Why the two methods give different answers?

Question

Question

If Kinetic energy of the body is increased by $300\text{%}$, its momentum will increase by:

Method 1:Using proportionality

$$Ke=\frac{P^2}{2m}$$where Ke is Kinetic energy and P is momentum

Since mass is constant, $$\frac{Ke_i}{Ke_f}=\frac{P_i^2}{P_f^2}$$ $$\frac{Ke_i}{4Ke_i}=\frac{P_i^2}{P_f^2}$$ $$P_f=2P_i$$ $$∆P=2P_i-P_i=P_i$$ Change in momentum is 100%

METHOD 2:Using differentiation

$$Ke=\frac{P^2}{2m}$$ Differentiating both sides $$dKe=\frac{2Pdp}{2m}$$ Approximating $$dKe=∆Ke$$ $$∆Ke=3Ke$$ $$3Ke=\frac{2Pdp}{2m}$$ $$\frac{3Ke*2m}{2P}=dP$$ $$Ke*2m=P^2$$ Substituting , $$\frac{3*P^2}{2P}=dP$$ $$(dP/P)*100=\frac32*100=150%$$ Change in momentum =150%

Why the two methods give different answers?

Answer 1

Notice, $dKe$ is infinitesimal small change in kinetic energy i.e. $dKe\to 0$ while $\Delta Ke$ is change in kinetic energy much much larger than $0$ i.e. $\Delta Ke>>0$ (in this case). Thus
$$dKe<<\Delta Ke$$ Therefore using $dKe=\Delta Ke$ gives a large difference in the results due to overestimation.

The first method computes $\text{%}\Delta P$ correctly by taking actual difference $\Delta Ke$ (without any approximation) while second method computes $\text{%}\Delta P$ incorrectly due to overestimation by taking $dKe=\Delta Ke$

Answer 2

The first one is correct. As for the second one, you can't say $dKe=\Delta Ke$, because in this case $\Delta Ke $is very large.Had it been a very small increase in Kinetic energy (say, $0.1\%$), this substitution ($dKe=\Delta Ke$) would have given the correct answer. However, even in that case, only the first method gives the exact answer, while the second one is just an approximation,to make calculations easier.