The approximate value of $ f(5.001) $, $\text{where} $ $f(x) = x^3 - 7x^2 + 10 $
We are given the function \( f(x) = x^3 - 7x^2 + 10 \) and need to approximate \( f(5.001) \).
The first step is to approximate the value of the function using a linear approximation. We use the fact that the linear approximation to a function around a point \( x_0 \) is given by:
\[ f(x) \approx f(x_0) + f'(x_0)(x - x_0) \]
Step 1: Calculate \( f(5) \)
\[ f(5) = 5^3 - 7(5^2) + 10 = 125 - 7 \cdot 25 + 10 = 125 - 175 + 10 = -40 \]
Step 2: Find the derivative \( f'(x) \)
\[ f'(x) = 3x^2 - 14x \]
Step 3: Calculate \( f'(5) \)
\[ f'(5) = 3(5^2) - 14(5) = 3 \cdot 25 - 70 = 75 - 70 = 5 \]
Step 4: Apply the linear approximation
Now, use the approximation formula to estimate \( f(5.001) \): \[ f(5.001) \approx f(5) + f'(5)(5.001 - 5) = -40 + 5 \cdot 0.001 = -40 + 0.005 = -39.995 \] Thus, the approximate value of \( f(5.001) \) is \( -39.995 \).
To approximate the value of \( f(5.001) \) where \( f(x) = x^3 - 7x^2 + 10 \), we'll use linear approximation.
1. Set Up the Approximation:
We break \( 5.001 \) into \( x = 5 \) and \( \Delta x = 0.001 \), and use the formula:
\[
f(x + \Delta x) \approx f(x) + \Delta x \cdot f'(x)
\]
2. Compute the Derivative:
First find the derivative of \( f(x) \):
\[
f'(x) = \frac{d}{dx}(x^3 - 7x^2 + 10) = 3x^2 - 14x
\]
3. Apply the Approximation Formula:
Substitute into the approximation formula:
\[
f(5.001) \approx (5^3 - 7 \cdot 5^2 + 10) + 0.001 \cdot (3 \cdot 5^2 - 14 \cdot 5)
\]
4. Calculate Each Component:
Compute each term step by step:
- \( 5^3 = 125 \)
- \( 7 \cdot 25 = 175 \)
- \( 3 \cdot 25 = 75 \)
- \( 14 \cdot 5 = 70 \)
5. Substitute and Simplify:
\[
f(5.001) \approx (125 - 175 + 10) + 0.001 \cdot (75 - 70)
\]
\[
= (-40) + 0.001 \cdot 5
\]
\[
= -40 + 0.005 = -39.995
\]
Final Answer:
The approximate value is \( \boxed{-39.995} \).
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Based upon the above information, answer the following questions:
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A solid cylinder of mass 2 kg and radius 0.2 m is rotating about its own axis without friction with angular velocity 5 rad/s. A particle of mass 1 kg moving with a velocity of 5 m/s strikes the cylinder and sticks to it as shown in figure.
The angular velocity of the system after the particle sticks to it will be: