the normal is $x=a$.

1. Calculate the coordinates of the point $A$:

$$
\left\{\begin{array}{l}
a=f\left(t_{0}\right) \\
y(a)=g\left(t_{0}\right)
\end{array}\right.
$$

2. Find the derivative $y^{\prime}$ at the point of tangency when $t=t_{0}$:

$$
y^{\prime}(a)=\frac{g^{\prime}\left(t_{0}\right)}{f^{\prime}\left(t_{0}\right)}
$$

3. Substitute the obtained values into the equations of the tangent (1) and normal (2) and write the answer.