Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 14630 by Don sai last updated on 03/Jun/17

solve the eqn dr/dθ=[r(a^2 −r^2 )/a^2 +r^2 ]cotθ hint. let a^2 +r^2 =a^2 −r^2 +2r^2 .

$$\mathrm{solve}\:\mathrm{the}\:\mathrm{eqn} \\ $$$$\mathrm{dr}/\mathrm{d}\theta=\left[\mathrm{r}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} \right)/\mathrm{a}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right]\mathrm{cot}\theta \\ $$$$\mathrm{hint}.\:\mathrm{let}\:\mathrm{a}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} +\mathrm{2r}^{\mathrm{2}} . \\ $$

Commented by prakash jain last updated on 03/Jun/17

(dr/dθ)=((r(a^2 −r^2 ))/((a^2 +r^2 )))cot θ (((a^2 +r^2 ))/(r(a^2 −r^2 )))dr=cot θdθ ∫(((a^2 +r^2 ))/(r(a^2 −r^2 )))dr (using hint a^2 +r^2 =a^2 −r^2 +2r^2 =∫(((a^2 −r^2 +2r^2 ))/(r(a^2 −r^2 )))dr =∫(((a^2 −r^2 ))/(r(a^2 −r^2 )))dr+∫((2r^2 )/(r(a^2 −r^2 )))dr =∫(dr/r)+∫((2r)/((a^2 −r^2 )))dr sunstitute a^2 −r^2 =u in second part −2rdr=du =ln r+∫((−du)/u) +c =ln r−ln u=ln(r/(a^2 −r^2 ))+C ∫cot θdθ=ln sin θ+c ln (r/(a^2 −r^2 ))=ln sin θ+c (r/((a^2 −r^2 )sin θ))=e^c

$$\frac{{dr}}{{d}\theta}=\frac{{r}\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)}\mathrm{cot}\:\theta \\ $$$$\frac{\left({a}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)}{{r}\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{dr}=\mathrm{cot}\:\theta{d}\theta \\ $$$$\int\frac{\left({a}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)}{{r}\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{dr} \\ $$$$\left({using}\:{hint}\:{a}^{\mathrm{2}} +{r}^{\mathrm{2}} ={a}^{\mathrm{2}} −{r}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} \right. \\ $$$$=\int\frac{\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} \right)}{{r}\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{dr} \\ $$$$=\int\frac{\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{{r}\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{dr}+\int\frac{\mathrm{2}{r}^{\mathrm{2}} }{{r}\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{dr} \\ $$$$=\int\frac{{dr}}{{r}}+\int\frac{\mathrm{2}{r}}{\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}{dr} \\ $$$${sunstitute}\:{a}^{\mathrm{2}} −{r}^{\mathrm{2}} ={u}\:{in}\:{second} \\ $$$${part}\:−\mathrm{2}{rdr}={du} \\ $$$$=\mathrm{ln}\:{r}+\int\frac{−{du}}{{u}}\:+{c} \\ $$$$=\mathrm{ln}\:{r}−\mathrm{ln}\:{u}={ln}\frac{{r}}{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }+{C} \\ $$$$\int\mathrm{cot}\:\theta{d}\theta=\mathrm{ln}\:\mathrm{sin}\:\theta+{c} \\ $$$$\mathrm{ln}\:\frac{{r}}{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }=\mathrm{ln}\:\mathrm{sin}\:\theta+{c} \\ $$$$\frac{{r}}{\left({a}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)\mathrm{sin}\:\theta}={e}^{{c}} \\ $$

Commented by Don sai last updated on 06/Jun/17

Thank you

$$\mathrm{Thank}\:\mathrm{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com