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

Commented bymhmd last updated on 01/Jul/20

help me sir

Commented bysmridha 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))).

Commented bymhmd last updated on 01/Jul/20

thank you sir .very very thank

Commented bysmridha 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)

Commented bysmridha last updated on 01/Jul/20

welcome...

Commented bymhmd last updated on 01/Jul/20

thank you sir

Commented bysmridha 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

Commented bymhmd last updated on 01/Jul/20

sir can you help me in question 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)