AP Calculus BC - Tangent Planes and Linear Approximations

Find the linear approximation to $z=6+\frac{x^2}{4}+\frac{y^2}{16}$ at .

$L(x,y)\approx 4-x+\frac{1}{2}y$

$L(x,y)\approx 4+x-\frac{1}{2}y$

$L(x,y)\approx 4-x-\frac{1}{2}y$

$L(x,y)\approx -x+\frac{1}{2}y$

$L(x,y)\approx 4+\frac{1}{2}y$

Explanation

The question is really asking for a tangent plane, so lets first find partial derivatives and then plug in the point.

$f(x,y)=6+\frac{x^2}{4}+\frac{y^2}{16}$ , $f(-2,4)=6+\frac{(-2)^2}{4}+\frac{(4)^2}{16}=6+1+1=8$

$f_x(x,y)=\frac{x}{2}$ , $f_x(-2,4)=\frac{-2}{2}=-1$

$f_y(x,y)=\frac{y}{8}$ , $f_y(-2,4)=\frac{4}{8}=\frac{1}{2}$

Remember that we need to build the linear approximation general equation which is as follows.

$L(x,y)\approx f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

$L(x,y)\approx 8-(x+2)+\frac{1}{2}(y-4)$

$L(x,y)\approx 8-x-2+\frac{1}{2}y-2$

$L(x,y)\approx 4-x+\frac{1}{2}y$

Find the tangent plane to the function $f(x,y) = 2xy+e^{-x}$ at the point .

Explanation

To find the equation of the tangent plane, we use the formula

Taking partial derivatives, we have

$2y-e^{-x}$

Substituting our values into these, we get

Substituting our point into , and partial derivative values in the formula we get

Find the equation of the plane tangent to at the point .

Explanation

To find the equation of the tangent plane, we find: and evaluate at the point given. $f_x=\frac{\partial }{\partial x}(2x^2+4y^3)=4x$ , $f_y=\frac{\partial }{\partial y}(2x^2+4y^3)=12y^2$ , and . Evaluating at the point gets us . We then plug these values into the formula for the tangent plane: . We then get . The equation of the plane then becomes, through algebra,

Find the equation of the plane tangent to $z=3\sin(x)+2\cos(y)$ at the point $(\frac{\pi}{2},\frac{\pi}{6})$

$z=-y+\frac{\pi}{6}+3+\sqrt3$

$z=y-\frac{\pi}{3}+2+\sqrt3$

$z=x-\frac{\pi}{3}+3+\sqrt3$

$z=y-\frac{\pi}{6}$

Explanation

To find the equation of the tangent plane, we find: and evaluate at the point given. $f_x=\frac{\partial }{\partial x}(3\sin(x)+2(\cos(y))=3\cos(x)$ , $f_y=\frac{\partial }{\partial y}(3\sin(x)+2\cos(y))=-2\sin(y)$ , and $z_0=3\sin(\frac{\pi}{2})+2\cos(\frac{\pi}{6})=3+\sqrt3$ . Evaluating at the point $(\frac{\pi}{2},\frac{\pi}{6})$ gets us . We then plug these values into the formula for the tangent plane: . We then get $z-(3+\sqrt3)=0(x-\frac{\pi}{2})+-1(y-\frac{\pi}{6})$ . The equation of the plane then becomes, through algebra, $z=-y+\frac{\pi}{6}+3+\sqrt3$