Question

If $r=\sqrt{x^2+y^2+z^2}$ and $x=2\sin 3t,y=2\cos 3t,z=8t$ then $\frac{dr}{dt}=$

• A. $\frac{32t}{\sqrt{1+16t^2}}$
• B. $\frac{16t}{\sqrt{1+16t^2}}$
• C. $\frac{t}{\sqrt{1+16t^2}}$
• D. $\frac{4t}{\sqrt{1+16t^2}}$

Answer 1

The correct option is A $\frac{32t}{\sqrt{1+16t^2}}$

Given : $x=2\sin 3t,y=2\cos 3t,z=8t$

$r=\sqrt{x^2+y^2+z^2}$

$=\sqrt{(2\sin 3t{)}^2+(2\cos 3t{)}^2+(8t{)}^2}$

$=\sqrt{4{\sin }^{2}3t+4{\cos }^{2}3t+64t^2}$

$=\sqrt{4+64t^2}$

$⟹r^2=4+64t^2$

Differentiating above equation we get
$2r\frac{dr}{dt}=2\times 64\times t$

$⟹\frac{dr}{dt}=\frac{64\times t}{r}$

$⟹\frac{dr}{dt}=\frac{64\times t}{\sqrt{4+64t^2}}=\frac{64t}{2\sqrt{1+16t^2}}=\frac{32t}{\sqrt{1+16t^2}}$