Tangent Planes and Linear Approximations - AP Calculus BC

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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.

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

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

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

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

Compare your answer with the correct one above

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

.

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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.

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

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

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

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

Compare your answer with the correct one above

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

.

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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,

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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.

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

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

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

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

Compare your answer with the correct one above

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

.

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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,

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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.

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

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

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

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

Compare your answer with the correct one above

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

.

Compare your answer with the correct one above

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