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OLEGan [10]
3 years ago
8

Value of the derivative of g(x)=8-10Cosx at 'x=0' is?

Mathematics
1 answer:
VLD [36.1K]3 years ago
3 0

Answer:

g'(0) = 0

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Function Notation

<u>Pre-Calculus</u>

  • Unit Circle

<u>Calculus</u>

  • Derivatives
  • Derivative Notation
  • The derivative of a constant is equal to 0
  • Derivative Property: \frac{d}{dx} [cf(x)] = c \cdot f'(x)
  • Trig Derivative: \frac{d}{dx} [cos(x)] = -sin(x)

Step-by-step explanation:

<u>Step 1: Define</u>

g(x) = 8 - 10cos(x)

x = 0

<u>Step 2: Differentiate</u>

  1. Differentiate [Trig]:                    g'(x) = 0 - 10[-sin(x)]
  2. Simplify Derivative:                   g'(x) = 10sin(x)

<u>Step 3: Evaluate</u>

  1. Substitute in <em>x</em>:                    g'(0) = 10sin(0)
  2. Evaluate Trig:                      g'(0) = 10(0)
  3. Multiply:                               g'(0) = 0
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fenix001 [56]

Answer:

The angle between vector \vec{u} = 5\, \vec{i} - 8\, \vec{j} and \vec{v} = 5\, \vec{i} + \, \vec{j} is approximately 1.21 radians, which is equivalent to approximately 69.3^\circ.

Step-by-step explanation:

The angle between two vectors can be found from the ratio between:

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  • the product of their lengths.

To be precise, if \theta denotes the angle between \vec{u} and \vec{v} (assume that 0^\circ \le \theta < 180^\circ or equivalently 0 \le \theta < \pi,) then:

\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}.

<h3>Dot product of the two vectors</h3>

The first component of \vec{u} is 5 and the first component of \vec{v} is also

The second component of \vec{u} is (-8) while the second component of \vec{v} is 1. The product of these two second components is (-8) \times 1= (-8).

The dot product of \vec{u} and \vec{v} will thus be:

\begin{aligned} \vec{u} \cdot \vec{v} = 5 \times 5 + (-8) \times1 = 17 \end{aligned}.

<h3>Lengths of the two vectors</h3>

Apply the Pythagorean Theorem to both \vec{u} and \vec{v}:

  • \| u \| = \sqrt{5^2 + (-8)^2} = \sqrt{89}.
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<h3>Angle between the two vectors</h3>

Let \theta represent the angle between \vec{u} and \vec{v}. Apply the formula\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} to find the cosine of this angle:

\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} = \frac{17}{\sqrt{89}\cdot \sqrt{26}}\end{aligned}.

Since \theta is the angle between two vectors, its value should be between 0\; \rm radians and \pi \; \rm radians (0^\circ and 180^\circ.) That is: 0 \le \theta < \pi and 0^\circ \le \theta < 180^\circ. Apply the arccosine function (the inverse of the cosine function) to find the value of \theta:

\displaystyle \cos^{-1}\left(\frac{17}{\sqrt{89}\cdot \sqrt{26}}\right) \approx 1.21 \;\rm radians \approx 69.3^\circ .

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