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MTH 254 Calculus IV Winter 2026

Richard Zhao

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Worksheet Assigned Problems for Section 14.5

The problems listed below are assigned to be included in your problem set portfolio. Note that a specific selection of these problems will also form the written homework assignments. I recommend working through all of them to ensure a solid grasp of the material. Reach out to Richard for help if you get stuck or have any questions.
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The solutions will be posted after the written homework due dates. If you have any questions about your work, talk to Richard and he is happy to discuss the process with you.
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14.5.5.

Calculate the gradient of \(f(x,y) = \cos\lp x^2 + y \rp\text{.}\)
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Solution.
To find the gradient \(\nabla f(x,y)\text{,}\) we compute the partial derivatives with respect to \(x\) and \(y\) using the Chain Rule.
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Partial derivative with respect to \(x\text{:}\)
\begin{equation*} f_x(x,y) = \frac{\partial}{\partial x} \cos(x^2 + y) = -\sin(x^2 + y) \cdot \frac{\partial}{\partial x}(x^2 + y) = -2x\sin(x^2 + y) \end{equation*}
Partial derivative with respect to \(y\text{:}\)
\begin{equation*} f_y(x,y) = \frac{\partial}{\partial y} \cos(x^2 + y) = -\sin(x^2 + y) \cdot \frac{\partial}{\partial y}(x^2 + y) = -\sin(x^2 + y) \end{equation*}
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Thus, the gradient vector is:
\begin{equation*} \nabla f(x,y) = \la -2x\sin(x^2 + y), -\sin(x^2 + y) \ra \end{equation*}
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14.5.17.

Use the Chain Rule to calculate \(\dfrac{d}{dt} f(\v{r}(t))\text{,}\) where
\begin{equation*} f(x,y) = \ln(x) + \ln(y) \qquad \text{ and } \qquad \v{r}(t) = \la \cos(t), t^2 \ra \end{equation*}
at the point where \(t = \dfrac{\pi}{4}\text{.}\)
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Solution.
According to the Chain Rule for paths, \(\frac{d}{dt} f(\v{r}(t)) = \nabla f(\v{r}(t)) \cdot \v{r}'(t)\text{.}\)
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First, let’s find the gradient of \(f\text{:}\)
\begin{equation*} \nabla f(x,y) = \la \frac{\partial}{\partial x}(\ln x + \ln y), \frac{\partial}{\partial y}(\ln x + \ln y) \ra = \la \frac{1}{x}, \frac{1}{y} \ra \end{equation*}
Next, let’s find the derivative of the path \(\v{r}(t)\text{:}\)
\begin{equation*} \v{r}'(t) = \la \frac{d}{dt}\cos(t), \frac{d}{dt}t^2 \ra = \la -\sin(t), 2t \ra \end{equation*}
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Now we evaluate both at \(t = \frac{\pi}{4}\text{.}\)
\begin{equation*} \v{r}\lp \frac{\pi}{4} \rp = \la \cos\lp \frac{\pi}{4} \rp, \lp \frac{\pi}{4} \rp^2 \ra = \la \frac{\sqrt{2}}{2}, \frac{\pi^2}{16} \ra \end{equation*}
So at this point, \(x = \frac{\sqrt{2}}{2}\) and \(y = \frac{\pi^2}{16}\text{.}\)
\begin{equation*} \nabla f\lp \v{r}\lp \frac{\pi}{4} \rp \rp = \la \frac{1}{\sqrt{2}/2}, \frac{1}{\pi^2/16} \ra = \la \sqrt{2}, \frac{16}{\pi^2} \ra \end{equation*}
And the velocity vector is:
\begin{equation*} \v{r}'\lp \frac{\pi}{4} \rp = \la -\sin\lp \frac{\pi}{4} \rp, 2\lp \frac{\pi}{4} \rp \ra = \la -\frac{\sqrt{2}}{2}, \frac{\pi}{2} \ra \end{equation*}
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Finally, compute the dot product:
\begin{align*} \frac{d}{dt} f\lp \v{r}\lp \frac{\pi}{4} \rp \rp \amp= \la \sqrt{2}, \frac{16}{\pi^2} \ra \cdot \la -\frac{\sqrt{2}}{2}, \frac{\pi}{2} \ra \\ \amp= (\sqrt{2})\lp -\frac{\sqrt{2}}{2} \rp + \lp \frac{16}{\pi^2} \rp \lp \frac{\pi}{2} \rp \\ \amp= -1 + \frac{8}{\pi} \end{align*}
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14.5.25.

