The differential equation of the family of cu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the equation of the family of curves: $r^2 = a^2 \cos 2	heta$. Differentiating both sides with respect to $	heta$: $\frac{d}{d	heta}(r^2)

Vedclass · Accepted Answer

$\frac{dr}{d	heta} = -r 	an 2	heta$

Vedclass · Answer

(B) Given the equation of the family of curves: $r^2 = a^2 \cos 2	heta$. Differentiating both sides with respect to $	heta$: $\frac{d}{d	heta}(r^2) = \frac{d}{d	heta}(a^2 \cos 2	heta)$. Using the chain rule: $2r \frac{dr}{d	heta} = a^2 (-\sin 2	heta) \cdot 2$. Simplifying: $r \frac{dr}{d	heta} = -a^2 \sin 2	heta$. From the original equation,$a^2 = \frac{r^2}{\cos 2	heta}$. Substituting $a^2$ into the differentiated equation: $r \frac{dr}{d	heta} = -\left(\frac{r^2}{\cos 2	heta}ight) \sin 2	heta$. $r \frac{dr}{d	heta} = -r^2 	an 2	heta$. Dividing by $r$ (assuming $r 
eq 0$): $\frac{dr}{d	heta} = -r 	an 2	heta$.

The differential equation of the family of curves $r^2 = a^2 \cos 2\theta$,where '$a$' is an arbitrary constant,is:

Similar Questions

The order of the differential equation whose solution is $a e^{x} + b e^{2x} + c e^{3x} + d = 0$ is:

The differential equation of the family of curves $v = \frac{A}{r} + B$,where $A$ and $B$ are arbitrary constants,is

The differential equation whose solution is $y=c^2+\frac{c}{x}$,where $c$ is a constant,is

If $l$ and $m$ are the degree and the order respectively of the differential equation of the family of all circles in the $XY$ plane with radius $5$ units,then $2l + 3m =$

Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.

Vedclass Test Series

Exam Paper Generator

Online Exam Module