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{\bf Exam Question

Topic: Tangent Plane}

Find an equation for the plane which is tangent to the surface
whose equation is $z=x^2y^3-3(x+y^2)$ at the point $P(1,-1,7).$

Express the equation in the form $ax+by+cz=d,$ where $a,b,c,d$ are
constants

the line $L$ is a horizontal tangent line (parallel to the $x$-$y$
plane) to the surface at the point $P(1,-1,7).$

Find the angle which $L$ makes with the $x$-direction

Express your answer in radians correct to 3 decimal places using
your calculator.

 \vspace{0.5in}

{\bf Solution}

$z=f(x,y)=x^2y^3-3(x+y^2)$

$f_x=2xy^3-3;\ \ f_x(1,-1)=-5.$

$f_y=3x^2y^2-6y;\ \ f_y(1,-1)=9.$

The equation of the tangent plane is therefore

$z+7= -5(x-1)+9(y+1);\ \  z+7=-5x+5+9y+9;\ \ 5x-9y+z=7.$

the directional derivative at $P$ in direction $\theta$ is given
by $$D(\theta)=-5\cos\theta+9\sin\theta.$$ This is zero when
$\tan\theta=5/9;\ \ \theta=0.507$ radians (3 d.p.)


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