Calculate the directional derivative of \(f(x,y) = \tan^{-1}(xy)\) in the direction of \(\v{v} = \la 1,1 \ra\) at the point \(P = (3,4)\text{.}\) Remember to use a unit vector in your directional derivative computation.
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Solution.
First, find the gradient \(\nabla f(x,y)\text{:}\)
\begin{align*} f_x(x,y) \amp= \frac{1}{1+(xy)^2} \cdot y = \frac{y}{1+x^2y^2} \\ f_y(x,y) \amp= \frac{1}{1+(xy)^2} \cdot x = \frac{x}{1+x^2y^2} \end{align*}
Evaluate the gradient at \(P(3,4)\text{:}\)
\begin{align*} f_x(3,4) \amp= \frac{4}{1+(3^2)(4^2)} = \frac{4}{1+144} = \frac{4}{145} \\ f_y(3,4) \amp= \frac{3}{1+(3^2)(4^2)} = \frac{3}{1+144} = \frac{3}{145} \\ \nabla f(3,4) \amp= \la \frac{4}{145}, \frac{3}{145} \ra \end{align*}
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Next, find the unit vector \(\v{u}\) in the direction of \(\v{v} = \la 1,1 \ra\text{:}\)
\begin{align*} \|\v{v}\| \amp= \sqrt{1^2 + 1^2} = \sqrt{2} \\ \v{u} \amp= \frac{\v{v}}{\|\v{v}\|} = \la \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \ra \end{align*}
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Finally, calculate the directional derivative using the dot product:
\begin{align*} D_\v{u}f(3,4) \amp= \nabla f(3,4) \cdot \v{u} \\ \amp= \la \frac{4}{145}, \frac{3}{145} \ra \cdot \la \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \ra \\ \amp= \frac{4}{145\sqrt{2}} + \frac{3}{145\sqrt{2}} \\ \amp= \frac{7}{145\sqrt{2}} = \frac{7\sqrt{2}}{290} \end{align*}
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14.5.35.

Determine the direction in which \(f(x,y,z) = \dfrac{xy}{z}\) has maximum rate of increase from \(P = (1,-1,3)\text{,}\) and give the rate of change in that direction.
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Solution.
The direction of maximum increase is given by the gradient \(\nabla f(P)\text{,}\) and the rate is the magnitude \(\|\nabla f(P)\|\text{.}\)
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First, calculate the partial derivatives:
\begin{align*} f_x \amp= \frac{y}{z} \\ f_y \amp= \frac{x}{z} \\ f_z \amp= -\frac{xy}{z^2} \end{align*}
Evaluate at \(P(1,-1,3)\text{:}\)
\begin{align*} f_x(1,-1,3) \amp= \frac{-1}{3} = -\frac{1}{3} \\ f_y(1,-1,3) \amp= \frac{1}{3} \\ f_z(1,-1,3) \amp= -\frac{(1)(-1)}{3^2} = \frac{1}{9} \end{align*}
So the direction of maximum increase is:
\begin{equation*} \nabla f(1,-1,3) = \la -\frac{1}{3}, \frac{1}{3}, \frac{1}{9} \ra \end{equation*}
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The maximum rate of change is the magnitude:
\begin{align*} \|\nabla f\| \amp= \sqrt{\lp -\frac{1}{3} \rp^2 + \lp \frac{1}{3} \rp^2 + \lp \frac{1}{9} \rp^2} \\ \amp= \sqrt{\frac{1}{9} + \frac{1}{9} + \frac{1}{81}} \\ \amp= \sqrt{\frac{9}{81} + \frac{9}{81} + \frac{1}{81}} \\ \amp= \sqrt{\frac{19}{81}} = \frac{\sqrt{19}}{9} \end{align*}
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14.5.37.

