Multivariable Calculus Questions and Answers

Q: Let \(f : D \to \mathbf{R}\) be defined as \(f(x, y, z) = xyz\); where \(D = \{(x, y, z) \in \mathbf{R}^{3}/x^{2} + y^{2} + z^{2} \lt 1\}\). Let P denote the plane \(\{(x, y, z)/2x + 3y + 5z = 0\}\). Then \(\displaystyle\int_{D \cap P} f\, dx\, dy\, dz\) is :

The integral is 0.\(D \cap P\) is a flat disc: the intersection of the open unit ball with a plane through the origin. As a subset of \(\mathbf{R}^{3}\) it has three-dimensional volume zero, and the integral of a bounded function with respect to \(dx\,dy\,dz\) over a set of volume zero is 0.(Even as a surface integral over the disc the answer would be 0, since \(xyz\) is odd: \((x, y, z)\) and \(-(x, y, z)\) both lie in the disc and \(f\) changes sign between them.) The integral is well defined, so 'undefined' is wrong, and \(\pi\) and \(2\pi\) confuse it with an area.

Q: The vector normal to the gradient of the function \(f : \mathbf{R}^{3} \to \mathbf{R}\) defined by\(f(x, y, z) = xy + y^{2} + z\)at the point P = (1, -1, 1) is :

The vector is (3, 0, 3).\[ \nabla f = (y,\ x + 2y,\ 1) \]At P = (1, -1, 1) this is (-1, -1, 1). A vector normal to it has zero dot product with it:\[ (3, 0, 3) \cdot (-1, -1, 1) = -3 + 0 + 3 = 0 \]The others fail: (-3, -3, 0) gives 3 + 3 = 6, (-3, 0, 3) gives 3 + 3 = 6, and (1, 0, 0) gives -1. Note that (-3, -3, 3), parallel to the gradient, would be the normal to the level surface; the question asks for a vector perpendicular to the gradient.

Q: Let the first and second fundamental forms associated with a surface patch be respectively \(Edu^{2} + 2F\,dudv + Gdv^{2}\) and \(Ldu^{2} + 2M\,dudv + Ndv^{2}\). If \(S = \begin{bmatrix} E & F \\ F & G \end{bmatrix}\) and \(T = \begin{bmatrix} L & M \\ M & N \end{bmatrix}\), then :

The answer is S is non-singular and T may not be.The first fundamental form is \(E = \sigma_{u}\cdot\sigma_{u}\), \(F = \sigma_{u}\cdot\sigma_{v}\), \(G = \sigma_{v}\cdot\sigma_{v}\). For a regular patch \(\sigma_{u}\) and \(\sigma_{v}\) are linearly independent, so by Lagrange's identity \(EG - F^{2} = |\sigma_{u} \times \sigma_{v}|^{2} \gt 0\): S is positive definite and non-singular.The second fundamental form can be degenerate. For a plane, \(L = M = N = 0\), and for a cylinder \(LN - M^{2} = 0\), so T is singular there. Since \(\det T = K\,\det S\), T is singular exactly where the Gaussian curvature is 0. So T may be singular, but S never is.

Q: Let U be an open subset of \(\mathbf{R}^{2}\) and S be a smooth surface in \(\mathbf{R}^{3}\). If \(\sigma : U \to S\) is a regular surface patch such that\[ \det\begin{pmatrix} \sigma_{u} . \sigma_{u} & \sigma_{u} . \sigma_{v} \\ \sigma_{u} . \sigma_{v} & \sigma_{v} . \sigma_{v} \end{pmatrix} = 1 \]Then :

The answer is \(\sigma\) preserves area, but need not be conformal.The determinant is \(EG - F^{2}\), and the area element of the surface is \(\sqrt{EG - F^{2}}\,du\,dv\). If it equals 1, every region of U has the same area as its image, so \(\sigma\) is area-preserving (equiareal).Conformality needs \(E = G\) and \(F = 0\) (up to a common factor), and an isometry needs \(E = G = 1\), \(F = 0\). The condition \(EG - F^{2} = 1\) allows, for example, \(E = 4\), \(F = 0\), \(G = 1/4\), which preserves area but stretches one direction and shrinks the other. So neither conformality nor isometry follows.

