The geometric interpretation of the derivative.

Literature: Collection of problems in mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.

The value of the derivative $f'(x_0)$ of the function $y=f(x)$ at the point $x_0$ is equal to the slope $k=\tan\varphi$ of the tangent $TT'$ to the graph of this function drawn through the point $M_0(x_0, y_0)$, where $y_0=f(x_0)$ (geometric interpretation of the derivative).

The line $NN'$ passing through the point of tangency $M_0$ perpendicular to the tangent is called the normal to the graph of the function $y=f(x)$ at this point. The equation of the normal is$$(x-x_0)+f'(x_0)(y-y_0)=0.$$The equation of the tangent $TT'$ to the graph of the function $y=f(x)$ at its point $M_0(x_0, y_0)$ has the form: $$y-y_0=f'(x_0)(x-x_0)$$

The angle $\omega$ between the curves $y=f_1(x)$ and $y=f_2(x)$ at their common point $M_0(x_0, y_0)$ is called the angle between the tangents to these curves at point $M_0$. It can be calculated using the formula:$$tg\,\omega=\frac{f_2'(x_0)-f'_1(x_0)}{1+f'_1(x_0)f'_2(x_0)}.$$

Examples.

Write down the equations of the tangent and the normal to the graph of the function $y=f(x)$ at the given point if:

5.235. $y=x^2-5x+4$, $x_0=-1$.

Solution.

We will find the equation of the tangent using the formula $y-y_0=f'(x_0)(x-x_0)$ and the equation of the normal using the formula $(x-x_0)+f'(x_0)(y-y_0)=0$.

According to the conditions, $x_0=-1$.

$y_0=y(x_0)=(-1)^2-5(-1)+4=1+5+4=10$.

$y'(x)=2x-5\Rightarrow y'(x_0)=y'(-1)=2(-1)-5=-2-5=-7$.

Substitute all found values into the equation of the tangent:

$y-10=-7(x+1)\Rightarrow 7x+y-3=0$.

Now find the equation of the normal:

$(x+1)-7(y-10)=0\Rightarrow x-7y+71=0$.

Answer: Equation of the tangent: $7x+y-3=0$; equation of the normal: $ x-7y+71=0$.

5.237. $y=\sqrt x,$ $x_0=4.$

Solution.

We will find the equation of the tangent using the formula $y-y_0=f'(x_0)(x-x_0)$ and the equation of the normal using the formula $(x-x_0)+f'(x_0)(y-y_0)=0$.

According to the conditions, $x_0=4$.

$y_0=y(x_0)=\sqrt{4}=2$.

$y'(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\Rightarrow y'(x_0)=y'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4}$.

Substitute all found values into the equation of the tangent:

$y-2=\frac{1}{4}(x-4)\Rightarrow 4(y-2)=x-4\Rightarrow 4y-8=x-4\Rightarrow x-4y+4=0$.

Now find the equation of the normal:

$(x-4)+\frac{1}{4}(y-2)=0\Rightarrow 4(x-4)+(y-2)=0\Rightarrow 4x+y-18=0$.

Answer: Equation of the tangent: $x-4y+4=0$; equation of the normal: $4x+y-18=0$.

5.241. Write down the equations of the tangent and the normal at the point

$M_0(2, 2)$ on the curve $x=\frac{1+t}{t^3},$ $y=\frac{3}{2t^2}+\frac{1}{2t},\,\, t\neq 0.$

Solution.

Let's find the value of $t_0$ by substituting the coordinates of the point $M_0$ into the equation of the curve:$2=\frac{1+t}{t^3},$ $2=\frac{3}{2t^2}+\frac{1}{2t}.$

$\left\{\begin{array}{rcl} 2=\frac{1+t}{t^3},\\ 2=\frac{3}{2t^2}+\frac{1}{2t},\end{array}\right.\Rightarrow$ $2=\frac{1+t}{t^3}=\frac{3}{2t^2}+\frac{1}{2t}$

Let's solve the equation:

$\frac{1+t}{t^3}=\frac{3}{2t^2}+\frac{1}{2t}$

$2(1+t)=3t+t^2\Rightarrow$

$t^2+t-2=0\Rightarrow t_1=1, t_2=-2.$

Let's substitute the obtained solutions into the equation: $\frac{1+t}{t^3}=\frac{3}{2t^2}+\frac{1}{2t}:$

$t_1=1: \frac{1+1}{1}=\frac{3}{2}+\frac{1}{2}=2$

$t_2=-2: \frac{1-2}{-8}=\frac{3}{8}-\frac{1}{4}=\frac{1}{8}\neq 2$ -- does not satisfy our system.

Let's find the derivative of the function defined parametrically, $y'_x$.

