QA111mth

#$&*

course Mth 277

qa 11.1`q001. If f(x, y) = x^2 / 4 - y^2 / 25, then what curve do you get for each of the following?

f(x, y) = 0?

f(x, y) = 1?

f(x, y) = 4?

f(x, y) = 9?

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Your solution:

x^2 / 4 - y^2 / 25=0

x^2 / 4 = y^2 / 25

y=sqrt(25x^2/4)

assymptote at y=+-2.5x

(2,5)

This is in standard form

x^2 / 4 - y^2 / 25=1

y=sqrt(25x^2/4-25)

x^2 / 4 - y^2 / 25=4

x^2-4y^2/25=1

y=sqrt(25x^2/4-25/4)

x^2 / 4 - y^2 / 25=9

x^2/36-y^2/225=1

y=sqrt(225x^2/36-225)

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Given Solution: For example, the curve f(x, y) = 16 would be

x^2 / 4 - y^2 / 25 = 16,

which is a hyperbola. Its standard form is

x^2 / 64 - y^2 / 400 = 1.

It is asymptotic to the lines y = 5/2 x and y = -5/2 x.

If x = 0 then y^2 would have to be negative, so there is no vertex on the y axis.

If y = 0 then x = +- 8 so the vertices are (8, 0) and (-8, 0).

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Self-critique (if necessary):OK

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Question: `q002. Plot each of the curves obtained in the preceding.

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Your solution:

I have plotted the curves based on asymptotes and major intercepts.

confidence rating #$&*:

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Given Solution:

The hyperbola is asymptotic to the lines y = 5/2 x and y = -5/2 x.

If x = 0 then y^2 would have to be negative, so there is no vertex on the y axis.

If y = 0 then x = +- 8 so the vertices are (8, 0) and (-8, 0).

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Self-critique (if necessary):OK

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Question: `q003. If f(x, y, z) = x^2 / 4 + y^2 / 25 - z^2 / 16, then what equation corresponds to the condition f(x, y, z) = 9? What would the plot of this equation look like in 3-dimensional space?

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Your solution:

x^2/4+y^2/25-z^2/16=9

In the X-y plane, we can tell the shape based on the equation given

x^2/4+y^2/25=9. This is the equation for an ellipse.vertices (+-6,0) and (0,+-15)

X-Z plane

x^2/4+-z^2/16=9. Since the z is negative, this is a hyperbola.vertices (+-6,0)

Y-Z plane

y^2/25-z^2/16=9. This is also an equation for a hyperbola.vertices (0,+-15)

confidence rating #$&*:

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Given Solution:

f(x, y, z) = x^2 / 2 + y^2 / 25 - z^2 / 16, so the condition f(x,y,z) = 9 gives us

x^2 / 4 + y^2 / 25 - z^2 / 16 = 9.

The xy trace would be x^2 / 4 + y^2 / 25 = 9, an ellipse centered at the origin with vertices at (+-6, 0) and (+- 15, 0).

The x z trace would be x^2 / 4 - z^2 / 16 = 9, a hyperbola asymptotic to the lines z = +- x / 4 with vertices at (+- 6, 0).

The y z trace would be y^2 / 25 - z^2 / 16 = 9, a hyperbola asymptotic to the lines z = +- 5/4 * y with vertices at (+- 15, 0).

If z = c, a constant, we get

x^2 / 4 + y^2 / 25 = 9 + c^2 / 16.

an ellipse in the plane z = c with vertices at (+-2 sqrt(9 + c^2 / 16), 0) and (0, +-5 sqrt(9 + c^2 / 16)).

sqrt(9 + c^2 / 16) increases as c increases, and for large values of c is nearly equal to c / 4. The ellipses grow accordingly as you move up or down from the y axis.

The parts of the ellipses above and near the x axis, for c = 0, 2, 4, 6, 8, 10, are depicted below:

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Self-critique (if necessary):

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In the yz plane y goes first.

In the equation the y^2 term is positive; only the z term can be zero.

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Question: `q004. If f(x, y) = x * y then what curve do you get for each of the following? Sketch each curve.

f(x, y) = 0?

f(x, y) = 1?

f(x, y) = 4?

f(x, y) = 9?

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Your solution:

x*y=0

y=0/x, x cannot equal 0

Therefore, this line is the x axis.

x*y=1

y=1/x

From this we know that x and y have the same signs at the same time. Therefore, the graph is defined in the 1 and 3 quadrants. Also, there is a vertical assymptote on the yaxis and a horiz assympt at the xaxis. Also, we know that as abs(x) gets smaller y gets either more positive or neg based on the sign of x.

Since the rest of the graphs are the same I will tell you what will happen. As the number gets larger that the expression is equal to, the curve will gradually come closer to the vertex in both the 1 and 3 quadrant. Therefore, 9 is closer to the vertex than 4 and so on.

confidence rating #$&*:

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Given Solution:

f(x, y) = c corresponds to x * y = c, so that y = c / x.

For c = 0, all y values are 0 so the graph is just the x axis, excluding the point where x = 0 (at which the function is undefined).

For c = 1 the function is y = 1 / x, the graph of which should be familiar.

For c = 4 the function is y = 4 / x, the graph of which is just the c = 1 graph, stretched vertically by factor 4.

The graphs for c = 0, 1, 4, 9 are depicted below.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q005. If f(x, y, z) = x * y / z then what equation expresses the condition f(x, y, z) = 1? What would the plot of this equation look like in 3-dimensional space?

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Your solution:

x*y/z=1

X*y and x-z plane

since these are divided wouldnt the only plane that would work is the Y-Z plane.

y=z

Would be linear in this plane.

confidence rating #$&*:

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Given Solution:

The equation would be x * y / z = 1.

