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Question Number 101293 by mhmd last updated on 01/Jul/20

Commented by mhmd last updated on 01/Jul/20

help me sir

$${help}\:{me}\:{sir} \\ $$

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))).

$$\boldsymbol{{unit}}\:\boldsymbol{{vector}}\:\boldsymbol{{along}}\:\overset{\rightarrow} {\boldsymbol{{A}}}\:\boldsymbol{{is}}\hat {\boldsymbol{{A}}}=\frac{\mathrm{1}}{\sqrt{\mathrm{21}}}\left(\mathrm{4}\hat {\boldsymbol{{i}}}−\mathrm{2}\hat {\boldsymbol{{j}}}−\hat {\boldsymbol{{k}}}\right) \\ $$$$\boldsymbol{{gradient}}\:\boldsymbol{{of}}\:\boldsymbol{{scaler}}\:\boldsymbol{{field}}\:\boldsymbol{{at}}\left(−\mathrm{1},\mathrm{1},\mathrm{1}\right) \\ $$$$\left[\bigtriangledown\boldsymbol{\emptyset}\right]_{\left(−\mathrm{1},\mathrm{1},\mathrm{1}\right)} =\left[\left(\mathrm{8}\boldsymbol{{x}}+\boldsymbol{{z}}\right)\hat {\boldsymbol{{i}}}−\boldsymbol{{z}}^{\mathrm{2}} \hat {\boldsymbol{{j}}}+\left(\boldsymbol{{x}}−\mathrm{2}\boldsymbol{{zy}}\right)\hat {\boldsymbol{{k}}}\right]_{\left(−\mathrm{1},\mathrm{11}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[−\mathrm{7}\hat {\boldsymbol{{i}}}−\hat {\boldsymbol{{j}}}−\mathrm{3}\hat {\boldsymbol{{k}}}\right] \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{the}}\:\boldsymbol{{directional}}\:\boldsymbol{{derivative}}\:\boldsymbol{{is}} \\ $$$$=\left[\bigtriangledown\boldsymbol{\emptyset}\right]_{\left(−\mathrm{1},\mathrm{1},\mathrm{1}\right)} .\hat {\boldsymbol{{A}}}=\frac{\mathrm{1}}{\sqrt{\mathrm{21}}}\left(−\mathrm{28}+\mathrm{2}+\mathrm{3}\right) \\ $$$$=\frac{−\mathrm{23}}{\sqrt{\mathrm{21}}}. \\ $$

Commented by mhmd last updated on 01/Jul/20

thank you sir .very very thank

$${thank}\:{you}\:{sir}\:.{very}\:{very}\:{thank}\: \\ $$

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)

$$\int_{\mathrm{0}} ^{\mathrm{2}} \boldsymbol{{dx}}\left[\boldsymbol{{ysin}}\left(\boldsymbol{{x}}\right)+\boldsymbol{{sin}}\left(\boldsymbol{{y}}\right)\right]_{\mathrm{0}} ^{\boldsymbol{\pi}} \\ $$$$=\boldsymbol{\pi}\int_{\mathrm{0}} ^{\mathrm{2}} \boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}=−\boldsymbol{\pi}\left[\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} =\boldsymbol{\pi}\left(\mathrm{1}−\boldsymbol{{cos}}\mathrm{2}\right) \\ $$$$\boldsymbol{{or}}\:\mathrm{2}\boldsymbol{\pi{sin}}^{\mathrm{2}} \left(\mathrm{1}\right) \\ $$

Commented by smridha last updated on 01/Jul/20

welcome...

Commented by mhmd last updated on 01/Jul/20

thank you sir

$${thank}\:{you}\:{sir}\: \\ $$

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

$$\boldsymbol{{dy}}−\mathrm{32}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{x}}\right)\boldsymbol{{dx}}=\mathrm{0} \\ $$$$\boldsymbol{{integrating}}\:\boldsymbol{{both}}\:\boldsymbol{{sides}}\:... \\ $$$$\boldsymbol{{y}}−\mathrm{32}\left[\boldsymbol{{x}}^{\mathrm{2}} \int\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}+\int\boldsymbol{{x}}.\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}\right)\right]=\boldsymbol{{C}} \\ $$$$\boldsymbol{{y}}−\mathrm{32}\left[−\frac{\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}}{\mathrm{2}}+\frac{\boldsymbol{{xsin}}\left(\mathrm{2}\boldsymbol{{x}}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\right]=\boldsymbol{{C}} \\ $$$$\boldsymbol{{y}}+\mathrm{16}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}−\mathrm{16}\boldsymbol{{xsin}}\mathrm{2}\boldsymbol{{x}}−\mathrm{8}\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}=\boldsymbol{{C}} \\ $$

Commented by mhmd last updated on 01/Jul/20

sir can you help me in question 4 pleas

$${sir}\:{can}\:{you}\:{help}\:{me}\:{in}\:{question}\:\mathrm{4}\:{pleas}\: \\ $$

Answered by mr W last updated on 01/Jul/20

Q4  see Q88758    Method I  4x^2 =8−4x^2   8x^2 −8=0  a=8, b=0, c=−8  Δ=b^2 −4ac=0^2 −4×8×(−8)=256  bounded area=((Δ(√Δ))/(6a^2 ))=((256(√(256)))/(6×8^2 ))=((32)/3)    METHOD II  bounded area=2×(2/3)×2×4=((32)/3)

$${Q}\mathrm{4} \\ $$$${see}\:{Q}\mathrm{88758} \\ $$$$ \\ $$$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} =\mathrm{8}−\mathrm{4}{x}^{\mathrm{2}} \\ $$$$\mathrm{8}{x}^{\mathrm{2}} −\mathrm{8}=\mathrm{0} \\ $$$${a}=\mathrm{8},\:{b}=\mathrm{0},\:{c}=−\mathrm{8} \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{0}^{\mathrm{2}} −\mathrm{4}×\mathrm{8}×\left(−\mathrm{8}\right)=\mathrm{256} \\ $$$${bounded}\:{area}=\frac{\Delta\sqrt{\Delta}}{\mathrm{6}{a}^{\mathrm{2}} }=\frac{\mathrm{256}\sqrt{\mathrm{256}}}{\mathrm{6}×\mathrm{8}^{\mathrm{2}} }=\frac{\mathrm{32}}{\mathrm{3}} \\ $$$$ \\ $$$$\boldsymbol{{METHOD}}\:\boldsymbol{{II}} \\ $$$${bounded}\:{area}=\mathrm{2}×\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{2}×\mathrm{4}=\frac{\mathrm{32}}{\mathrm{3}} \\ $$

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