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Calculus 3 : Calculus 3

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Example Questions

Example Question #1 : Relative Minimums And Maximums

Find the relative maxima and minima of f(x,y)=(x+4)^2-2(y-1)^2+3x^2y+17 .

Possible Answers:

(1.429,-0.757) is a saddle point.

(1.429,-0.757) is a relative minimum.

(1.429,-0.757) is a relative maximum.

(0,0) and (1.429,-0.757) are relative minima.

Correct answer:

(1.429,-0.757) is a saddle point.

Explanation:

To find the relative maxima and minima, we must find all the first order and second order partial derivatives. We will use the first order partial derivative to find the critical points, then use the equation

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=2(x+4)+6xy=2x+8+6xy$

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

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2+6y$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=-4$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=6x$

To find the critical points, we will set the first derivatives equal to

$f_x(x,y)=2x+8+6xy=0\Rightarrow 2x+6xy=-8$

$2x+6xy=-8\Rightarrow x+3xy=-4\Rightarrow x(1+3y)=-4$

$x=\frac{-4}{1+3y}$

$f_y(x,y)=3x^2-4y+4=0\Rightarrow 3\left ( \frac{-4}{1+3y} \right )^2-4y+4=0$

$3\left ( \frac{16}{(1+3y)^2} \right )-4y+4=0$

$\frac{48}{(1+3y)^2}=4y-4$

48=(4y-4)(1+3y)^2

(4y-4)(1+3y)^2-48=0

(4y-4)(1+6y+9y^2)-48=0

4y+24y^2+36y^3-4-24y-36y^2-48=0

36y^3-12y^2-20y-52=0

There is only one real value of ; $y\approx1.429$

We find the corresponding value of using $x=\frac{-4}{1+3y}$ (found by rearranging the first derivative)

$x(1.429)=\frac{-4}{1+3(1.429)}\approx -0.757$

There is a critical point at (1.429,-0.757) . We need to determine if the critical point is a maximum or minimum using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

D(x,y)=(2+6y)(-4)-(6x)^2

D(x,y)=-8-24y-36x^2

At (1.429,-0.757) ,

D(x,y)=-8-24(-0.757)-36(1.429)^2=-63.35<0

Since D(x,y)<0 , (1.429,-0.757) is a saddle point.

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Example Question #121 : Applications Of Partial Derivatives

Find the relative maxima and minima of f(x,y)=y^4+2x^3-2x^2y-7xy+9 .

Possible Answers:

(0,0) and (-1.6,-1.786) are relative maxima

(0,0) is a relative maxima, (-1.6,-1.786) is a relative minima

(0,0) and (-1.6,-1.786) are saddle points

(0,0) is a relative minima, (-1.6,-1.786) is a relative maxima

Correct answer:

(0,0) and (-1.6,-1.786) are saddle points

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=6x^2-4xy-7y$

$\frac{\partial f }{\partial y}=f_y(x,y)=4y^3-2x^2-7x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=12x-4y$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=12y^2$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=-4x-7$

To find the critical points, we will set the first derivatives equal to

$f_x(x,y)=6x^2-4xy-7y=0\Rightarrow 6x^2-y(4x-7)=0$

$6x^2=y(4x-7)\Rightarrow y=\frac{6x^2}{4x-7}$

$f_y(x,y)=4y^3-2x^2-7x=0\Rightarrow 4\left ( \frac{6x^2}{4x-7} \right )^3-2x^2-7x=0$

4(6x^2)^3-2x^2(4x-7)^3-7x(4x-7)^3=0(4x-7)^3

4(6x^2)^3-(2x^2+7x)(4x-7)^3=0

864x^6-128x^5+224x^4+1176x^3-3430x^2+2401x=0

The real values of are x=0 and x=-1.6

We find the corresponding value of using $y=\frac{6x^2}{4x-7}$ (found by rearranging the first derivative)

$y(0)=\frac{6(0)^2}{4(0)-7}=0$

$y(-1.6)=\frac{6(-1.6)^2}{4(-1.6)-7}=\frac{6*2.56}{-1.6-7}\approx -1.786$

There are critical points at (0,0) and (-1.6,-1.786) . We need to determine if the critical points are maxima or minima using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

D(x,y)=(12x-4y)(12y^2)-(-4x-7)^2

At (0,0) ,

D(x,y)=(12(0)-4(0))(12(0)^2)-(-4(0)-7)^2=0-(-7)^2=-49<0

Since D(x,y)<0 , (0,0) is a saddle point.