The equation is undefined if z = 0. So the graph does not intersect the x-y plane.

If x = 0 or y = 0 the equation cannot be satisfied, since in either case we would get 0 = 36.

If z = c then we have x * y / c = 1, so that

y = c / x.

For positive values of c the graphs in the z = c plane would be the same as in the preceding. For negative values the graphs would be 'flipped' over the x axis.

The surface in the first octant, and its intersection with the plane z = 1, are depicted below.

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Self-critique (if necessary):

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Question: `q006. Find the x and y derivatives of the following functions:

f(x, y) = x^2 sqrt(y)

f(x, y) = cos(x y)

f(x, y) = e^-(x^2 + y^2)

f(x, y) = sqrt(x) * e^y

f(x, y) = cos(x y) + e^(x^2 y)

f(x, y) = x y / (x^2 + y^2)

f(x, y) = sqrt(x^2 / 4 - y^2 / 9)

f(x, y) = y e^(-k x^2) + x y cos(x + y)

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Your solution:

f'(x)=2xsqrt(y)

f'(y)=.5x^2/sqrt(y)

f'(x)=-ysin(xy)

f'(y)=-xsin(xy)

f'(x)=-2x*e^-(x^2 + y^2)

f'(y)=-2y*e^-(x^2 + y^2)

f'(x)=.5e^y/sqrt(x)

f'(y)=sqrt(x)e^y

f'(x)=-ysin(xy)+2xe^(x^2 y)

f'(y)=-xsin(xy)+e^(x^2 y)

f'(x)=((x^2 + y^2)y-(xy)2x)/(x^2+y^2)^2

f'(y)=((x^2 + y^2)x-(xy)2y)/(x^2+y^2)^2

f'(x)=(x/2)/sqrt(x^2 / 4 - y^2 / 9)

f'(y)=(2y/9)/sqrt(x^2 / 4 - y^2 / 9)

f'(x)=2kxye^(-kx^2)+ycos(3x+7y)-3xysin(3x+7y)

f'(y)=e^(-kx^2)+xcos(3x+7y)-7xysin(3x+7y)

confidence rating #$&*:

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Given Solution:

f(x, y) = x^2 sqrt(y)

f_x(x, y) = 2 x sqrt(y)

f_y(x, y) = x^2 / (2 sqrt(y))

f(x, y) = cos(x y^2)

f_x(x, y) = (x y^2) ' * (-sin(x y^2)), where ' denotes derivative with respect to x, so f_x = - y^2 sin(x y^2)

f_y(x, y) = (x y^2) ' * (-sin(x y^2), where ' denotes derivative with respect to y, so f_y = - 2 x y sin(x y^2)

f(x, y) = e^-(x^2 + y^2)

f_x(x, y) = -(x^2 + y^2) ' e^(x^2 + y^2), where ' denotes derivative with respect to x, so f_x = -2 x e^(-x^2 + y^2)

f_y(x, y) = -(x^2 + y^2) ' e^(x^2 + y^2), where ' denotes derivative with respect to y, so f_y = -2 y e^(-x^2 + y^2)

f(x, y) = sqrt(x) * e^y

f_x(x, y) = 1 / (2 sqrt(x)) * e^y. e^y is a function of y, hence constant when y is held constant, so we treat e^y as a constant when taking the derivative with

respect to x.

f_y(x, y) = sqrt(x) * e^y. sqrt(x) is a function of x, hence constant when x is held constant, so we treat sqrt(x) as a constant when taking the derivative with

respect to y.

f(x, y) = cos(x y) + e^(x^2 y)

f_x(x, y) = y sin(x y) + 2 x y e^(x^2 y)

f_y(x, y) = x sin(x y) + x^2 e^(x^2 y)

f(x, y) = x y / (x^2 + y^2)

f_x(x, y) = ( (x y) ' ( x^2 + y^2) - x y ( x^2 + y^2) ' ) / (x^2 + y^2)^2, where ' denotes derivative with respect to x, so that f_x = ( y ( x^2 + y^2) - x y ( 2

x^) ) / (x^2 + y^2)^2. The numerator simplifies further.

f_y(x, y) = ( (x y) ' ( x^2 + y^2) - x y ( x^2 + y^2) ' ) / (x^2 + y^2)^2, where ' denotes derivative with respect to y, so that f_y = ( x ( x^2 + y^2) - x y ( 2

y) ) / (x^2 + y^2)^2. This simplifies further.

f(x, y) = sqrt(x^2 / 4 - y^2 / 9)

f_x(x, y) = (x^2 / 4 - y^2 / 9) ' * 1 / (2 sqrt( x^2 / 4 - y^2 / 9), where ' denotes derivative with respect to x, so that f_x = ( 2 x / 4 ) / (2 sqrt(x^2 / 4 -

y^2 / 9), which simplifies further.

f_y(x, y) = (x^2 / 4 - y^2 / 9) ' * 1 / (2 sqrt( x^2 / 4 - y^2 / 9), where ' denotes derivative with respect to y, so that f_y = ( 2 y / 9 ) / (2 sqrt(x^2 / 4 -

y^2 / 9), which simplifies further.

f(x, y) = y e^(-k x^2) + x y cos(3x + 7y)

f_x(x, y) = 2 k x y e^(- k x^2 ) + y cos(3 x + 7 y) - 3 x y sin(3 x + 7 y)

f_y(x, y) = e^(- k x^2 ) + x cos(3 x + 7 y) - 7 x y sin(3 x + 7 y)

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Self-critique (if necessary):OK

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&#This looks good. See my notes. Let me know if you have any questions. &#