Differential Calculus Questions and Answers

Answer: C

Dividing out leaves a linear quotient plus a remainder that dies away, so the curve approaches a slant line. Equal degrees give a horizontal asymptote at the ratio of leading coefficients, a smaller numerator gives \( y = 0 \), and an excess of two or more gives a parabolic or higher approach with no straight asymptote at all

Answer: B

The envelope is tangent to each curve of the family at some point, and is found by eliminating the parameter between \( F(x,y,c) = 0 \) and \( \partial F/\partial c = 0 \). The orthogonal-crossing curve is a member of the orthogonal family instead, and the locus of centres of curvature is the evolute

Answer: A

Differentiating with respect to \( x \) treats \( y \) as a constant, so the factor \( y^{2} \) comes along unchanged:

\[ f_{x} \;=\; \frac{\partial}{\partial x}\left(x^{3} y^{2}\right) \;=\; 3x^{2} y^{2}. \]

The form \( 2x^{3}y \) is \( f_{y} \) instead, and \( 6x^{2}y \) is the mixed derivative \( f_{xy} \), got by differentiating once with respect to each variable

Answer: D

When the mixed second derivatives are continuous the order of differentiation does not matter, so \( f_{xy} = f_{yx} \). This is what allows a second-order expression to be written without keeping track of which variable came first, and it is the condition under which a differential \( M\,dx + N\,dy \) is exact

Answer: C

The total derivative sums the two routes by which \( t \) reaches \( u \):

\[ \frac{du}{dt} \;=\; \frac{\partial u}{\partial x}\frac{dx}{dt} + \frac{\partial u}{\partial y}\frac{dy}{dt} \;=\; 2x(1) + 2y(2t). \]

Substituting \( x = t \) and \( y = t^{2} \) gives \( 2t + 4t^{3} \). Substituting first and then differentiating \( t^{2} + t^{4} \) confirms it

Answer: B

Homogeneity means \( f(tx, ty) = t^{n} f(x,y) \). Differentiating with respect to \( t \) and then setting \( t = 1 \) gives

\[ x f_{x} + y f_{y} \;=\; n f. \]

The weighting by \( x \) and \( y \) is essential: without it the left side is not even of the same degree as the right

Answer: A

The function is homogeneous of degree three, so Euler's theorem applies with \( n = 3 \). Checking it directly,

\[ x f_{x} + y f_{y} \;=\; x\left(3x^{2}\right) + y\left(3y^{2}\right) \;=\; 3\left(x^{3} + y^{3}\right) \;=\; 3f. \]

The degree, not the number of variables, sets the multiplier, which is why a quadratic form would return \( 2f \)

Answer: D

The Jacobian is

\[ \frac{\partial(u,v)}{\partial(x,y)} \;=\; \begin{vmatrix} u_{x} & u_{y} \\ v_{x} & v_{y} \end{vmatrix}, \]

a determinant of first derivatives. It measures the local factor by which area is stretched, which is why it appears when a double integral changes variables, and its vanishing signals that the transformation is locally degenerate

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