At (-1.6,-1.786) ,

D(x,y)=(12(-1.6)-4(-1.786))(12(-1.786)^2)-(-4(-1.6)-7)^2

D(x,y)=(-19.2-7.144)(38.278)-(6.4-7)^2=-1008.7<0

Since D(x,y)<0 , (-1.6,-1.786) is a saddle point.

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Example Question #3 : Relative Minimums And Maximums

Find the relative maxima and minima of f(x,y)=x^4+y^4+3xy .

Possible Answers:

(0,0) is a saddle point, $(-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ are relative minima.

(0,0) is a relative minima, $(-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ are relative maxima.

(0,0) is a saddle point, $(-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ are saddle points.

(0,0) , $(-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ are relative maxima.

Correct answer:

(0,0) is a saddle point, $(-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ are relative minima.

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=4x^3+3y$

$\frac{\partial f }{\partial y}=f_y(x,y)=4y^3+3x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=12x^2$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=12y^2$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=3$

To find the critical points, we will set the first derivatives equal to

$f_x(x,y)=4x^3+3y=0\Rightarrow 4x^3=-3y \Rightarrow y=\frac{-4x^3}{3}$

$f_y(x,y)=4y^3+3x=0\Rightarrow 4\left ( \frac{-4x^3}{3} \right )^3+3x=0$

$4\left ( \frac{-64x^9}{27} \right )+3x=0\Rightarrow \frac{-256}{27}x^9+3x=0$

$x\left (\frac{-256}{27}x^8+3 \right )=0$

Setting each factor in the expression equal to gives us

x=0 and $\frac{-256}{27}x^8+3=0\Rightarrow\frac{-256}{27}x^8=-3$

$x=\sqrt[8]{-3*\frac{27}{-257}}=\pm \sqrt3/2$

The real values of are x=0 , $x=-\sqrt3/2$ and $x=\sqrt3/2$

We find the corresponding value of using $y=\frac{-4x^3}{3}$ (found by rearranging the first derivative)

$y(0)=\frac{-4(0)^3}{3}=0$

$y(-\sqrt3/2)=\frac{-4(-\sqrt3/2)^3}{3}=\sqrt3/2$

$y(\sqrt3/2)=\frac{-4(-\sqrt3/2)^3}{3}=-\sqrt3/2$

There are critical points at (0,0) , $(-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ . We need to determine if the critical points are maxima or minima using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

D(x,y)=(12x^2)(12y^2)-(3)^2=144x^2y^2-9

At (0,0) ,

D(x,y)=144(0)^2(0)^2-9=-9<0

Since D(x,y)<0 , (0,0) is a saddle point.

At $(-\sqrt3/2,\sqrt3/2)$ ,

$D(x,y)=144\left ( \frac{-\sqrt3}{2} \right )^2\left ( \frac{\sqrt3}{2} \right )^2-9$

$D(x,y)=144\left ( \frac{3}{4} \right )\left ( \frac{3}{4} \right )-9=72>0$

$f_{xx}(x,y)=12x^2=12\left ( \frac{-\sqrt3}{2} \right )^2=12\left ( \frac{3}{4} \right )=9>0$

Since D(x,y)>0 and $f_{xx}(x,y)>0$ , $(-\sqrt3/2,\sqrt3/2)$ is a minimum.

