JEE Main 2021 (Online) 4th August Evening ShiftQuestion 1
If $\dfrac{dw(t)}{dt}=t\sqrt{121-t^{2}}$ and $w(0)=0$, then the value of $3w\left(\sqrt{21}\right)$ is $\_\_\_\_\_\_\_\_$ .
Differential Equation
Answer$331$
Solution
Concept. Integrate directly with $u=121-t^2$. Formulas used. (1) $\int t\sqrt{121-t^2}\,dt=-\frac13\left(121-t^2\right)^{3/2}+C$Step 1 - Integrate. By (1) and $w(0)=0$: $$w(t)=\frac13\left(121^{3/2}-\left(121-t^2\right)^{3/2}\right)=\frac13\left(1331-\left(121-t^2\right)^{3/2}\right).$$ Step 2 - At $t=\sqrt{21}$. $121-21=100$ and $100^{3/2}=1000$, so $$3w\left(\sqrt{21}\right)=1331-1000=331.$$ Answer: 331