Q: Let \(S = \{(x, y, f(x, y)) \in \mathbf{R}^{3} \mid (x, y) \in U\}\) where U is open in \(\mathbf{R}^{2}\) and \(f : U \to \mathbf{R}\) is smooth. If \(f\) admits a local maximum at \(p\) and a local minimum at \(q\), then the Gaussian curvatures \(k_{1}\) and \(k_{2}\) to S at the points \((p, f(p))\) and \((q, f(q))\) respectively satisfy :

The curvatures satisfy \(k_{1}, k_{2} \ge 0\).For a graph \(z = f(x, y)\), the Gaussian curvature at a critical point (where \(f_{x} = f_{y} = 0\)) is\[ K = f_{xx}f_{yy} - f_{xy}^{2} \]the determinant of the Hessian. At a local maximum the Hessian is negative semi-definite and at a local minimum positive semi-definite; in both cases its two eigenvalues have the same sign (or one is 0), so the determinant is \(\ge 0\). Both surfaces bend the same way in every direction, like a cap or a bowl.A negative curvature would mean a saddle, which is neither a maximum nor a minimum, so the options with \(\le 0\) are wrong.

Q: Consider the function \(f : \mathbf{R}^{2} \to \mathbf{R}\) given by \(f(x, y) = x^{2} + y^{2} - 4y\). Then in a neighbourhood of (0, 0) :

Near (0, 0), \(y\) can be expressed as a function of \(x\).Consider the level curve \(f(x, y) = f(0, 0) = 0\). At the origin, \(f_{x} = 2x = 0\) and \(f_{y} = 2y - 4 = -4 \neq 0\). By the implicit function theorem, the non-zero \(f_{y}\) lets us solve for \(y\) as a smooth function of \(x\) near 0. Indeed \(y = 2 - \sqrt{4 - x^{2}}\) near the origin.\(x\) cannot be written as a function of \(y\): solving gives \(x = \pm\sqrt{4y - y^{2}}\), two values for each small \(y \gt 0\), and none for \(y \lt 0\). So A and D fail, and C fails because \(y\) does work.

Q: Consider the function \(f : \mathbf{R}^{2} \to \mathbf{R}^{2}\) defined as \(f(x, y) = (x^{2} - y^{2}, 2xy)\). Then the image of the ball \(B(P, \frac{1}{2})\), where P = (1, 0) is the centre and \(\frac{1}{2}\) is radius, is :

The image is an open set.In complex notation \(f(z) = z^{2}\), since \((x + iy)^{2} = x^{2} - y^{2} + 2ixy\). The ball centred at 1 with radius 1/2 does not contain 0, so \(f'(z) = 2z \neq 0\) there. A non-constant analytic function is an open map (open mapping theorem), so the image of an open ball is open.It is bounded, since \(|z| \lt 3/2\) gives \(|z^{2}| \lt 9/4\), so D is wrong. An open, non-empty, bounded proper subset of the plane cannot be closed, which rules out A and C.

21.

Let \(f : D \to \mathbf{R}\) be defined as \(f(x, y, z) = xyz\); where \(D = \{(x, y, z) \in \mathbf{R}^{3}/x^{2} + y^{2} + z^{2} \lt 1\}\). Let P denote the plane \(\{(x, y, z)/2x + 3y + 5z = 0\}\). Then \(\displaystyle\int_{D \cap P} f\, dx\, dy\, dz\) is :

A.Undefined
B.0
C.\(\pi\)
D.\(2\pi\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The integral is 0.

\(D \cap P\) is a flat disc: the intersection of the open unit ball with a plane through the origin. As a subset of \(\mathbf{R}^{3}\) it has three-dimensional volume zero, and the integral of a bounded function with respect to \(dx\,dy\,dz\) over a set of volume zero is 0.