$y'_t=\left(\frac{3}{2}t^{-2}+\frac{1}{2}t^{-1}\right)'=\frac{3}{2}\cdot (-2)t^{-3}+\frac{1}{2}\cdot (-1)t^{-2}=-3t^{-3}-\frac{1}{2}t^{-2}$

$y'_t|_{t=1}=-3-1/2=-3,5;$

$x'_t=\left(\frac{1+t}{t^3}\right)'=\frac{(1+t)'t^3-(1+t)(t^3)'}{t^6}=\frac{t^3-(1+t)3t^2}{t^6}=\frac{t^3-3t^2-3t^3}{t^6}=\frac{-3t^2-2t^3}{t^6}.$

$x'_t|_{t=1}=-3-2=-5;$

$y'_x=\frac{y'_t}{x_t}.$

$y'_x|_{t=1}=\frac{-3,5}{-5}=\frac{7}{10}.$

Substitute all the found values into the equation of the tangent:

$y-y_0=f'(x_0)(x-x_0)\Rightarrow$ $y-2=\frac{7}{10}(x-2)\Rightarrow 10(y-2)=7(x-2)\Rightarrow 10y-20=7x-14\Rightarrow$ $7x-10y+6=0.$

Now let's find the equation of the normal:

$(x-x_0)+f'(x_0)(y-y_0)=0\Rightarrow$ $(x-2)+\frac{7}{10}(y-2)=0\Rightarrow 10(x-2)+7(y-2)=0\Rightarrow 10x+7y-34=0.$

Answer: Equation of the tangent: $7x-10y+6=0$; equation of the normal: $10x+7y-34=0$.

Find the angles at which the given curves intersect:

5.254. $y=x^2$ и $y=x^3.$

Solution.

We find the angle between the curves using the formula $$tg\,\omega=\frac{f_2'(x_0)-f'_1(x_0)}{1+f'_1(x_0)f'_2(x_0)}.$$

Let's find the coordinates of the intersection points of the given curves. We solve the system of equations:

$\left\{\begin{array}{rcl} y=x^2,\\ y=x^3,\end{array}\right.\Rightarrow$ $\left\{\begin{array}{rcl} y=x^2,\\ x^2=x^3,\end{array}\right.\Rightarrow$ $\left\{\begin{array}{rcl} y=x^2,\\ x_1=0\\x_2=1,\end{array}\right.$ Thus, the curves intersect at points $M_1(0, 0)$ and $M_2(1, 1).$

Next, let's find the values of the derivatives of the given functions at the intersection points.

$f_1(x)=x^2\Rightarrow f_1'(x)=2x$

$f_2(x)=x^3\Rightarrow f_2'(x)=3x^2$

$f_1'(0)=0;$

$f_2'(0)=0;$

$f_1'(1)=2;$

$f_2'(1)=3.$

Substituting the found values into the formula for finding the angle:

$$tg\,\omega_1=\frac{f_2'(0)-f'_1(0)}{1+f'_1(0)f'_2(0)}=\frac{0-0}{1+0}=0.$$

Thus, $\omega_1=0.$

$$tg\,\omega_2=\frac{f_2'(1)-f'_1(1)}{1+f'_1(1)f'_2(1)}=\frac{3-2}{1+2\cdot 3}=\frac{1}{7}.$$

Thus, $\omega_2=arctg\frac{1}{7}.$

Answer: At point $M_1(0, 0)$ the angle is 0 (i.e., the tangents coincide), at point $M_2(1, 1)$ the angle is $arctan\frac{1}{7}$.

Homework.

Write down the equations of the tangent and normal lines to the graph of the function $y=f(x)$ at the given point, if:

5.236. $y=x^3+2x^2-4x-3,$ $x_0=-2.$

Answer: Equation of tangent: $y-5=0;$ equation of normal: $x+2=0.$

5.238. $y=\tan(2x),,,, x_0=0.$

Answer: Equation of tangent: $y-2x=0;$ equation of normal: $2y+x=0.$

5.239. $y=\ln(x),,,, x_0=1.$

Answer: Equation of tangent: $x-y-1=0;$ equation of normal: $x+y-1=0.$

5.242. Write down the equations of the tangents to the curve $$x=t\cos t, \,\,\, y=t\sin t,\,\,\, t\in(-\infty,\,\, +\infty),$$ at the origin and at the point $t=\pi/4.$

Answer: $y=0,$ $(\pi+4)x+(\pi-4)y-\pi^2\frac{\sqrt 2}{4}=0$

5.244. To write down the equation of the tangent to the curve $x^5+y^5-2xy=0$ at the point $M_0(1, 1)$.

Answer: $ x+y-2=0.$

Find the angles at which the given curves intersect.

5.255. $y=(x-2)^2$ и $y=4x-x^2+4.$

Answer: $arctg\frac{8}{15}.$

5.256. $y=\sin x$ и $y=\cos x,\,\, x\in[0, 2\pi].$

Answer: $arctg2\sqrt 2.$

5.260. Find the distance from the origin to the normal line to the curve $y=e^{2x}+x^2$ drawn at the point with abscissa $x=0$.

Answer: $\frac{2}{\sqrt 5}.$

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