Calc Help: Moving Particle from Origin & Integral of 5^ radical x

  • Thread starter parwana
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In summary: So the work from any two positions would just be the line integral for the straight line between any two points over this vector field.
  • #1
parwana
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When particle is at distance meters from the origin, there is a force of Newtons (N) pulling it toward the origin. How much work is needed to move the particle from the position to the position ?



and




What is integral of 5^ radical x ?
 
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  • #2
That's a very strangely worded question. Let's call the distance from the origin, d, the force, F. Now what on Earth do you mean "from the position to the position?" What position to what position? Very strange?!...

[tex]\int 5^{\root n\of x}dx[/tex]

Let [itex]u = x^{1/n}[/itex], therefore [itex]du = \frac{1}{n}x^{1/n - 1}dx[/itex].

[tex]dx = \frac{ndu}{x^{1/n - 1}}[/tex]

[tex]n\int \frac{5^udu}{u^{1 - n}}[/tex]

Hmmm... maybe this is a start, I don't know where to go with this.

EDIT: Look up integral tables, see if they have ways to solve things like [itex]\int u5^udu[/itex], you might be able to solve this for the first few n (n = 1, 2, 3), and if you find a pattern, you might be able to use induction to prove it generally.
 
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  • #3
Your integral is way beyond me. But http://integrals.wolfram.com/index.en.cgi says:

[tex]\int 5^{x^{\frac{1}{n}}}dx = -nx \left(-x^{\frac{1}{n}} \right)^{-n} \Gamma \left( n, -x^{\frac{1}{n}} \ln 5 \right) \left(\ln 5\right)^{-n} + C[/tex]
 
  • #4
Zurtex said:
[tex]\int 5^{x^{\frac{1}{n}}}dx = -nx \left(-x^{\frac{1}{n}} \right)^{-n} \Gamma \left( n, -x^{\frac{1}{n}} \ln 5 \right) \left(\ln 5\right)^{-n} + C[/tex]
Hah! Yes, I knew I was close :wink:.
 
  • #5
Wouldn't the constant force towards the origin indicate a vector field of constant magnitude with all vectors pointing toward the origin? So the work from any two positions would just be the line integral for the straight line between any two points over this vector field.
 
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  • #6
µ³ said:
Wouldn't the constant force towards the origin indicate a vector field of constant magnitude with all vectors pointing toward the origin? So the work from any two positions would just be the line integral for the straight line between any two points over this vector field.

The line integral of the component of force parallel to the straight line over the line segment.
 

1. What is the formula for calculating the position of a moving particle from the origin?

The formula for calculating the position of a moving particle from the origin is given by x=√(5t), where x represents the position of the particle and t represents time.

2. How do you find the velocity of a particle from the origin?

The velocity of a particle from the origin can be found by taking the derivative of the position formula, which is given by dx/dt = √5.

3. What is the meaning of the integral of 5^(√x)?

The integral of 5^(√x) represents the area under the curve of the function 5^(√x) when graphed on a coordinate plane. It can also be interpreted as the accumulation of the function over a given interval.

4. How do you solve for the integral of 5^(√x)?

To solve for the integral of 5^(√x), you can use integration by substitution. Let u=√x, then the integral becomes ∫5^u * 2u du. This can be further simplified and solved using integration techniques.

5. Can you use the integral of 5^(√x) to find the area under the curve from x=1 to x=5?

Yes, the integral of 5^(√x) can be used to find the area under the curve from x=1 to x=5. The definite integral can be evaluated by plugging in the upper and lower limits of integration, 5 and 1 respectively, into the indefinite integral formula.

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