At $(\sqrt3/2,-\sqrt3/2)$ ,

$D(x,y)=144\left ( \frac{\sqrt3}{2} \right )^2\left ( \frac{-\sqrt3}{2} \right )^2-9$

$D(x,y)=144\left ( \frac{3}{4} \right )\left ( \frac{3}{4} \right )-9=72>0$

$f_{xx}(x,y)=12x^2=12\left ( \frac{\sqrt3}{2} \right )^2=12\left ( \frac{3}{4} \right )=9>0$

Since D(x,y)>0 and $f_{xx}(x,y)>0$ , $(-\sqrt3/2,\sqrt3/2)$ is a minimum.

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Example Question #2 : Relative Minimums And Maximums

Find the relative maxima and minima of f(x,y)=sin(x)-cos(y)+4xy .

Possible Answers:

(0.0617,-0.2495) is a relative maxima

(0.0617,-0.2495) is a saddle point

(0,0) and (0.0617,-0.2495) is a relative maxima

(0,0) and (0.0617,-0.2495) is a relative minima

Correct answer:

(0.0617,-0.2495) is a saddle point

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

f(x,y)=sin(x)-cos(y)+4xy

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=cos(x)+4y$

$\frac{\partial f }{\partial y}=f_y(x,y)=sin(y)+4x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=-sin(x)$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=cos(y)$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=4$

To find the critical points, we will set the first derivatives equal to

$f_x(x,y)=cos(x)+4y=0\Rightarrow cos(x)=-4y$

y=-cos(x)/4

$f_y(x,y)=sin(y)+4x=0\Rightarrow sin\left (\frac{-cos(x)}{4} \right )+4x=0$

Using a TI-83 or other software to find the root, we find that $x\approx 0.0617$ ,

We find the corresponding value of using y=-cos(x)/4 (found by rearranging the first derivative)

$y(0.0617)=-cos(0.0617)/4 \approx-0.2495$

There is a critical points at (0.0617,-0.2495) . We need to determine if the critical point is a maximum or minimum using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

D(x,y)=-sin(x)cos(y)-4^2

At (0.0617,-0.2495) ,

$D(x,y)=-sin(0.0617)cos(-0.2495)-4^2\approx -16.06<0$

Since D(x,y)<0 , (0.0617,-0.2495) is a saddle point.

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Example Question #9 : Relative Minimums And Maximums

Find the relative maxima and minima of f(x,y)=sin^2x+cos^2y-16 .

Possible Answers:

(0,0) , $(0,3\pi/2)$ , $(\pi/2,\pi/2)$ , $(\pi/2,\pi)$ , $(\pi,\pi/2)$ , $(\pi,\pi)$ , $(3\pi/2,0)$ and $(3\pi/2,3\pi/2)$ are relative maxima

$(0,\pi/2)$ , $(0,\pi)$ , $(3\pi/2,\pi/2)$ and $(3\pi/2,\pi)$ are saddle points

$(\pi/2,0)$ , $(\pi/2,3\pi/2)$ , $(\pi,0)$ and $(\pi,3\pi/2)$ are relative minima

(0,0) , $(0,3\pi/2)$ , $(\pi/2,\pi/2)$ , $(\pi/2,\pi)$ , $(\pi,\pi/2)$ , $(\pi,\pi)$ , $(3\pi/2,0)$ and $(3\pi/2,3\pi/2)$ are saddle points

$(0,\pi/2)$ , $(0,\pi)$ , $(3\pi/2,\pi/2)$ and $(3\pi/2,\pi)$ are relative minima

$(\pi/2,0)$ , $(\pi/2,3\pi/2)$ , $(\pi,0)$ and $(\pi,3\pi/2)$ are relative maxima

(0,0) , $(0,3\pi/2)$ , $(\pi/2,\pi/2)$ , $(\pi/2,\pi)$ , $(\pi,\pi/2)$ , $(\pi,\pi)$ , $(3\pi/2,0)$ and $(3\pi/2,3\pi/2)$ are relative minima

$(0,\pi/2)$ , $(0,\pi)$ , $(3\pi/2,\pi/2)$ and $(3\pi/2,\pi)$ are relative maxima

