Query11_3mth

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course Mth 277

Question: `q001. Find f_x and f_y when f(x,y) = xy^4*arctan(y).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

f_x=y^4*arctan(y)

f_y=x*y^4*(1/(1+ y^2)+y*4y^3=y^4(x/(1+y^2)+4)

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Given Solution:

f_x = y^4 arctan(y)

f_y = x * ( (y^4) ' arcTan(y) - y^4 * (arcTan(y)) ' ) = x * 4 y^3 - y^4 * 1 / (1 + y^2) = y^3 ( 4 x - (y / (1 + y^2) ). Simplification should be continued until the result is a single fraction with numerator y^3 ( 4 x + 4 x y^2 - y) and denominator (1 + y^2).

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Self-critique (if necessary):OK

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Question: `q002. Determine z_x and z_y by differentiating the expression 4x^2 + 2y^2 + 3z^2 = 9 implicitly.

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Your solution:

f_x = 8x f_y = 4y f_z = 6z

z_x = 8x/6z=(4x/3z)

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Differentiating with respect to x, considering z as a function of x and y, we get

8 x + 4 y dy/dx + 6 z dz/dx = 0.

Since dy/dx = 0, this gives us

8 x + 6 z dz/dx = 0

so that

dz/dx = -8x / (6z) = -4x /(3 z).

We can verify this by solving the equation explicitly for z and taking the derivative with respect to x:

z = 1/sqrt(3) * sqrt( 9 - 4 x^2 - 2 y^2 )

so

z_x = 1/sqrt(3) * (-8 x) * 1 / (2 sqrt( 9 - 4 x^2 - 2 y^2) )

and since z = 1/sqrt(3) sqrt( 9 - 4 x^2 - 2 y^2) our expression becomes

z_x = -4 x / (3 z).

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z_y = 4y/6z=2y/3z

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Given Solution:

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Self-critique (if necessary):OK

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Question: `q003. Let f(x,y) = (x^2 + y^2)/(xy), P = (2, -1, -5/2)

Find the slope of the tangent line of the graph of f parallel to the xz-plane at the point P.

Find the slope of the tangent line of the graph of f parallel to the yz-plane at the point P.

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Your solution:

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

f(x,-1)=(x^2-1)/(-x)

f_x=-1+1/x^2

f_x(2)=-1+1/4=-3/4

f(2,y)=(4+y^2)/2y

f_y=-2/y^2+(1/2)

f_y(-1)=3/2

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Given Solution:

In a plane parallel to the xz plane, y is constant and z is a function of x only. If y = -1, then the function becomes -(x^2 + 1) / x, and its derivative is -1 + 1 / x^2. At x = 2 the slope is therefore -3/4.

Alternatively, f_x = 1/y - y / x^2. At P = (2, -1, -5/2) we get f_x = -1 + 1/2^2 = -3/4.

Analogous analysis of the slope in a plane parallel to the yz plane indicates a slope at (2, -1, -5/2) of 2 - 1/2 = 3/2.

Thus the vectors i - 3/4 k and j + 3/2 k are tangent to the plane at (2, -1, -5/2). We can calculate a cross product to get a normal vector, and knowing the coordinates of P we can then easily find the equation of the tangent plane.

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Self-critique (if necessary):OK

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Question: `q004. For the two following functions, show that f_xy = f_yx.

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

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

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Your solution:

Prove f_xy = f_yx

f_x = -2 x sin(y x^2)

f_xy = -2 x^3 cos(y x^2)

f_y = -x^2 sin(y x^2)

f_yx = -2 x^3 cos(y x^2)

As you can see, both f_xy and f_yx are equal to each other.

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Given Solution:

For the first function

f_x = -2 x sin(y x^2) so f_xy = -2 x^3 cos(y x^2)

f_y = -x^2 sin(y x^2) and f_yx = -2 x^3 cos(y x^2)

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Self-critique (if necessary):

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Self-critique rating:OK

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Question: `q005. In physics the wave equation is given by z_tt = c^2 * z_xx and the heat equation is given by z_t = c^2 * z_xx. In the two following cases, see if z satisfies the wave equation, the heat equation, or neither.

z = sin(2t)*sin(2cx).

z = (e^(-t))(sin(x/c)+cos(x/c).

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Your solution:

z_t=2cos(2t)sin(2cx)

z_tt=-4sin(2t)sin(2cx)

z_x=2csin(2t)*cos(2cx)

z_xx=-4c^2sin(2t)sin(2cx)

Obviously, Z does not satisy the equations.

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Given Solution:

For the first function

z_t = (sin(2t)) ' sin( 2 c x) = 2 cos (2 t) sin( 2 c x), and z_tt = -4 sin(2 t) sin(2 c x)

z_x = 2 c sin(2 t) * cos(2 c x) and z_xx = -4 c^2 sin(2 t) sin(2 c x).

The two second partials are identical except for the c^2 in z_xx.

So we see that z_xx = z_tt * c^2.

This is close, but not quite, of the same form as the wave equation. However the wave equation has the c^2 on the z_xx term, not the z_tt term.

Had the function been z = sin(2 x) * sin( 2 c t), it would have satisfied the wave equation.

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Self-critique (if necessary):OK

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Question: `q002.

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Your solution:

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Given Solution:

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