(Even as a surface integral over the disc the answer would be 0, since \(xyz\) is odd: \((x, y, z)\) and \(-(x, y, z)\) both lie in the disc and \(f\) changes sign between them.) The integral is well defined, so 'undefined' is wrong, and \(\pi\) and \(2\pi\) confuse it with an area.

Discuss in Forum

22.

The vector normal to the gradient of the function \(f : \mathbf{R}^{3} \to \mathbf{R}\) defined by

\(f(x, y, z) = xy + y^{2} + z\)

at the point P = (1, -1, 1) is :

A.(-3, -3, 0)
B.(3, 0, 3)
C.(-3, 0, 3)
D.(1, 0, 0)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The vector is (3, 0, 3).

\[ \nabla f = (y,\ x + 2y,\ 1) \]

At P = (1, -1, 1) this is (-1, -1, 1). A vector normal to it has zero dot product with it:

\[ (3, 0, 3) \cdot (-1, -1, 1) = -3 + 0 + 3 = 0 \]

The others fail: (-3, -3, 0) gives 3 + 3 = 6, (-3, 0, 3) gives 3 + 3 = 6, and (1, 0, 0) gives -1. Note that (-3, -3, 3), parallel to the gradient, would be the normal to the level surface; the question asks for a vector perpendicular to the gradient.

Discuss in Forum

23.

Let the first and second fundamental forms associated with a surface patch be respectively \(Edu^{2} + 2F\,dudv + Gdv^{2}\) and \(Ldu^{2} + 2M\,dudv + Ndv^{2}\). If \(S = \begin{bmatrix} E & F \\ F & G \end{bmatrix}\) and \(T = \begin{bmatrix} L & M \\ M & N \end{bmatrix}\), then :

A.S and T are non-singular
B.S is non-singular and T may not be
C.T is not-singular and S may not be
D.S and T are singular

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The answer is S is non-singular and T may not be.

The first fundamental form is \(E = \sigma_{u}\cdot\sigma_{u}\), \(F = \sigma_{u}\cdot\sigma_{v}\), \(G = \sigma_{v}\cdot\sigma_{v}\). For a regular patch \(\sigma_{u}\) and \(\sigma_{v}\) are linearly independent, so by Lagrange's identity \(EG - F^{2} = |\sigma_{u} \times \sigma_{v}|^{2} \gt 0\): S is positive definite and non-singular.

The second fundamental form can be degenerate. For a plane, \(L = M = N = 0\), and for a cylinder \(LN - M^{2} = 0\), so T is singular there. Since \(\det T = K\,\det S\), T is singular exactly where the Gaussian curvature is 0. So T may be singular, but S never is.

Discuss in Forum

24.

Let U be an open subset of \(\mathbf{R}^{2}\) and S be a smooth surface in \(\mathbf{R}^{3}\). If \(\sigma : U \to S\) is a regular surface patch such that

\[ \det\begin{pmatrix} \sigma_{u} . \sigma_{u} & \sigma_{u} . \sigma_{v} \\ \sigma_{u} . \sigma_{v} & \sigma_{v} . \sigma_{v} \end{pmatrix} = 1 \]

Then :

A.\(\sigma\) is an isometry
B.\(\sigma\) is conformal, but not an isometry
C.\(\sigma\) preserves area, but need not be conformal
D.\(\sigma\) preserves area and conformal

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The answer is \(\sigma\) preserves area, but need not be conformal.

The determinant is \(EG - F^{2}\), and the area element of the surface is \(\sqrt{EG - F^{2}}\,du\,dv\). If it equals 1, every region of U has the same area as its image, so \(\sigma\) is area-preserving (equiareal).

Conformality needs \(E = G\) and \(F = 0\) (up to a common factor), and an isometry needs \(E = G = 1\), \(F = 0\). The condition \(EG - F^{2} = 1\) allows, for example, \(E = 4\), \(F = 0\), \(G = 1/4\), which preserves area but stretches one direction and shrinks the other. So neither conformality nor isometry follows.