$(\pi/2,0)$ , $(\pi/2,3\pi/2)$ , $(\pi,0)$ and $(\pi,3\pi/2)$ are saddle points

(0,0) , $(0,3\pi/2)$ , $(\pi/2,\pi/2)$ , $(\pi/2,\pi)$ , $(\pi,\pi/2)$ , $(\pi,\pi)$ , $(3\pi/2,0)$ and $(3\pi/2,3\pi/2)$ are relative maxima

$(0,\pi/2)$ , $(0,\pi)$ , $(3\pi/2,\pi/2)$ and $(3\pi/2,\pi)$ , $(\pi/2,0)$ , $(\pi/2,3\pi/2)$ , $(\pi,0)$ and $(\pi,3\pi/2)$ are saddle points

Correct answer:

(0,0) , $(0,3\pi/2)$ , $(\pi/2,\pi/2)$ , $(\pi/2,\pi)$ , $(\pi,\pi/2)$ , $(\pi,\pi)$ , $(3\pi/2,0)$ and $(3\pi/2,3\pi/2)$ are saddle points

$(0,\pi/2)$ , $(0,\pi)$ , $(3\pi/2,\pi/2)$ and $(3\pi/2,\pi)$ are relative minima

$(\pi/2,0)$ , $(\pi/2,3\pi/2)$ , $(\pi,0)$ and $(\pi,3\pi/2)$ are relative maxima

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=2sin(x)cos(x)$

$\frac{\partial f }{\partial y}=f_y(x,y)=2cos(y)(-sin(y))=-2sin(y)cos(y)$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2\left [ sin(x)(-sin(x))+cos(x)cos(x)\right ]$

$f_{xx}(x,y)=2\left [ -sin^2(x)+cos^2(x)\right ]=2cos^2(x)-2sin^2(x)$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=-2\left [ sin(y)(-sin(y))+cos(y)cos(y)\right ]$

$f_{yy}(x,y)=-2\left [ -sin^2(y)+cos^2(y)\right ]=2sin^2(y)-2cos^2(y)$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=0$

To find the critical points, we will set the first derivatives equal to

f_x(x,y)=2sin(x)cos(x)=0

$sin(x)=0\Rightarrow x=0, \pi$

$cos(x)=0\Rightarrow x= \pi/2, 3\pi/2$

f_y(x,y)=-2sin(y)cos(y)=0

$sin(y)=0\Rightarrow y=0, \pi$

$cos(y)=0\Rightarrow y= \pi/2, 3\pi/2$

Our derivatives equal when $x=0, \pi/2, \pi, 3\pi/2$ and $y=0, \pi/2, \pi, 3\pi/2$ . Every linear combination of these points is a critical point. The critical points are

(0,0) , $(0,\pi/2)$ , $(0,\pi)$ , $(0,3\pi/2)$

$(\pi/2,0)$ , $(\pi/2,\pi/2)$ , $(\pi/2,\pi)$ , $(\pi/2,3\pi/2)$

$(\pi,0)$ , $(\pi,\pi/2)$ , $(\pi,\pi)$ , $(\pi,3\pi/2)$

$(3\pi/2,0)$ , $(3\pi/2,\pi/2)$ , $(3\pi/2,\pi)$ , $(3\pi/2,3\pi/2)$

We need to determine if the critical point is a maximum or minimum using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

D(x,y)=(2cos^2x-2sin^2x)(2sin^2y-2cos^2y)-0^2

$f_{xx}(x,y)=2cos^2(x)-2sin^2(x)$

(0,0) , $(0,\pi/2)$ , $(0,\pi)$ , $(0,3\pi/2)$

D(0,0)=(2cos^2(0)-2sin^2(0))(2sin^2(0)-2cos^2(0))-0^2

=(2(1)-0)(2(0)-2(1))=(2)(-2)=-4<0

Saddle point

$D(0,\pi/2)=(2cos^2(0)-2sin^2(0))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

=(2(1)-2(0))(2(1)-2(0))=(2)(2)=4>0

$f_{xx}(0,\pi/2)=2cos^2(0)-2sin^2(0)=2(1)-2(0)=2>0$

minimum

$D(0,\pi)=(2cos^2(0)-2sin^2(0))(2sin^2(\pi)-2cos^2(\pi))-0^2$

=(2(1)-0)(2(0)-2(-1))=(2)(2)=4>0

$f_{xx}(0,\pi)=2cos^2(0)-2sin^2(0)=2(1)-2(0)=2>0$

minimum

$D(0,3\pi/2)=(2cos^2(0)-2sin^2(0))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

=(2(1)-2(0))(2(-1)-2(0))=(2)(-2)=-4<0

Saddle point

$(\pi/2,0)$ , $(\pi/2,\pi/2)$ , $(\pi/2,\pi)$ , $(\pi/2,3\pi/2)$

$D(\pi/2,0)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(0)-2cos^2(0))-0^2$

=(2(0)-2(1))(2(0)-2(1))=(-2)(-2)=4>0

$f_{xx}(\pi/2,0)=2cos^2(\pi/2)-2sin^2(\pi/2)=2(0)-2(1)=-2<0$

maximum

$D(\pi/2,\pi/2)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

=(2(0)-2(1))(2(1)-2(0))=(-2)(2)=-4<0

saddle point

$D(\pi/2,\pi)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(\pi)-2cos^2(\pi))-0^2$

=(2(0)-2(1))(2(0)-2(-1))=(-2)(2)=-4<0

saddle point

$D(\pi/2,3\pi/2)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

=(2(0)-2(1))(2(-1)-2(0))=(-2)(-2)=4>0

$f_{xx}(\pi/2,2\pi/2)=2cos^2(\pi/2)-2sin^2(\pi/2)=2(0)-2(1)=-2<0$

maximum

$(\pi,0)$ , $(\pi,\pi/2)$ , $(\pi,\pi)$ , $(\pi,3\pi/2)$

$D(\pi,0)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(0)-2cos^2(0))-0^2$

=(2(-1)-0)(2(0)-2(1))=(-2)(-2)=4>0

$f_{xx}(\pi,\pi/2)=2cos^2(\pi)-2sin^2(\pi)=2(-1)-2(0)=-2<0$

maximum

$D(\pi,\pi/2)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

=(2(-1)-2(0))(2(1)-2(0))=(-2)(2)=-4<0

saddle point

$D(\pi,\pi)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(\pi)-2cos^2(\pi))-0^2$

=(2(-1)-0)(2(0)-2(-1))=(-2)(2)=-4<0

saddle point

$D(\pi,3\pi/2)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

=(2(-1)-2(0))(2(-1)-2(0))=(-2)(-2)=4>0

$f_{xx}(\pi,3\pi/2)=2cos^2(\pi)-2sin^2(\pi)=2(-1)-2(0)=-2<0$

maximum

$(3\pi/2,0)$ , $(3\pi/2,\pi/2)$ , $(3\pi/2,\pi)$ , $(3\pi/2,3\pi/2)$

$D(3\pi/2,0)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(0)-2cos^2(0))-0^2$

=(2(0)-2(-1))(2(0)-2(1))=(2)(-2)=-4<0

saddle point

$D(3\pi/2,\pi/2)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

=(2(0)-2(-1))(2(1)-2(0))=(2)(2)=4>0

$f_{xx}(3\pi/2,\pi/2)=2cos^2(3\pi/2)-2sin^2(3\pi/2)=2(0)-2(-1)=2>0$

minimum

$D(3\pi/2,\pi)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(\pi)-2cos^2(\pi))-0^2$

=(2(0)-2(-1))(2(0)-2(-1))=(2)(2)=4>0

$f_{xx}(3\pi/2,\pi)=2cos^2(3\pi/2)-2sin^2(3\pi/2)=2(0)-2(-1)=2>0$

minimum

$D(3\pi/2,3\pi/2)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

=(2(0)-2(-1))(2(-1)-2(0))=(2)(-2)=-4<0

saddle point

Report an Error

Example Question #2 : Relative Minimums And Maximums

Find the relative maxima and minima of f(x,y)=x^2+y^2 .

Possible Answers:

(1,1) is a relative maximum

(0,0) is a relative minimum

(1,1) is a relative minimum

(0,0) is a relative maximum

Correct answer:

(0,0) is a relative minimum

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=2x$

$\frac{\partial f }{\partial y}=f_y(x,y)=2y$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=2$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=0$

To find the critical points, we will set the first derivatives equal to

$f_x(x,y)=2x=0\Rightarrow x=0$

$f_y(x,y)=2y=0\Rightarrow y=0$

There is a critical point at (0,0) . We need to determine if the critical point is a maximum or minimum using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

D(x,y)=2(2)-0^2=4

At (0,0) ,

D(x,y)=4>0

$f_{xx}(x,y)=2>0$

Since D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at (0,0) .

Report an Error

Example Question #11 : Relative Minimums And Maximums

Find the relative maxima and minima of f(x,y)=3x^2y^2 .

Possible Answers:

(-1,0) and (1,0) are saddle points

(-1,0) is a relative minima and (1,0) is a relative maxima

(-1,0) and (1,0) are relative minima

(-1,0) is a relative maxima and (1,0) is a relative minima

Correct answer:

(-1,0) and (1,0) are saddle points

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

f(x,y)=3x^2y^2-3x^2

The first order partial derivatives are

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

$\frac{\partial f }{\partial y}=f_y(x,y)=6x^2y$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=6y^2-6$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=6x^2$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=12xy$

To find the critical points, we will set the first derivatives equal to

$f_{xx}(x,y)=6y^2-6=0\Rightarrow 6y^2=6\Rightarrow y=\pm1$

$f_{yy}(x,y)=6x^2=0 \Rightarrow x=0$

The critical points are (-1,0) and (1,0) . We need to determine if the critical point is a maximum or minimum using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

D(x,y)=(6y^2-6)(6x^2)-(12xy)^2=36x^2y^2-36x^2-144x^2y^2

D(x,y)=-108x^2y^2-36x^2

At (-1,0) ,

D(x,y)=-108(-1)^2(0)^2-36(-1)^2=-36<0

Since D(x,y)<0 , (-1,0) is a saddle point.

At (1,0) ,

D(x,y)=-108(-1)^2(0)^2-36(1)^2=-36<0

Since D(x,y)<0 , (1,0) is a saddle point.

Report an Error

Example Question #132 : Applications Of Partial Derivatives

Find the relative maxima and minima of $f(x,y)=e^{5x^3+5y^2-2x}$ .

Possible Answers:

(0,0) is a saddle point

(0,0) is a relative minima

(5/4,0) is a relative minima

(5/4,0) is a saddle point

Correct answer:

(5/4,0) is a saddle point

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=e^{5x^3+5y^2-2x}(15x^2-2)=(15x^2-2)e^{5x^3+5y^2-2x}$

$\frac{\partial f }{\partial y}=f_y(x,y)=e^{5x^3+5y^2-2x}(10y)=(10y)e^{5x^3+5y^2-2x}$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=(15x^2-2)e^{5x^3+5y^2-2x}(15x^2-2)+e^{5x^3+5y^2-2x}(30x)$

$f_{xx}(x,y)=((15x^2-2)^2+30x)e^{5x^3+5y^2-2x}$

$f_{xx}(x,y)=(225x^4-60x^2+30x+4)e^{5x^3+5y^2-2x}$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=(10y)e^{5x^3+5y^2-2x}(10y)+e^{5x^3+5y^2-2x}(10)$

$f_{yy}(x,y)=(100y^2)e^{5x^3+5y^2-2x}+(10)e^{5x^3+5y^2-2x}$

$f_{yy}(x,y)=(100y^2+10)e^{5x^3+5y^2-2x}$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=(100y^2+10)e^{5x^3+5y^2-2x}(15x^2-2)$

$f_{xy}(x,y)=(100y^2+10)(15x^2-2)e^{5x^3+5y^2-2x}$

$f_{xy}(x,y)=(1500x^2y^2+150x^2-200y^2-20)e^{5x^3+5y^2-2x}$

To find the critical points, we will set the first derivatives equal to

$f_x(x,y)=(15x^2-2)e^{5x^3+5y^2-2x}$

$f_y(x,y)=(10y)e^{5x^3+5y^2-2x}$

The exponential part of each expression cannot equal , so each derivative is only when (15x^2-2)=0 and 10y=0 . That is $x=\pm \sqrt{2/15}$ and y=0 .

The critical points are $(-\sqrt{2/15},0)$ and $(\sqrt{2/15},0)$ . We need to determine if the critical points are maxima or minima using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$D(x,y)=(225x^4-60x^2+30x+4)e^{5x^3+5y^2-2x}(100y^2+10)e^{5x^3+5y^2-2x} -((1500x^2y^2+150x^2-200y^2-20)e^{5x^3+5y^2-2x})^2$

$D(x,y)=(225x^4-60x^2+30x+4)(100y^2+10)(e^{5x^3+5y^2-2x})^2 -(1500x^2y^2+150x^2-200y^2-20)^2(e^{5x^3+5y^2-2x})^2$

At (5/4,0) ,

$D(x,y)=(225(5/4)^4-60(5/4)^2+30(5/4)+4)(0+10)(e^{5(5/4)^3+5(0)^2-2(5/4)})^2 -(1500(5/4)^2(0)^2+150(5/4)^2-200(0)^2-20)^2(e^{5(5/4)^3+5(0)^2-2(5/4)})^2$

$D(x,y)=(225(5/4)^4-60(5/4)^2+30(5/4)+4)(10)(e^{5(5/4)^3-2(5/4)})^2 -(150(5/4)^2-20)^2(e^{5(5/4)^3-2(5/4)})^2$

$D(x,y)\approx(497)(10)(2.0457) -(45956.6)(2.0457)$

$D(x,y)\approx -83846.3<0$

Since D(x,y)<0 , (5/4,0) is a saddle point.

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Example Question #13 : Relative Minimums And Maximums

Find the relative maxima and minima of $f(x,y)=e^{5x^2-7xy}$ .

Possible Answers:

(0,0) is a relative maximum

(0,0) is a saddle point

(5,-7) is a relative maximum

(5,-7) is a saddle point

Correct answer:

(0,0) is a saddle point

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If D(x,y)>0 and $f_{xx}(x,y)>0$ , then there is a relative minimum at this point.

If D(x,y)>0 and $f_{xx}(x,y)<0$ , then there is a relative maximum at this point.

If D(x,y)<0 , then this point is a saddle point.

If D(x,y)=0 , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=e^{5x^2-7xy}(10x-7y)$

$\frac{\partial f }{\partial y}=f_y(x,y)=e^{5x^2-7xy}(-7x)$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=e^{5x^2-7xy}(10)+(10x-7y)e^{5x^2-7xy}(10x-7y)$

$f_{xx}(x,y)=(10+(10x-7y)^2)e^{5x^2-7xy}$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=e^{5x^2-7xy}(-7x)(-7x)$

$f_{yy}(x,y)=90x^2*e^{5x^2-7xy}$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=e^{5x^2-7xy}(-7)+(-7x)e^{5x^2-7xy}(10x-7y)$

$f_{xy}(x,y)=(-7+(-7x)(10x-7y))e^{5x^2-7xy}$

$f_{xy}(x,y)=(-7-70x^2+49xy)e^{5x^2-7xy}$

To find the critical points, we will set the first derivatives equal to

$\frac{\partial f }{\partial x}=f_x(x,y)=e^{5x^2-7xy}(10x-7y)$

$\frac{\partial f }{\partial y}=f_y(x,y)=e^{5x^2-7xy}(-7x)$

The exponential part of each expression cannot equal , so each derivative is only when (10x-7y)=0 and -7x=0 . That is x=0 and y=0 .

The critical point is (0,0) . We need to determine if the critical points are maxima or minima using D(x,y) and $f_{xx}(x,y)$ .

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$D(x,y)=(10+(10x-7y)^2)e^{5x^2-7xy}*90x^2*e^{5x^2-7xy}-((-7-70x^2+49xy)e^{5x^2-7xy})^2$

$D(x,y)=(90x^2)(10+(10x-7y)^2)(e^{5x^2-7xy})^2 -(-7-70x^2+49xy)^2(e^{5x^2-7xy})^2$

$D(x,y)=\left [(90x^2)(10+(10x-7y)^2)-(-7-70x^2+49xy)^2 \right ](e^{5x^2-7xy})^2$

$D(x,y)=(4100x^4-5740x^3y+2009x^2y^2-80x^2+686xy-49)(e^{5x^2-7xy})^2$

At (0,0) ,

$D(x,y)=(4100(0)-5740(0)+2009(0)-80(0)+686(0)-49)(e^{5(0)-7(0)})^2$

D(x,y)=(-49)(1)^2<0

Since D(x,y)<0 , (0,0) is a saddle point.

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Example Question #1 : Double Integration With Change Of Variables

Evaluate $\int \int_R 3x+2y \ dA$ , where is the trapezoidal region with vertices given by (0,0) , (5,0) , $(\frac{5}{2}, \frac{5}{2})$ , and $(\frac{5}{2}, -\frac{5}{2})$ ,

using the transformation x=2u+3v , and y=2u-3v .

Possible Answers:

$\frac{375}{2}$

$\frac{35}{4}$

$\frac{375}{4}$

$\frac{37}{4}$

Correct answer:

$\frac{375}{4}$

Explanation:

The first thing we have to do is figure out the general equations for the lines that create the trapezoid.

y=x

y=-x

y=-x+5

y=x-5

Now we have the general equations for out trapezoid, now we need to plug in our transformations into these equations.

y=x

2u+3v=2u-3v

$-6v=0\rightarrow v=0$

y=-x

2u+3v=-2u+3v

$4u=0\rightarrow u=0$

y=-x+5

2u+3v=-2u+3v+5

$4u=5\rightarrow u=\frac{5}{4}$

y=x-5

2u+3v=2u-3v-5

$-6v=-5\rightarrow v=\frac{5}{6}$

So our region is a rectangle given by $0 \leq u \leq \frac{5}{4}$ , $0 \leq v \leq \frac{5}{6}$

Next we need to calculate the Jacobian.

$\begin{vmatrix} 2 & 3\\ 2 & -3 \end{vmatrix} =-6-6=-12$

Now we can put the integral together.

$\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} (3(2u+3v)+2(2u-3v))) \cdot \left | -12\right | \ du\ dv$

$=\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} 120u+36v \ du\ dv$

$=\int_{0}^{\frac{5}{6}} 60u^2+36uv \Big|_{0}^{\frac{5}{4}}\ dv$

$=\int_{0}^{\frac{5}{6}} 60( \frac{5}{4})^2+36(\frac{5}{4})v-0 \ dv$

$=\int_{0}^{\frac{5}{6}} \frac{375}{4}+45v \ dv$

$=\frac{375}{4}v+\frac{45}{2}v^2\Big|_{0}^{\frac{5}{6}} \ dv$

$=\frac{375}{4}\cdot\frac{5}{6}+\frac{45}{2}\cdot\Big(\frac{5}{6}\Big)^2 -0=\frac{375}{4}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Question #1 : Relative Minimums And Maximums

Example Question #121 : Applications Of Partial Derivatives

Example Question #3 : Relative Minimums And Maximums

Example Question #2 : Relative Minimums And Maximums

Example Question #9 : Relative Minimums And Maximums

Example Question #2 : Relative Minimums And Maximums

Example Question #11 : Relative Minimums And Maximums

Example Question #132 : Applications Of Partial Derivatives

Example Question #13 : Relative Minimums And Maximums

Example Question #1 : Double Integration With Change Of Variables

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