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Question Number 65797 by Souvik Ghosh last updated on 04/Aug/19
find the constant a,b and c so that the direction derivative of Φ=axy^2 +byz+cz^2 x^3 at (1,2,−1) has a maximum of magnitude 64 jn a direction parallel to the z axis.
$${find}\:{the}\:{constant}\:\:{a},{b}\:{and}\:\:{c}\:\:{so} \\ $$$${that}\:{the}\:{direction}\:{derivative}\:{of} \\ $$$$\Phi={axy}^{\mathrm{2}} +{byz}+{cz}^{\mathrm{2}} {x}^{\mathrm{3}} \:{at}\:\left(\mathrm{1},\mathrm{2},−\mathrm{1}\right) \\ $$$${has}\:{a}\:{maximum}\:{of}\:{magnitude} \\ $$$$\mathrm{64}\:{jn}\:{a}\:{direction}\:{parallel}\:{to}\:{the} \\ $$$${z}\:{axis}. \\ $$
Answered by Tanmay chaudhury last updated on 04/Aug/19
▽Φ (i(∂/∂x)+j(∂/∂y)+k(∂/∂z))(axy^2 +byz+cz^2 x^3 ) i(ay^2 +3x^2 cz^2 )+j(2axy+bz)+k(by+2cx^3 z)
$$\bigtriangledown\Phi \\ $$$$\left({i}\frac{\partial}{\partial{x}}+{j}\frac{\partial}{\partial{y}}+{k}\frac{\partial}{\partial{z}}\right)\left({axy}^{\mathrm{2}} +{byz}+{cz}^{\mathrm{2}} {x}^{\mathrm{3}} \right) \\ $$$${i}\left({ay}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} {cz}^{\mathrm{2}} \right)+{j}\left(\mathrm{2}{axy}+{bz}\right)+{k}\left({by}+\mathrm{2}{cx}^{\mathrm{3}} {z}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tanmay chaudhury last updated on 04/Aug/19
eait i have corrected my error detected by you ..thank you wait iam trying
$${eait}\:{i}\:{have}\:{corrected}\:{my}\:{error}\:{detected}\:{by}\:{you}\:..{thank}\:{you} \\ $$$${wait}\:{iam}\:{trying} \\ $$
Commented by Souvik Ghosh last updated on 04/Aug/19
but sir hear ▽^→ Φ=(ay^2 +3cz^2 x^2 )i+(2axy+bz)j +(by+2czx^3 )k
$${but}\:{sir}\:{hear}\: \\ $$$$\overset{\rightarrow} {\bigtriangledown}\Phi=\left({ay}^{\mathrm{2}} +\mathrm{3}{cz}^{\mathrm{2}} {x}^{\mathrm{2}} \right){i}+\left(\mathrm{2}{axy}+{bz}\right){j} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({by}+\mathrm{2}{czx}^{\mathrm{3}} \right){k} \\ $$
Commented by Tanmay chaudhury last updated on 04/Aug/19
yes you are right...in hurry
$${yes}\:{you}\:{are}\:{right}…{in}\:{hurry} \\ $$

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