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Calculus 3 : Applications of Partial Derivatives

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Example Questions

Example Question #1 : Applications Of Partial Derivatives

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

Possible Answers:

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

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

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

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

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

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 #1 : 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

-x+z=1

x+y=1

x-z=2

y-z=0

Correct answer:

-x+z=1

Explanation:

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

None of the Above

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

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

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

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 #3 : 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=4x+300y-1002

z=x+30y-10

z=2x+300y-100

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 . Evaluating at 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:

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

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

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

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

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 $z_0=3\sin(\frac{\pi}{2})+2\cos(\frac{\pi}{6})=3+\sqrt3$ . Evaluating at 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=x+36y

z=48x+12y-96

z=48x-12

z=5x+50y

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

Using the equation of the tangent plane

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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Example Question #1 : Applications Of Partial Derivatives

Find the absolute minimums and maximums of f(x,y)=10x^2-3y^2+10y on the disk of radius , $x^2+y^2\leq16$ .

Possible Answers:

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ , $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ , $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,4)

Absolute Maximum: $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Correct answer:

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ , $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Explanation:

The first thing we need to do is find the partial derivative in respect to , and .

$\frac{\partial}{\partial x}=20x$ , $\frac{\partial}{\partial y}=-6y+10$

We need to find the critical points, so we set each of the partials equal to .

$0=20x\rightarrow x=0$

$0=-6y+10 \rightarrow y=\frac{5}{3}$

We only have one critical point at $f\Big(0, \frac{5}{3}\Big)$ , now we need to find the function value in order to see if it is inside or outside the disk.

$f\Big(0, \frac{5}{3}\Big)=10(0)^2-3\Big(\frac{5}{3}\Big)^2+10\Big(\frac{5}{3}\Big)=\frac{25}{3}\approx8.33333$

This is within our disk.

We now need to take a look at the boundary, x^2+y^2=16 . We can solve for x^2 , and plug it into .

x^2+y^2=16

x^2=16-y^2

g(y)=10(16-y^2)-3y^2+10y=-13y^2+10y+160

We will need to find the absolute extrema of this function on the range $-4\leq y \leq 4$ . We need to find the critical points of this function.

g'(y)=-26y+10

$0=-26y+10 \rightarrow y=\frac{5}{13}$

The function value at the critical points and end points are:

g(-4)=-13(-4)^2+10(-4)+160=-88

g(4)=-13(4)^2+10(4)+160=-8

$g\Big(\frac{5}{13}\Big)=-13\Big(\frac{5}{13}\Big)^2+10\Big(\frac{5}{13}\Big)+160=\frac{2105}{13}\approx161.92$

Now we need to figure out the values of these correspond to.

$y=-4: x^2=(-4)^2-16 \rightarrow x^2=0\rightarrow x=0$

$y=4: x^2=4^2-16 \rightarrow x^2=0\rightarrow x=0$

$y=\frac{5}{13}: x^2=\Big(\frac{5}{13}\Big)^2-16 \rightarrow x^2=\frac{2679}{169}\rightarrow x\approx\pm\ 3.98$

Now lets summarize our results as follows:

$g(-4)=-88 \rightarrow f(0,-4)=-88$

$g(4)=-8 \rightarrow f(0,4)=-8$

$g\Big(\frac{5}{13}\Big)=\frac{2105}{13} \rightarrow f\Big(-\sqrt{\frac{2679}{169}},\frac{5}{13}\Big)=\frac{2105}{13}$

$g\Big(\frac{5}{13}\Big)=\frac{2105}{13} \rightarrow f\Big(\sqrt{\frac{2679}{169}},\frac{5}{13}\Big)=\frac{2105}{13}$

From this we can conclude that there is an absolute minimum at (0,-4) , and two absolute maximums at $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ and $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ .

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Example Question #1 : Applications Of Partial Derivatives

Find the minimum and maximum of f(x,y)=2x-5y , subject to the constraint x^2+y^2=144 .