Discuss in Forum

25.

Let \(S = \{(x, y, f(x, y)) \in \mathbf{R}^{3} \mid (x, y) \in U\}\) where U is open in \(\mathbf{R}^{2}\) and \(f : U \to \mathbf{R}\) is smooth. If \(f\) admits a local maximum at \(p\) and a local minimum at \(q\), then the Gaussian curvatures \(k_{1}\) and \(k_{2}\) to S at the points \((p, f(p))\) and \((q, f(q))\) respectively satisfy :

A.\(k_{1}, k_{2} \ge 0\)
B.\(k_{1} \ge 0, k_{2} \le 0\)
C.\(k_{1} \le 0, k_{2} \ge 0\)
D.\(k_{1}, k_{2} \le 0\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The curvatures satisfy \(k_{1}, k_{2} \ge 0\).

For a graph \(z = f(x, y)\), the Gaussian curvature at a critical point (where \(f_{x} = f_{y} = 0\)) is

\[ K = f_{xx}f_{yy} - f_{xy}^{2} \]

the determinant of the Hessian. At a local maximum the Hessian is negative semi-definite and at a local minimum positive semi-definite; in both cases its two eigenvalues have the same sign (or one is 0), so the determinant is \(\ge 0\). Both surfaces bend the same way in every direction, like a cap or a bowl.

A negative curvature would mean a saddle, which is neither a maximum nor a minimum, so the options with \(\le 0\) are wrong.

Discuss in Forum

26.

Consider the function \(f : \mathbf{R}^{2} \to \mathbf{R}\) given by \(f(x, y) = x^{2} + y^{2} - 4y\). Then in a neighbourhood of (0, 0) :

A.\(x\) variable can be expressed as a function of \(y\)
B.\(y\) variable can be expressed as a function of \(x\)
C.none of \(x\) and \(y\) be expressed as a function of other variable
D.both \(x\) and \(y\) can be expressed as a function of the other variable

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

Near (0, 0), \(y\) can be expressed as a function of \(x\).

Consider the level curve \(f(x, y) = f(0, 0) = 0\). At the origin, \(f_{x} = 2x = 0\) and \(f_{y} = 2y - 4 = -4 \neq 0\). By the implicit function theorem, the non-zero \(f_{y}\) lets us solve for \(y\) as a smooth function of \(x\) near 0. Indeed \(y = 2 - \sqrt{4 - x^{2}}\) near the origin.

\(x\) cannot be written as a function of \(y\): solving gives \(x = \pm\sqrt{4y - y^{2}}\), two values for each small \(y \gt 0\), and none for \(y \lt 0\). So A and D fail, and C fails because \(y\) does work.

Discuss in Forum

27.

Consider the function \(f : \mathbf{R}^{2} \to \mathbf{R}^{2}\) defined as \(f(x, y) = (x^{2} - y^{2}, 2xy)\). Then the image of the ball \(B(P, \frac{1}{2})\), where P = (1, 0) is the centre and \(\frac{1}{2}\) is radius, is :

A.a closed set
B.an open set
C.neither open nor closed set
D.an unbounded set

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The image is an open set.

In complex notation \(f(z) = z^{2}\), since \((x + iy)^{2} = x^{2} - y^{2} + 2ixy\). The ball centred at 1 with radius 1/2 does not contain 0, so \(f'(z) = 2z \neq 0\) there. A non-constant analytic function is an open map (open mapping theorem), so the image of an open ball is open.

It is bounded, since \(|z| \lt 3/2\) gives \(|z^{2}| \lt 9/4\), so D is wrong. An open, non-empty, bounded proper subset of the plane cannot be closed, which rules out A and C.

Multivariable Calculus Questions and Answers

Topics

Other Subjects

Tags: Multivariable Calculus Questions and Answers

About Us

Multivariable Calculus Questions and Answers

Topics

Other Subjects

Tags: Multivariable Calculus Questions and Answers

About Us

Stay informed about all important Dates & Notifications