The deflection, $$ y $$, at the centre of a rod under constant load, varies directly as the cube of the length, $$ L $$, and indirectly as the fourth power of the diameter, $$ d $$, of the rod. Calculate the percentage change in $$ y $$ if $$ L $$ decreases by $$ 1 \% $$ and $$ d $$ increases by $$ 2 \% $$.

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We start with the relation
$$
y = k \frac{L^3}{d^4},
$$
where $$ k $$ is a constant of proportionality.

If $$ L $$ decreases by $$1\%$$, the new length is
$$
L' = 0.99\,L.
$$
If $$ d $$ increases by $$2\%$$, the new diameter is
$$
d' = 1.02\,d.
$$

Substituting these into the expression for $$ y $$, we have
$$
y' = k \frac{(0.99\,L)^3}{(1.02\,d)^4} = k \frac{0.99^3\,L^3}{1.02^4\,d^4}.
$$

The ratio of the new deflection to the original deflection is
$$
\frac{y'}{y} = \frac{0.99^3}{1.02^4}.
$$

Calculating the numerical values:
$$
0.99^3 \approx 0.9703,
$$
$$
1.02^4 \approx 1.0824.
$$
Thus,
$$
\frac{y'}{y} \approx \frac{0.9703}{1.0824} \approx 0.896.
$$

This means that $$ y $$ becomes approximately $$ 89.6\% $$ of its original value. Therefore, the percentage change in $$ y $$ is
$$
0.896 - 1 = -0.104,
$$
or about a $$-10.4\%$$ decrease in the deflection $$ y $$.
The deflection $$ y $$ decreases by approximately $$ 10.4\% $$.

The deflection, , at the centre of a rod under constant load, varies directly as the cube of
the length, , and indirectly as the fourth power of the diameter, , of the rod.
Calculate the percentage change in if decreases by and increases by .

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The deflection, , at the centre of a rod under constant load, varies directly as the cube of the length, , and indirectly as the fourth power of the diameter, , of the rod. Calculate the percentage change in if decreases by and increases by .