She multiplied through:
Thus, the velocity profile was:
[ \frac{dv}{dr} + \frac{v}{r} = 3r^2 ]
Kael nodded grimly. "That’s the energy. If you release a counter-vortex with exactly that integrated strength, shaped like ( u(r) = 48 - \frac{3}{4}r^3 ), the sum of the two integrals will be zero. The Churnheart will still itself." Integral calculus including differential equations
[ 4^4 = 256, \quad \frac{3}{16} \times 256 = 3 \times 16 = 48 ]
Integrating both sides with respect to ( r ):
[ \int_{0}^{4} \frac{3}{4} r^3 , dr = \frac{3}{4} \cdot \left[ \frac{r^4}{4} \right]_{0}^{4} = \frac{3}{16} \left( 4^4 - 0 \right) ] She multiplied through: Thus, the velocity profile was:
"Here," said her master, old Kael, handing her a data slate. "This equation models how the spin changes with radius. The whirlpool’s total destructive potential is the area under the velocity curve from ( r=0 ) to ( r=R ). Solve for ( v(r) ), then integrate it. That area is the energy you must dissipate."
Now came the integral calculus. The total destructive potential ( P ) was the integral of velocity across the whirlpool’s radius ( R ) (which was 4 meters):
[ v(r) = \frac{3}{4} r^3 + \frac{C}{r} ] The Churnheart will still itself
Lyra, a young apprentice, faced her final trial: to tame the , a rogue whirlpool deep beneath the city that pulsed with erratic, destructive energy. If she failed, Aethelburg would be torn apart by the year's first monsoon.
[ r \frac{dv}{dr} + v = 3r^3 ]
Lyra recognized the form. It was a first-order linear ODE. She rewrote it:
