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Calculus 3 : Tangent Planes and Linear Approximations

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Example Questions

Example Question #1 :Tangent Planes And Linear Approximations

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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.

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

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

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

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

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Example Question #2 :Tangent Planes And Linear Approximations

Find the tangent plane to the function $f(x,y) = 2xy+e^{-x}$ at the point (0,1,1) .

Possible Answers:

y-z=-1

y-z=0

x-z=2

x+y=1

-x+z=1

Correct answer:

-x+z=1

Explanation:

为了找到切平面的方程,我们使用the formula

z- z_0 = f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) .

Taking partial derivatives, we have

f_x(x,y) = $2y-e^{-x}$

f_y(x,y) = 2x

Substituting our x, y values into these, we get

f_x(0,1) =1

f_y(0,1) = 0

Substituting our point into x_0, y_0, z_0 , and partial derivative values in the formula we get

z - 1 = 1(x-0)+0(y-1)

z - 1 = x

-x+z =1 .

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Example Question #3 :Tangent Planes And Linear Approximations

Find the Linear Approximation to $z=2+\frac {x^2}{9}+\frac {y^2}{4}$ at (-3,2) .

Possible Answers:

$-4+\frac {2}{3}(x+3)+(y-1)$

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

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

None of the Above

Correct answer:

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

Explanation:

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

Step 1: Find f(x,y)

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

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 (x,y) :

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

Step 5: Evaluate the partial derivative at y=2

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

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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Example Question #4 :Tangent Planes And Linear Approximations

Find the equation of the plane tangent to z=2x^2+4y^3 at the point (1,5) .

Possible Answers:

z=x+30y-10

z=2x+300y-100

z=4x+300y-1002

z=4x^2+3y^3-1002

Correct answer:

z=4x+300y-1002

Explanation:

To find the equation of the tangent plane, we find: f_x,f_y,z_0 and evaluate f_x, f_y 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 z_0=2(1^2)+4(5^3)=502 . Evaluatingat the point (1,5) gets us f_x=4, f_y=300 . We then plug these values into the formula for the tangent plane: z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) . We then get z-502=4(x-1)+300(y-5) . The equation of the plane then becomes, through algebra, z=4x+300y-1002

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Example Question #5 :Tangent Planes And Linear Approximations

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the equation of the tangent plane, we find: f_x,f_y,z_0 and evaluate f_x, f_y 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 $z_0=3\sin(\frac{\pi}{2})+2\cos(\frac{\pi}{6})=3+\sqrt3$ . Evaluatingat the point $(\frac{\pi}{2},\frac{\pi}{6})$ gets us f_x=0, f_y=-1 . We then plug these values into the formula for the tangent plane: z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) . 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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Example Question #6 :Tangent Planes And Linear Approximations

Find the equation of the tangent plane to z=3x^2y at the point (2,4)

Possible Answers:

z=48x-12

z=x+36y

z=5x+50y

z=48x+12y-96

Correct answer:

z=48x+12y-96

Explanation:

To find the equation of the tangent plane, we need 5 things: f_x,f_y,z(x,y),f_x(x,y),f_y(x,y)

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

f_x(2,4)=6*2*4=48

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

f_y(2,4)=3(2)^2=12

z(2,4)=3(2)^2(4)=48

使用切平面的方程

z-z(x,y)=f_x(x,y)(x-x_0)+f_y(x,y)(y-y_0) , we get

z-48=48(x-2)+12(y-4)

Through algebraic manipulation to get z by itself, we get

z=48x+12y-96

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Calculus 3 : Tangent Planes and Linear Approximations

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 :Tangent Planes And Linear Approximations

Example Question #2 :Tangent Planes And Linear Approximations

Example Question #3 :Tangent Planes And Linear Approximations

Example Question #4 :Tangent Planes And Linear Approximations

Example Question #5 :Tangent Planes And Linear Approximations

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