Tangent Planes and Linear Approximations - Calculus 3

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Question

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

Answer

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.

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

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

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

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

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Question

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

Answer

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

.

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Question

Find the Linear Approximation to $z=2+\frac {x^2}{9}+\frac {y^2}{4}$ at .

Answer

We are just asking for the equation of the tangent plane:

Step 1: Find

$f(x,y)=2+\frac {(-3)^2}{9}+\frac {2^2}{4}=2+\frac {9}{9}+\frac {4}{4}=2+1+1=4$

Step 2: Take the partial derivative of $\frac {x^2}{9}$ with respect with (x,y):

$\partial(\frac {x^2}{9})=\frac {2x}{9}$

Step 3: Evaluate the partial derivative of x at

$\frac {2(3)}{9}=\frac {6}{9}=\frac {2}{3}$

Step 4: Take the partial derivative of $\frac {y^2}{4}$ with respect to :

$\partial (\frac{y^2}{4})=\frac {2y}{4}$

Step 5: Evaluate the partial derivative at

$\frac {2(2)}{4}=\frac {4}{4}=1$ .

Step 6: Convert (x,y) back into binomials:

$x\rightarrow (x+3)$
$y\rightarrow (y-2)$

Step 7: Write the equation of the tangent line:

$4+\frac {2}{3}(x+3)+(y-2)$

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Question

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

Answer

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,

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Question

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

Answer

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

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Question

Find the equation of the tangent plane to at the point

Answer

To find the equation of the tangent plane, we need 5 things:

$f_x=\frac{\partial }{\partial x}(3x^2y)=6xy$

$f_y=\frac{\partial }{\partial y}(3x^2y)=3x^2$

Using the equation of the tangent plane

, we get

Through algebraic manipulation to get z by itself, we get

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