Question

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3. Newton-Raphson algorithm Solve the following problem (Problem 6.30 of the Edition textbook 6 or Problem 6.17 of the textbook Edition 7): You are designing a spherical tank to hold water for a small village in a developing country. The volume of liquid it can hold can be computed as: where V=volume (m3, h-depth of water in tank (m), and R the tank radius (m). If R-3m, what depth must the tank be filled to so that it holds 30m3? Use three iterations of the Newton Raphson method to determine you answer. Determine the approximate percent relative error after each iteration. Note that an initial guess of R will always converge.

Answer 1

Given:

$V(h)=\pi h^2\frac{3R-h}{3}-30$

To be find out value of so that V=30 m³ in 3 iteration.

$V^{'}(h)=2 \pi *h*R-\pi*h^2\\$

$V^{'}(h)= \pi *h(2*R-h)\\$

Let the intial value of that is $h_{0}=2$ m

Firts iteration :

$h_{1}=h_{0}-\frac{V(h_{0})}{V^{'}(h_{0})}$

$h_{1}=2-\frac{-0.6785}{25.1327}$

$h_{1}=2.0270$

$er_{1}=\left \| (2.0270-2)/2.0270 \right \|$

$er_{1}=0.0133$

Second Iteration :

$h_{2}=h_{1}-\frac{V(h_{1})}{V^{'}(h_{1})}$

$h_{2}=2.0270-\frac{0.0024}{25.3001}$

$h_{2}=2.0271$

$er_{2}=\left \| (2.0271-2.0270)/2.0271 \right \|$

$er_{2}=4.9332e-05$

Third Iteration:

$h_{3}=h_{2}-\frac{V(h_{2})}{V^{'}(h_{2})}$

$h_{3}=2.0271-\frac{0.0049}{25.3007}$

$h_{3}=2.0273$

$er_{3}=\left \| (2.0273-2.0271)/2.0273 \right \|$

$er_{3}=9.8653e-05$