MA 26100 - Spring 2026                          Exam 1                          Test/Quiz #: 11

From the handwritten notes, let the surface equation be defined as the function \(f(x, y, z)\):

\[ f(x, y, z) = 4x^3 - 12x + 6y + z^3 + 2 = 0 \]

The normal vector to the tangent plane at any point on the surface is given by \(\vec{n} = \nabla f\). The \(xz\)-plane is defined by the equation \(y = 0\), so its normal vector is \(\vec{n} = \langle 0, 1, 0 \rangle\).

We want to find a point \((x, y, z)\) such that \(\nabla f \parallel \langle 0, 1, 0 \rangle\).

First, calculate the gradient of \(f\):

\[ \nabla f = \langle 12x^2 - 12, 6, 3z^2 \rangle \]

For \(\nabla f\) to be parallel to \(\langle 0, 1, 0 \rangle\), its \(x\) and \(z\) components must be equal to zero:

\[ 12x^2 - 12 = 0 \rightarrow x = \pm 1 \] \[ 3z^2 = 0 \Rightarrow z = 0 \]

Next, substitute these candidate values back into the original surface equation to solve for \(y\):

Case 1: \(x = 1, z = 0 \rightsquigarrow\)

\[ 4 - 12 + 6y + 0 + 2 = 0 \Rightarrow 6y - 6 = 0 \Rightarrow y = 1 \]

This provides the point \((1, 1, 0)\).

Case 2: \(x = -1, z = 0 \rightsquigarrow\)

\[ -4 + 12 + 6y + 2 = 0 \Rightarrow 6y + 10 = 0 \Rightarrow y = -10/6 \]

This provides the point \((-1, -10/6, 0)\).

The point \((1, 1, 0)\) corresponds to Option E.

Handwritten Solution:

The direction of most rapid increase is given by the gradient vector \( \vec{\nabla} f \).

\[ \vec{\nabla} f = \langle 2xy, x^2 + 3y^2 \rangle \] \[ \vec{\nabla} f(1, 2) = \langle 4, 13 \rangle \]

max. Rate of change is \( |\vec{\nabla} f(1, 2)| = \sqrt{4^2 + 13^2} = \sqrt{185} \)

Solution:

\(\vec{r}(t) = \langle x(t), y(t), z(t) \rangle\)

\(\text{Velocity} = \vec{r}'(t)\),   \(\text{Speed} = |\vec{r}'(t)|\)

\(\text{Acceleration} = \vec{r}''(t)\)

\(\vec{r}'(t) = \langle e^t, e^t, 1 \rangle \Rightarrow |\vec{r}'(t)|\) dependent on \(t\).

\(\vec{r}''(t) = \langle e^t, e^t, 0 \rangle\)

\(\vec{r}''(0) \cdot \vec{r}'(0) = \langle 1, 1, 0 \rangle \cdot \langle 1, 1, 1 \rangle = 1 + 1 + 0 = 2 \neq 0\)

To determine if the motion lies in a plane, we look at the parametric equations of the components:

\(x = e^t\)
\(y = e^t\)
\(z = t\)

Eliminating the parameter \(t\) from the equations for \(x\) and \(y\), we find:

\(x = y\)

Since the motion is constrained by the relationship \(x - y = 0\), the entire path lies in a plane. Therefore, statement C is true.

Solution:

Using FTC (Fundamental Theorem of Calculus) & Chain Rule:

\[ F(y) = \int_{f_1(y)}^{f_2(y)} g(t) \, dt \] \[ F'(y) = g(f_2(y)) f_2'(y) - g(f_1(y)) f_1'(y) \]

Applying this formula to \( f(x,y) \), where we differentiate with respect to \( y \) and treat \( x \) as a constant:

\[ \frac{\partial f}{\partial y} = g(x-y) \cdot \left( \frac{\partial}{\partial y}(x-y) \right) - g(x+y) \cdot \left( \frac{\partial}{\partial y}(x+y) \right) \]

The derivatives of the limits of integration with respect to \( y \) are:

• \( \frac{\partial}{\partial y}(x-y) = -1 \)
• \( \frac{\partial}{\partial y}(x+y) = 1 \)

Substituting these values back into the partial derivative expression:

\[ \frac{\partial f}{\partial y} = g(x-y) \cdot (-1) - g(x+y) \cdot (1) \] \[ \frac{\partial f}{\partial y} = -g(x-y) - g(x+y) \]

ln has only +ve numbers as inputs

Want \( |x^2 - y^2| > 0 \)
only points to remove are \( x^2 - y^2 = 0 \)
\( \Rightarrow x^2 = y^2 \rightsquigarrow |x| = |y| \)

Solution:

The unit tangent vector is defined as:

\[ \vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} \]

First, find the derivative of the position vector \(\vec{r}(t)\):

\[ \vec{r}'(t) = \left\langle 2t, 3t^2, \frac{1}{t+1} \right\rangle \]

Evaluate the velocity vector at \(t = 1\):

\[ \vec{r}'(1) = \langle 2, 3, 0.5 \rangle \]

Calculate the magnitude of the velocity vector at \(t = 1\):

\[ |\vec{r}'(1)| = \sqrt{4 + 9 + 0.25} = \sqrt{13.25} \]

Substitute these into the formula for the unit tangent vector at \(t = 1\):

\[ \vec{T}(1) = \frac{1}{\sqrt{13.25}} \langle 2, 3, 0.5 \rangle \]

A. The limit does not exist.

B. \( \frac{1}{2} \)

C. 0

D. \( -\frac{1}{2} \)

E. 1

Worked Solution:

To determine if the limit exists, we test the limit along different paths approaching \((0,0)\).

Path 1: Let \( x = 0 \)

\[ \lim_{y \to 0} \frac{0 \cdot y^2}{0^2 + y^4} = \lim_{y \to 0} \frac{0}{y^4} = 0 \]

Path 2: Let \( x = y^2 \)

\[ \lim_{y \to 0} \frac{y^2 \cdot y^2}{(y^2)^2 + y^4} = \lim_{y \to 0} \frac{y^4}{y^4 + y^4} = \lim_{y \to 0} \frac{y^4}{2y^4} = \frac{1}{2} \]

Distance travelled = Arc length = 52

Find \(t_2 - t_1\) such that \(52 = \int_{t_1}^{t_2} |\vec{r}'(t)| dt\)

\[ \vec{r}'(t) = \langle -5 \sin t, 12, 5 \cos t \rangle \] \[ |\vec{r}'(t)| = \sqrt{5^2 + 12^2} = 13 \]

\[ 52 = \int_{t_1}^{t_2} 13 dt = 13(t_2 - t_1) \]

\[ t_2 - t_1 = 4 \]

\[ \frac{\partial}{\partial x} \left[ e^{xz} + yz^2 \right] = \frac{\partial}{\partial x} 3 \] \[ e^{xz} \left[ 1 \cdot z + x \cdot \frac{\partial z}{\partial x} \right] + 2yz \cdot \frac{\partial z}{\partial x} = 0 \]

Substitute \( x = 0, y = 2, z = 1 \):

\[ 1 [1 + 0] + 4 \frac{\partial z}{\partial x} = 0 \implies \frac{\partial z}{\partial x} = -1/4 \]

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OR

\( F(x, y, z) = e^{xz} + yz^2 \)

\[ F_x + F_z \cdot \frac{\partial z}{\partial x} = 0 \implies \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{[z e^{xz}]}{x e^{xz} + 2yz} \]

Evaluating at \( (0, 2, 1) \):

\(\vec{n} = \vec{QR} = \langle 6, -1, 1 \rangle\)

equation of plane through \((1, 5, -3)\)
with \(\vec{n} = \langle 6, -1, 1 \rangle\)

\[ 6(x - 1) - 1(y - 5) + 1(z + 3) = 0 \] \[ 6x - 6 - y + 5 + z + 3 = 0 \longrightarrow \boxed{6x - y + z + 2 = 0} \]