Differential Calculus Questions and Answers

Answer: C

The four first derivatives are \( u_{x} = 1 \), \( u_{y} = 1 \), \( v_{x} = 1 \), \( v_{y} = -1 \), so

\[ \frac{\partial(u,v)}{\partial(x,y)} \;=\; \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} \;=\; (1)(-1) - (1)(1) \;=\; -2. \]

The magnitude two is the area stretch of this rotation and scaling, and the sign records that the map reverses orientation

Answer: B

The chain rule for Jacobians gives

\[ \frac{\partial(u,v)}{\partial(x,y)} \cdot \frac{\partial(x,y)}{\partial(u,v)} \;=\; 1, \]

so the two are reciprocals. A vanishing Jacobian therefore marks precisely the points where the inverse fails to exist, since no finite number has reciprocal zero

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

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

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

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 \( y = -x \)

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

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

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 \( x \)

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