A vanishing Jacobian \( \partial(u,v)/\partial(x,y) = 0 \) indicates that
Answer: A
If the Jacobian vanishes identically there is a relation \( \phi(u,v) = 0 \) between the two functions, so they carry no independent information and the map collapses the plane onto a curve. Area is preserved when the Jacobian is \( \pm 1 \), and orientation is reversed when it is negative rather than zero
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For \( f(x,y) \) with \( f_{x} = f_{y} = 0 \) at a point, write \( D = f_{xx}f_{yy} - f_{xy}^{2} \). A minimum is indicated by
Answer: D
A positive discriminant means the surface curves the same way in every direction, and the sign of \( f_{xx} \) then says which way: positive for a minimum, negative for a maximum. A negative \( D \) gives a saddle, curving up one way and down another, and \( D = 0 \) leaves the test with nothing to say
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The stationary point of \( f(x,y) = x^{2} + y^{2} - 4x - 6y + 20 \) lies at
Answer: C
Set both first derivatives to zero:
\[ f_{x} \;=\; 2x - 4 \;=\; 0, \qquad f_{y} \;=\; 2y - 6 \;=\; 0, \]
giving \( x = 2 \) and \( y = 3 \). Since \( f_{xx} = f_{yy} = 2 \) and \( f_{xy} = 0 \), the discriminant is \( D = 4 > 0 \) with \( f_{xx} > 0 \), so the point is a minimum, of value seven
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For \( f(x,y) = xy \) the origin is
Answer: B
Both first derivatives vanish at the origin, since \( f_{x} = y \) and \( f_{y} = x \). The second derivatives give \( f_{xx} = f_{yy} = 0 \) and \( f_{xy} = 1 \), so
\[ D \;=\; (0)(0) - 1^{2} \;=\; -1 \;<\; 0, \]
which is the saddle condition. The surface rises along \( y = x \) and falls along \( y = -x \)
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The method of Lagrange multipliers finds the extrema of a function subject to a constraint by
Answer: A
At a constrained extremum the level curve of \( f \) touches the constraint curve, so their normals are parallel and \( \nabla f = \lambda \nabla g \). Elimination is the alternative route, workable only when the constraint can be solved explicitly, and demanding that both gradients vanish imposes far more than is needed
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The maximum of \( f = xy \) subject to \( x + y = 10 \) is
Answer: D
With \( \nabla f = \lambda \nabla g \) we get \( y = \lambda \) and \( x = \lambda \), so \( x = y \). The constraint then gives \( x = y = 5 \) and
\[ f_{\max} \;=\; xy \;=\; 5 \times 5 \;=\; 25. \]
Substituting \( y = 10 - x \) and maximising \( 10x - x^{2} \) confirms the same vertex, so a fixed perimeter is enclosed most efficiently by the square
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Match the theorem with its conclusion. 1. Rolle 2. Lagrange 3. Cauchy P. A tangent parallel to the chord Q. A vanishing derivative R. A ratio of derivatives equal to a ratio of increments
Answer: C
Rolle's equal endpoints force a horizontal tangent, so the derivative vanishes. Lagrange drops that restriction and concludes only that some tangent matches the chord's slope. Cauchy compares two functions at once, equating the ratio of their derivatives to the ratio of their increments, and reduces to Lagrange when the second function is \( x \)
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Match the quantity with what it measures. 1. Curvature 2. Jacobian 3. Discriminant of the second-derivative test P. The local stretching of area under a change of variables Q. The rate at which the tangent direction turns R. Whether a stationary point curves the same way in every direction
Answer: B
Curvature is the turning rate of the tangent per unit arc length. The Jacobian determinant is the local area factor of a transformation, which is why it appears in a change of variables. The discriminant \( f_{xx}f_{yy} - f_{xy}^{2} \) decides between a genuine extremum and a saddle by testing whether the curvature keeps one sign in all directions
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Match the indeterminate form with the first step it calls for. 1. \( 0/0 \) 2. \( 0 \times \infty \) 3. \( 1^{\infty} \) P. Invert one factor to make a quotient Q. Differentiate numerator and denominator R. Take logarithms
Answer: A
Only \( 0/0 \) and \( \infty/\infty \) are ready for L'Hopital's rule, so a quotient is differentiated at once. A product is turned into a quotient by inverting one factor, and a power is turned into a product by taking logarithms, after which the same rule finishes the work
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Put these steps of locating and classifying the extrema of a function of two variables in order. 1. Evaluate the second derivatives at each stationary point 2. Solve the pair of equations for the stationary points 3. Read off the nature of each point from the discriminant and \( f_{xx} \) 4. Set both first partial derivatives to zero
Answer: D
The first derivatives are set to zero to obtain the equations, which are then solved for the candidate points. Only at those points are the second derivatives worth computing, and the discriminant formed from them together with the sign of \( f_{xx} \) gives the classification. Evaluating second derivatives before knowing where to evaluate them wastes the work
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