#$&* course Mth 277 Question: `q001. Find f_x and f_y when f(x,y) = xy^4*arctan(y).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: f_x = y^4 arctan(y) f_y = x * ( (y^4) ' arcTan(y) - y^4 * (arcTan(y)) ' ) = x * 4 y^3 - y^4 * 1 / (1 + y^2) = y^3 ( 4 x - (y / (1 + y^2) ). Simplification should be continued until the result is a single fraction with numerator y^3 ( 4 x + 4 x y^2 - y) and denominator (1 + y^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Determine z_x and z_y by differentiating the expression 4x^2 + 2y^2 + 3z^2 = 9 implicitly. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f_x = 8x f_y = 4y f_z = 6z z_x = 8x/6z=(4x/3z)
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. Let f(x,y) = (x^2 + y^2)/(xy), P = (2, -1, -5/2) Find the slope of the tangent line of the graph of f parallel to the xz-plane at the point P. Find the slope of the tangent line of the graph of f parallel to the yz-plane at the point P. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: z=f(x,y)=(x^2+y^2)/(xy) f(x,-1)=(x^2-1)/(-x) f_x=-1+1/x^2 f_x(2)=-1+1/4=-3/4 f(2,y)=(4+y^2)/2y f_y=-2/y^2+(1/2) f_y(-1)=3/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: ********************************************* 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). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For the first function f_x = -2 x sin(y x^2) so f_xy = -2 x^3 cos(y x^2) f_y = -x^2 sin(y x^2) and f_yx = -2 x^3 cos(y x^2) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q005. In physics the wave equation is given by z_tt = c^2 * z_xx and the heat equation is given by z_t = c^2 * z_xx. In the two following cases, see if z satisfies the wave equation, the heat equation, or neither. z = sin(2t)*sin(2cx). z = (e^(-t))(sin(x/c)+cos(x/c). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: z_t=2cos(2t)sin(2cx) z_tt=-4sin(2t)sin(2cx) z_x=2csin(2t)*cos(2cx) z_xx=-4c^2sin(2t)sin(2cx) Obviously, Z does not satisy the equations. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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