Section 14.1 : Tangent Planes and Linear Approximations
2. Find the equation of the tangent plane to \(z = x\sqrt {{x^2} + {y^2}} + {y^3}\) at \(\left( { - 4,3} \right)\) .
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Start SolutionFirst, we know we’ll need the two 1st order partial derivatives. Here they are,
\[{f_x} = \sqrt {{x^2} + {y^2}} + \frac{{{x^2}}}{{\sqrt {{x^2} + {y^2}} }}\hspace{0.5in}{f_y} = \frac{{xy}}{{\sqrt {{x^2} + {y^2}} }} + 3{y^2}\] Show Step 2Now we also need the two derivatives from the first step and the function evaluated at \(\left( { - 4,3} \right)\) . Here are those evaluations,
\[f\left( { - 4,3} \right) = 7\hspace{0.5in}{f_x}\left( { - 4,3} \right) = \frac{{41}}{5}\hspace{0.5in}{f_y}\left( { - 4,3} \right) = \frac{{123}}{5}\] Show Step 3The tangent plane is then,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{z = 7 + \frac{{41}}{5}\left( {x + 4} \right) + \frac{{123}}{5}\left( {y - 3} \right) = \frac{{41}}{5}x + \frac{{123}}{5}y - 34}}\]