Suppose that \(\nabla f_P = \la 2,-4,4 \ra\text{.}\) Is \(f\) increasing or decreasing at \(P\) in the direction of \(\v{v} = \la 2,1,3 \ra\text{?}\)
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Solution.
To determine if \(f\) is increasing or decreasing, we check the sign of the directional derivative \(D_\v{u}f(P)\text{.}\) The sign of \(D_\v{u}f\) is the same as the sign of the dot product \(\nabla f_P \cdot \v{v}\) (since dividing by \(\|\v{v}\|\) to get \(\v{u}\) is always positive).
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Calculate the dot product:
\begin{align*} \nabla f_P \cdot \v{v} \amp= \la 2, -4, 4 \ra \cdot \la 2, 1, 3 \ra \\ \amp= (2)(2) + (-4)(1) + (4)(3) \\ \amp= 4 - 4 + 12 = 12 \end{align*}
Since the dot product is positive (\(12 > 0\)), the directional derivative is positive. Therefore, \(f\) is increasing in the direction of \(\v{v}\text{.}\)
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14.5.39.

Let \(f(x,y,z) = \sin(xy + z)\) and \(P = (0,-1, \pi)\text{.}\) Calculate \(D_\v{u} f(P)\text{,}\) where \(\v{u}\) is a unit vector making an angle \(\theta = 30^\circ\) with \(\nabla f_P\text{.}\)
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Solution.
We use the formula \(D_\v{u} f(P) = \|\nabla f_P\| \cos(\theta)\text{.}\)
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First, compute the gradient \(\nabla f\text{:}\)
\begin{align*} f_x \amp= y\cos(xy+z) \\ f_y \amp= x\cos(xy+z) \\ f_z \amp= \cos(xy+z) \end{align*}
Evaluate at \(P(0,-1,\pi)\text{:}\)
\begin{equation*} \text{Argument } = (0)(-1) + \pi = \pi \end{equation*}
\begin{align*} f_x(0,-1,\pi) \amp= -1\cos(\pi) = (-1)(-1) = 1 \\ f_y(0,-1,\pi) \amp= 0\cos(\pi) = 0 \\ f_z(0,-1,\pi) \amp= \cos(\pi) = -1 \end{align*}
So \(\nabla f_P = \la 1, 0, -1 \ra\text{.}\)
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Next, find the magnitude of the gradient:
\begin{equation*} \|\nabla f_P\| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2} \end{equation*}
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Finally, calculate the directional derivative using \(\theta = 30^\circ\text{:}\)
\begin{align*} D_\v{u} f(P) \amp= \|\nabla f_P\| \cos(30^\circ) \\ \amp= \sqrt{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2} \end{align*}
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14.5.43.

