Question

The slope (????) of the beam is given by:

Answer 1

Answer:

Step-by-step explanation:

From the given question:

The slope is expressed as;

$\thteta (x) = \dfrac{dy}{dx} - - (1)$

$\thteta (x) = \dfrac{dy}{dx}\left \{ {{\dfrac{y(x+b)-y(x)}{h} \ \ \to approximate} \atop {\dfrac{1}{100}}(-4+4x-3x^2+2x^3) \ \to \ exact } \right.$

∴

$\theta (x) _{approximate} = \dfrac{y(1+0.3)-y(1)}{0.3}$

$\theta (x) _{approximate} = \dfrac{y(1.3)-y(1)}{0.3}$

$\theta (x) _{approximate} = -2.965 \times 10^{-5}$

$\theta(x)_{exact} = \dfrac{1}{100}(-4+4-3+2)$

$\theta(x)_{exact} = -0.01$

Finally, the true percent relative error TPRE is:

$TPRE= \dfrac{ | \theta(x)_{approximate} -\theta (x)_{exact} }{\theta (x)_{exact} }\times 100\%$

$TPRE= \dfrac{ 2.965 \times 10^5 --0.01}{-0.01}\times 100\%$

TPRE = 70..35%