Question Number 101293 by mhmd last updated on 01/Jul/20
Commented by mhmd last updated on 01/Jul/20
$${help}\:{me}\:{sir} \\ $$
Commented by smridha last updated on 01/Jul/20
![unit vector along A^→ isA^ =(1/( (√(21))))(4i^ −2j^ −k^ ) gradient of scaler field at(−1,1,1) [▽∅]_((−1,1,1)) =[(8x+z)i^ −z^2 j^ +(x−2zy)k^ ]_((−1,11)) =[−7i^ −j^ −3k^ ] so the directional derivative is =[▽∅]_((−1,1,1)) .A^ =(1/( (√(21))))(−28+2+3) =((−23)/( (√(21)))).](https://www.tinkutara.com/question/Q101298.png)
$$\boldsymbol{{unit}}\:\boldsymbol{{vector}}\:\boldsymbol{{along}}\:\overset{\rightarrow} {\boldsymbol{{A}}}\:\boldsymbol{{is}}\hat {\boldsymbol{{A}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{21}}}\left(\mathrm{4}\hat {\boldsymbol{{i}}}−\mathrm{2}\hat {\boldsymbol{{j}}}−\hat {\boldsymbol{{k}}}\right) \\ $$$$\boldsymbol{{gradient}}\:\boldsymbol{{of}}\:\boldsymbol{{scaler}}\:\boldsymbol{{field}}\:\boldsymbol{{at}}\left(−\mathrm{1},\mathrm{1},\mathrm{1}\right) \\ $$$$\left[\bigtriangledown\boldsymbol{\emptyset}\right]_{\left(−\mathrm{1},\mathrm{1},\mathrm{1}\right)} =\left[\left(\mathrm{8}\boldsymbol{{x}}+\boldsymbol{{z}}\right)\hat {\boldsymbol{{i}}}−\boldsymbol{{z}}^{\mathrm{2}} \hat {\boldsymbol{{j}}}+\left(\boldsymbol{{x}}−\mathrm{2}\boldsymbol{{zy}}\right)\hat {\boldsymbol{{k}}}\right]_{\left(−\mathrm{1},\mathrm{11}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[−\mathrm{7}\hat {\boldsymbol{{i}}}−\hat {\boldsymbol{{j}}}−\mathrm{3}\hat {\boldsymbol{{k}}}\right] \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{the}}\:\boldsymbol{{directional}}\:\boldsymbol{{derivative}}\:\boldsymbol{{is}} \\ $$$$=\left[\bigtriangledown\boldsymbol{\emptyset}\right]_{\left(−\mathrm{1},\mathrm{1},\mathrm{1}\right)} .\hat {\boldsymbol{{A}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{21}}}\left(−\mathrm{28}+\mathrm{2}+\mathrm{3}\right) \\ $$$$=\frac{−\mathrm{23}}{\:\sqrt{\mathrm{21}}}. \\ $$
Commented by mhmd last updated on 01/Jul/20
$${thank}\:{you}\:{sir}\:.{very}\:{very}\:{thank}\: \\ $$
Commented by smridha last updated on 01/Jul/20
![∫_0 ^2 dx[ysin(x)+sin(y)]_0 ^𝛑 =𝛑∫_0 ^2 sin(x)dx=−𝛑[cos(x)]_0 ^2 =𝛑(1−cos2) or 2𝛑sin^2 (1)](https://www.tinkutara.com/question/Q101300.png)
$$\int_{\mathrm{0}} ^{\mathrm{2}} \boldsymbol{{dx}}\left[\boldsymbol{{ysin}}\left(\boldsymbol{{x}}\right)+\boldsymbol{{sin}}\left(\boldsymbol{{y}}\right)\right]_{\mathrm{0}} ^{\boldsymbol{\pi}} \\ $$$$=\boldsymbol{\pi}\int_{\mathrm{0}} ^{\mathrm{2}} \boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}=−\boldsymbol{\pi}\left[\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} =\boldsymbol{\pi}\left(\mathrm{1}−\boldsymbol{{cos}}\mathrm{2}\right) \\ $$$$\boldsymbol{{or}}\:\mathrm{2}\boldsymbol{\pi{sin}}^{\mathrm{2}} \left(\mathrm{1}\right) \\ $$
Commented by smridha last updated on 01/Jul/20
welcome...
Commented by mhmd last updated on 01/Jul/20
$${thank}\:{you}\:{sir}\: \\ $$
Commented by smridha last updated on 01/Jul/20
![dy−32x^2 sin(2x)dx=0 integrating both sides ... y−32[x^2 ∫sin2x+∫x.cos(2x)]=C y−32[−((x^2 cos2x)/2)+((xsin(2x))/2)+(1/4)cos2x]=C y+16x^2 cos2x−16xsin2x−8cos2x=C](https://www.tinkutara.com/question/Q101305.png)
$$\boldsymbol{{dy}}−\mathrm{32}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{x}}\right)\boldsymbol{{dx}}=\mathrm{0} \\ $$$$\boldsymbol{{integrating}}\:\boldsymbol{{both}}\:\boldsymbol{{sides}}\:… \\ $$$$\boldsymbol{{y}}−\mathrm{32}\left[\boldsymbol{{x}}^{\mathrm{2}} \int\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}+\int\boldsymbol{{x}}.\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}\right)\right]=\boldsymbol{{C}} \\ $$$$\boldsymbol{{y}}−\mathrm{32}\left[−\frac{\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}}{\mathrm{2}}+\frac{\boldsymbol{{xsin}}\left(\mathrm{2}\boldsymbol{{x}}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\right]=\boldsymbol{{C}} \\ $$$$\boldsymbol{{y}}+\mathrm{16}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}−\mathrm{16}\boldsymbol{{xsin}}\mathrm{2}\boldsymbol{{x}}−\mathrm{8}\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}=\boldsymbol{{C}} \\ $$
Commented by mhmd last updated on 01/Jul/20
$${sir}\:{can}\:{you}\:{help}\:{me}\:{in}\:{question}\:\mathrm{4}\:{pleas}\: \\ $$
Answered by mr W last updated on 01/Jul/20
$${Q}\mathrm{4} \\ $$$${see}\:{Q}\mathrm{88758} \\ $$$$ \\ $$$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} =\mathrm{8}−\mathrm{4}{x}^{\mathrm{2}} \\ $$$$\mathrm{8}{x}^{\mathrm{2}} −\mathrm{8}=\mathrm{0} \\ $$$${a}=\mathrm{8},\:{b}=\mathrm{0},\:{c}=−\mathrm{8} \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{0}^{\mathrm{2}} −\mathrm{4}×\mathrm{8}×\left(−\mathrm{8}\right)=\mathrm{256} \\ $$$${bounded}\:{area}=\frac{\Delta\sqrt{\Delta}}{\mathrm{6}{a}^{\mathrm{2}} }=\frac{\mathrm{256}\sqrt{\mathrm{256}}}{\mathrm{6}×\mathrm{8}^{\mathrm{2}} }=\frac{\mathrm{32}}{\mathrm{3}} \\ $$$$ \\ $$$$\boldsymbol{{METHOD}}\:\boldsymbol{{II}} \\ $$$${bounded}\:{area}=\mathrm{2}×\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{2}×\mathrm{4}=\frac{\mathrm{32}}{\mathrm{3}} \\ $$