Find the two points on the ellipsoid
\begin{equation*} \frac{x^2}{4} + \frac{y^2}{9} + z^2 = 1 \end{equation*}
where the tangent plane is normal to \(\v{v} = \la 1,1,-2 \ra\text{.}\)
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Solution.
Let \(F(x,y,z) = \frac{x^2}{4} + \frac{y^2}{9} + z^2\text{.}\) The normal vector to the tangent plane is given by the gradient \(\nabla F\text{.}\)
\begin{equation*} \nabla F = \la \frac{x}{2}, \frac{2y}{9}, 2z \ra \end{equation*}
We want the tangent plane to be normal to \(\v{v} = \la 1,1,-2 \ra\text{,}\) which means the gradient must be parallel to \(\v{v}\text{.}\) So, \(\nabla F = k \v{v}\) for some scalar constant \(k\text{.}\)
\begin{equation*} \la \frac{x}{2}, \frac{2y}{9}, 2z \ra = k \la 1, 1, -2 \ra = \la k, k, -2k \ra \end{equation*}
Equating components gives us expressions for \(x, y, z\) in terms of \(k\text{:}\)
\begin{gather*} \frac{x}{2} = k \implies x = 2k \\ \frac{2y}{9} = k \implies y = \frac{9k}{2} \\ 2z = -2k \implies z = -k \end{gather*}
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Since the points must lie on the ellipsoid, substitute these into the ellipsoid equation:
\begin{align*} \frac{(2k)^2}{4} + \frac{(9k/2)^2}{9} + (-k)^2 \amp= 1 \\ \frac{4k^2}{4} + \frac{81k^2/4}{9} + k^2 \amp= 1 \\ k^2 + \frac{9k^2}{4} + k^2 \amp= 1 \\ 2k^2 + \frac{9k^2}{4} \amp= 1 \\ \frac{8k^2 + 9k^2}{4} \amp= 1 \\ \frac{17k^2}{4} \amp= 1 \implies k^2 = \frac{4}{17} \implies k = \pm \frac{2}{\sqrt{17}} \end{align*}
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Now find the two points by plugging \(k\) back into our expressions for \(x,y,z\text{:}\) Case 1: \(k = \frac{2}{\sqrt{17}}\)
\begin{equation*} P_1 = \lp 2\lp\frac{2}{\sqrt{17}}\rp, \frac{9}{2}\lp\frac{2}{\sqrt{17}}\rp, -\frac{2}{\sqrt{17}} \rp = \lp \frac{4}{\sqrt{17}}, \frac{9}{\sqrt{17}}, -\frac{2}{\sqrt{17}} \rp \end{equation*}
Case 2: \(k = -\frac{2}{\sqrt{17}}\)
\begin{equation*} P_2 = \lp -\frac{4}{\sqrt{17}}, -\frac{9}{\sqrt{17}}, \frac{2}{\sqrt{17}} \rp \end{equation*}
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14.5.47.

Find an equation of the tangent plane to the surface
\begin{equation*} xz + 2x^2y + y^2z^3 = 11 \end{equation*}
at the point \(P = (2,1,1)\text{.}\)
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Solution.
Let \(F(x,y,z) = xz + 2x^2y + y^2z^3\text{.}\) The surface is the level set \(F=11\text{.}\) First, find the gradient \(\nabla F\text{:}\)
\begin{align*} F_x \amp= z + 4xy \\ F_y \amp= 2x^2 + 2yz^3 \\ F_z \amp= x + 3y^2z^2 \end{align*}
Evaluate at \(P(2,1,1)\) to find the normal vector:
\begin{align*} F_x(2,1,1) \amp= 1 + 4(2)(1) = 9 \\ F_y(2,1,1) \amp= 2(2)^2 + 2(1)(1)^3 = 8 + 2 = 10 \\ F_z(2,1,1) \amp= 2 + 3(1)^2(1)^2 = 2 + 3 = 5 \end{align*}
The normal vector is \(\v{n} = \la 9, 10, 5 \ra\text{.}\)
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The equation of the tangent plane is:
\begin{align*} 9(x - 2) + 10(y - 1) + 5(z - 1) \amp= 0 \\ 9x - 18 + 10y - 10 + 5z - 5 \amp= 0 \\ 9x + 10y + 5z \amp= 33 \end{align*}
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14.5.53.

