The differential equation of the family of cu | vedclass

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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:

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