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Calculus arc length problem
- Thread starter rir0302
- Start date
\(\displaystyle \text{The function $y(x) = 3e^{-x^2}$ makes a curve in the $xy$ plane. $\\$
Over the interval $0\leq x \leq 2$ that curve has an arc length, i.e. the distance you $\\$
would travel if you walked that curve. There is an integral formula for the arc length of $\\$
a curve over a given interval. The want you to numerically evaluate this integral as noted.}\)
Over the interval $0\leq x \leq 2$ that curve has an arc length, i.e. the distance you $\\$
would travel if you walked that curve. There is an integral formula for the arc length of $\\$
a curve over a given interval. The want you to numerically evaluate this integral as noted.}\)
HallsofIvy
Elite Member
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- Jan 27, 2012
- Messages
- 7,763
You say you don't know what it "means"? What is unclear? Do you know what "arclength" means? Do you know what "Simpson's rule" is? Given a function of the form y= f(x), taking a very short length, from \(\displaystyle (x_0, f(x_0))\) to \(\displaystyle (x_0+ dx, f(x_0+ dx))\) we can approximate the graph by the tangent line at x. That is \(\displaystyle y= f'(x_0)(x- x_0)+ f(x_0)\). The distance from \(\displaystyle (x_0, f(x_0)\) to \(\displaystyle (x_0+ dx, f'(x_0)dx+ f(x_0))\) is \(\displaystyle \sqrt{dx^2+ (f'(x_0))^2dx^2}= \sqrt{(f'(x_0)^2+ 1}dx\). The arclength from x= a to x= b is \(\displaystyle \int_a^b \sqrt{(f'(x)^2+ 1}dx\). (Surely, if your text has a question like this it has already covered that formula?)
And "Simpson's method" for numerical integration approximates a function by taking an odd number of points and using the piecewise quadratic function through sets of three consecutive points. That is, given f(x) and points \(\displaystyle (x_1,f(x_1)\), \(\displaystyle (x_2,f(x_2)\), \(\displaystyle (x_3,f(x_3)\), \(\displaystyle (x_4,f(x_4)\), and \(\displaystyle (x_5,f(x_5)\) we approximate the f, for \(\displaystyle x_1\) to \(\displaystyle x_3\), by the quadratic that passes through \(\displaystyle (x_1,f(x_1)\), \(\displaystyle (x_2,f(x_2)\), \(\displaystyle (x_3,f(x_3)\) and, for \(\displaystyle x_3\) to \(\displaystyle x_5\), by the quadratic that passes through \(\displaystyle (x_3,f(x_3)\), \(\displaystyle (x_4,f(x_4)\), and \(\displaystyle (x_5,f(x_5)\). Again, surely your textbook has the formula for that. (Or you could have looked it up on Wikipedia: https://en.wikipedia.org/wiki/Simpson's_rule.)
And "Simpson's method" for numerical integration approximates a function by taking an odd number of points and using the piecewise quadratic function through sets of three consecutive points. That is, given f(x) and points \(\displaystyle (x_1,f(x_1)\), \(\displaystyle (x_2,f(x_2)\), \(\displaystyle (x_3,f(x_3)\), \(\displaystyle (x_4,f(x_4)\), and \(\displaystyle (x_5,f(x_5)\) we approximate the f, for \(\displaystyle x_1\) to \(\displaystyle x_3\), by the quadratic that passes through \(\displaystyle (x_1,f(x_1)\), \(\displaystyle (x_2,f(x_2)\), \(\displaystyle (x_3,f(x_3)\) and, for \(\displaystyle x_3\) to \(\displaystyle x_5\), by the quadratic that passes through \(\displaystyle (x_3,f(x_3)\), \(\displaystyle (x_4,f(x_4)\), and \(\displaystyle (x_5,f(x_5)\). Again, surely your textbook has the formula for that. (Or you could have looked it up on Wikipedia: https://en.wikipedia.org/wiki/Simpson's_rule.)