Find a function \(f(x,y,z)\) such that \(\nabla f = \la 2x,1,2 \ra\text{.}\)
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Solution.
We are given \(f_x = 2x\text{,}\) \(f_y = 1\text{,}\) and \(f_z = 2\text{.}\) We integrate each partial derivative to reconstruct \(f\text{.}\)
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Integrate \(f_x\) with respect to \(x\text{:}\)
\begin{equation*} f(x,y,z) = \int 2x \, dx = x^2 + C_1(y,z) \end{equation*}
Differentiate this result with respect to \(y\) and compare to the given \(f_y\text{:}\)
\begin{equation*} \frac{\partial}{\partial y}(x^2 + C_1(y,z)) = \frac{\partial C_1}{\partial y} = 1 \end{equation*}
Integrating \(1\) with respect to \(y\) gives \(C_1(y,z) = y + C_2(z)\text{.}\) So far, \(f(x,y,z) = x^2 + y + C_2(z)\text{.}\)
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Differentiate this with respect to \(z\) and compare to the given \(f_z\text{:}\)
\begin{equation*} \frac{\partial}{\partial z}(x^2 + y + C_2(z)) = C_2'(z) = 2 \end{equation*}
Integrating \(2\) with respect to \(z\) gives \(C_2(z) = 2z + K\) (where \(K\) is a constant).
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Thus, a possible function is:
\begin{equation*} f(x,y,z) = x^2 + y + 2z \end{equation*}
(Any constant \(K\) can be added, but usually \(K=0\) is sufficient for "Find a function" problems).
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14.5.65.

Let \(\c{C}\) be the curve of intersection of the spheres
\begin{equation*} x^3 + 2xy + yz = 7 \qquad \text{ and } \qquad 3x^2 - yz = 1 \end{equation*}
Find the parametric equations of the tangent line to \(\c{C}\) at \(P = (1,2,1)\text{.}\)
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Solution.
The curve \(\c{C}\) is the intersection of two surfaces. The tangent line to \(\c{C}\) must be tangent to both surfaces. This means the direction vector of the tangent line, \(\v{v}\text{,}\) must be perpendicular to the normal vectors of both surfaces.
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Let \(F(x,y,z) = x^3 + 2xy + yz\) and \(G(x,y,z) = 3x^2 - yz\text{.}\) First, find the gradients (normal vectors) at \(P(1,2,1)\text{.}\)
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Gradient of \(F\text{:}\)
\begin{equation*} \nabla F = \la 3x^2 + 2y, 2x + z, y \ra \end{equation*}
At \(P(1,2,1)\text{:}\)
\begin{equation*} \nabla F(1,2,1) = \la 3(1)^2 + 2(2), 2(1) + 1, 2 \ra = \la 7, 3, 2 \ra \end{equation*}
Gradient of \(G\text{:}\)
\begin{equation*} \nabla G = \la 6x, -z, -y \ra \end{equation*}
At \(P(1,2,1)\text{:}\)
\begin{equation*} \nabla G(1,2,1) = \la 6(1), -1, -2 \ra = \la 6, -1, -2 \ra \end{equation*}
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The direction vector \(\v{v}\) is the cross product of these two gradients:
\begin{align*} \v{v} \amp= \nabla F \times \nabla G \\ \amp= \begin{vmatrix} \v{i} \amp \v{j} \amp \v{k} \\ 7 \amp 3 \amp 2 \\ 6 \amp -1 \amp -2 \end{vmatrix} \\ \amp= \la (3)(-2) - (2)(-1), -((7)(-2) - (2)(6)), (7)(-1) - (3)(6) \ra \\ \amp= \la -6 + 2, -(-14 - 12), -7 - 18 \ra \\ \amp= \la -4, 26, -25 \ra \end{align*}
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The tangent line passes through \(P(1,2,1)\) with direction \(\la -4, 26, -25 \ra\text{.}\) The parametric equations are:
\begin{align*} x(t) \amp= 1 - 4t \\ y(t) \amp= 2 + 26t \\ z(t) \amp= 1 - 25t \end{align*}
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