Possible Answers:

f(4.46,-11.14) is a maximum

f(-4.46,11.14) is a minimum

f(4.46,11.14) is a maximum

f(-4.46,-11.14) is a minimum

There are no maximums or minimums

f(-4.46,11.14) is a maximum

f(4.46,-11.14) is a minimum

f(4.46,11.14) is a maximum

f(4.46,-11.14) is a minimum

Correct answer:

f(4.46,-11.14) is a maximum

f(-4.46,11.14) is a minimum

Explanation:

First we need to set up our system of equations.

$2=2\lambda x \rightarrow x=\frac{1}{\lambda}$

$-5=2\lambda y \rightarrow y=-\frac{5}{2 \lambda}$

x^2+y^2=144

Now lets plug in these constraints.

$(\frac{1}{\lambda})^2+(-\frac{5}{2\lambda})^2=144$

$\frac{1}{\lambda ^2}+\frac{25}{4\lambda^2}=144$

$\frac{29}{4\lambda^2}=144$

Now we solve for $\lambda$

$\frac{29}{4\lambda^2}=144\rightarrow \lambda^2=\frac{29}{576}\rightarrow\lambda=\pm\sqrt{\frac{29}{576}}$

$\lambda=\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx 4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx -11.14$

$\lambda=-\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx -4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx 11.14$

Now lets plug in these values of , and into the original equation.

f(4.46,-11.14)=2(4.46)-5(-11.14)=64.62

f(-4.46,11.14)=2(-4.46)-5(11.14)=-64.62

We can conclude from this that f(4.46,-11.14) is a maximum, and f(-4.46,11.14) is a minimum.

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Example Question #1 : Lagrange Multipliers

Find the absolute minimum value of the function f(x,y) = x^2+y^2 subject to the constraint x^2 +2y^2 = 1 .

Possible Answers:

$\frac{1}{2}$

$\sqrt2$

$2\sqrt2$

Correct answer:

$\frac{1}{2}$

Explanation:

Let g = x^2 +2y^2 To find the absolute minimum value, we must solve the system of equations given by

$\triangledown f = \lambda\triangledown g, g =1$ .

So this system of equations is

$f_x = \lambda g_x$ , $f_y = \lambda g_y$ , g =1 .

Taking partial derivatives and substituting as indicated, this becomes

$2x = \lambda(2x), 2y = \lambda(4y), x^2+2y^2 = 1$ .

From the left equation, we see either x=0 or $\lambda = 1$ . If x=0 , then substituting this into the other equations, we can solve for $y, \lambda$ , and get $y = \pm \sqrt{2}/2$ , $\lambda = 1/2$ , giving two extreme candidate points at $(0, \frac{\sqrt{2}}{2}), (0, -\frac{\sqrt{2}}{2})$ .

On the other hand, if instead $\lambda =1$ , this forces y = 0 from the 2nd equation, and $x = \pm 1$ from the 3rd equation. This gives us two more extreme candidate points; (-1,0),(1,0) .

Taking all four of our found points, and plugging them back into , we have

$f(-1,0) = 1, f(1,0) = 1, f(0,\frac{\sqrt{2}}{2}) = \frac{1}{2}, f(0,-\frac{\sqrt{2}}{2}) = \frac{1}{2}$ .

Hence the absolute minimum value is $\frac{1}{2}$ .

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Example Question #1 : Lagrange Multipliers

Find the dimensions of a box with maximum volume such that the sum of its edges is cm.

Possible Answers:

$6 \times 5 \times 4$

$5 \times 5 \times 5$

$5 \times 4 \times 3$

$6 \times 6 \times 6$

Correct answer:

$5 \times 5 \times 5$

Explanation:

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Calculus 3 : Applications of Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Applications Of Partial Derivatives

Example Question #1 : Tangent Planes And Linear Approximations

Example Question #3 : Tangent Planes And Linear Approximations

Example Question #3 : Tangent Planes And Linear Approximations

Example Question #5 : Tangent Planes And Linear Approximations

Example Question #6 : Tangent Planes And Linear Approximations

Example Question #1 : Applications Of Partial Derivatives

Example Question #1 : Applications Of Partial Derivatives

Example Question #1 : Lagrange Multipliers

Example Question #1 : Lagrange Multipliers

All Calculus 3 Resources

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