Geometric meaning of derivation

Solution:

Using the derivative of the function y = f(x) we can write the equation of the tangent or the equation of the normal to the graph of the function at the point T [x_T , y_T]

Equation of the Tangent Line:

$$y - y_T = k_t (x - x_T), \quad k_t = f'(x_T)$$

Equation of the Normal Line:

$$y - y_T = k_n (x - x_T), \quad k_n = -\frac{1}{f'(x_T)}$$

Point T [x_T, f(x_T)] is the common point of the tangent (or normal) with the graph of the function.

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Subtangent and Subnormal:

The tangent, the normal, and the x-axis form a right triangle Δ ABT (right angle at vertex T).

It holds:

• Point T lies on line t,

• Point A lies on line t ∩ x-axis,

• Point B lies on line n ∩ x-axis.

• Point T₁ [x_T; 0] is the orthogonal projection of point T onto the x-axis.

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Subtangent Sₜ is the length of segment $|AT_1|$:

$$S_t = \left| \frac{y}{y'} \right|$$

Subnormal Sₙ is the length of segment $|T_1B|$:

$$S_n = |y \cdot y'|$$

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Length of the Tangent Segment TA = t:

$$t^2 = y^{2} + S_t^{2$$

Length of the Normal Segment TB = n:

$$n^2 = y^{2} + S_n^{2}$$

Solution:

equation of tangent                     equation of normal

Solution:

equation of tangent                      equation of normal

Solution:

equation of tangent                              equation of normal

Solution:

equation of tangent                             equation of normal

Solution:

equation of tangent                        equation of normal

Solution:

equation of tangent               equation of normal does not exists

Solution:

equation of tangent 

Solution:

equation of tangent

Solution:

For horizontal tangents (parallel to the x-axis) the following holds:

 

The contact points are T₁[1 ; 10] and T₂[5; -22] 

Solution:

The angle between the tangents is φ = 62⁰

Solution:

Point T:

$$y_T = 2\sqrt{1} = 2,\quad T[1;2]$$

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a) Subtangent:

$$S_t = \left| \frac{y}{y'} \right| = \left| \frac{2\sqrt{x}}{2 \cdot \frac{1}{2\sqrt{x}}} \right| = 2x = 2 \cdot 1 = 2$$

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b) Subnormal:

$$S_n = |y \cdot y'| = \left| 2\sqrt{x} \cdot 2 \cdot \frac{1}{2\sqrt{x}} \right| = 2$$

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c) Tangent (length of the tangent segment):

$$t^2 = y_T^2 + S_t^2$$$$t^2 = 2^2 + 2^2$$$$t = \sqrt{8}$$$$t = 2\sqrt{2}$$

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d) Normal (length of the normal segment):

$$n^2 = y_T^2 + S_n^2$$$$n^2 = 2^2 + 2^2$$$$n = \sqrt{8}$$$$n = 2\sqrt{2}$$

Solution:

Point T : y_T = 2¹ = 2, T[1 ; 2 ]

a) Subtangent

Given the function:

$$y = 2^x$$

The subtangent is:

$$S_t = \left|\frac{y}{y'}\right|$$

Since:

$$y' = 2^x \ln 2$$

we get:

$$S_t = \left|\frac{2^x}{2^x \ln 2}\right| = \frac{1}{\ln 2}$$

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b) Subnormal

Given:

$$y = 2^x$$

The subnormal is:

$$S_n = |y \cdot y'|$$

Substituting:

$$S_n = |2^x \cdot 2^x \ln 2| = 2^{2x} \ln 2 = 2^{2\cdot 1} \ln 2 = 2^2 \ln 2$$

Thus:

$$S_n = 4 \ln 2$$

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c) Tangent Length

$$t^2 = y_T^2 + S_t^2$$

Given:

$$y_T = 2 \quad\text{and}\quad S_t = \frac{1}{\ln 2}$$

So:

$$t^2 = 2^2 + \left(\frac{1}{\ln 2}\right)^2$$$$t^2 = 4 + \frac{1}{(\ln 2)^2}$$$$t = \sqrt{4 + \frac{1}{(\ln 2)^2}}$$$$t = \frac{\sqrt{1 + 4(\ln 2)^2}}{\ln 2}$$

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d) Normal Length

$$n^2 = y_T^2 + S_n^2$$$$n^2 = 2^2 + (4 \ln 2)^2$$$$n^2 = 4 + 16(\ln 2)^2$$$$n = 2 \sqrt{1 + 4 (\ln 2)^2$$

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