Differential Calculus Questions and Answers

Answer: A

Each point of a curve has a centre of curvature, lying on the normal at a distance \( \rho \) on the concave side. As the point moves along the curve the centre traces the evolute, and the original curve is then one involute of it. Inflexion points, where the curvature vanishes, send the centre off to infinity

Answer: D

The denominator vanishes at \( x = 3 \), giving a vertical asymptote. For the horizontal one, divide through by \( x \):

\[ \lim_{x \to \pm\infty} \frac{2x+1}{x-3} \;=\; \lim_{x \to \pm\infty} \frac{2 + 1/x}{1 - 3/x} \;=\; 2. \]

So \( y = 2 \) is approached at both ends, the ratio of the leading coefficients

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