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MTH 254 Calculus IV Winter 2026

Richard Zhao

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Worksheet Assigned Problems for Section 14.3

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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14.3.3.

Use the Product Rule to compute \(f_x\) and \(f_y\) for \(f(x,y) = \lp x^2 - y\rp\lp x - y^2\rp\)
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Solution.
Using the Product Rule for \(f_x\text{:}\)
\begin{align*} f_x \amp = \frac{\partial}{\partial x}\lp x^2 - y\rp \cdot \lp x - y^2\rp + \lp x^2 - y\rp \cdot \frac{\partial}{\partial x}\lp x - y^2\rp \\ \amp = \lp 2x \rp \lp x - y^2 \rp + \lp x^2 - y \rp \lp 1 \rp \\ \amp = 3x^2 - 2xy^2 - y \end{align*}
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Using the Product Rule for \(f_y\text{:}\)
\begin{align*} f_y \amp = \frac{\partial}{\partial y}\lp x^2 - y\rp \cdot \lp x - y^2\rp + \lp x^2 - y\rp \cdot \frac{\partial}{\partial y}\lp x - y^2\rp \\ \amp = \lp -1 \rp \lp x - y^2 \rp + \lp x^2 - y \rp \lp -2y \rp \\ \amp = -x + y^2 - 2x^2y + 2y^2 \\ \amp = 3y^2 - 2x^2y - x \end{align*}
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14.3.5.

Use the Quotient Rule to compute \(\dfrac{\partial}{\partial y} \dfrac{y}{x + y}\)
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Solution.
\begin{align*} \frac{\partial}{\partial y} \frac{y}{x + y} \amp = \frac{\lp 1 \rp\lp x + y \rp - \lp y \rp \lp 1 \rp}{\lp x + y \rp^2} \\ \amp = \frac{x + y - y}{\lp x + y \rp^2} \\ \amp = \frac{x}{\lp x + y \rp^2} \end{align*}
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14.3.9.

The plane \(y = 1\) intersects the surface \(z = x^4 + 6xy - y^4\) in a certain curve. Find the slope of the tangent line to this curve at the point \(P = (1,1,6)\text{.}\)
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Solution.
The slope of the tangent line to the curve formed by the intersection of the surface and the plane \(y=1\) is given by the partial derivative with respect to \(x\) evaluated at the point \((1,1)\text{.}\)
\begin{align*} f_x(x,y) \amp = \frac{\partial}{\partial x} \lp x^4 + 6xy - y^4 \rp = 4x^3 + 6y \\ f_x(1,1) \amp = 4(1)^3 + 6(1) = 4 + 6 =10 \end{align*}
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Exercise Group.

For the following exercises, compute the first-order partial derivatives.
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14.3.17.
\(z = x^4y + xy^{-2}\)
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Solution.
\begin{align*} \frac{\partial z}{\partial x} \amp = \frac{\partial}{\partial x}\lp x^4y \rp + \frac{\partial}{\partial x}\lp xy^{-2} \rp \\ \amp = 4x^3y + y^{-2} \end{align*}
\begin{align*} \frac{\partial z}{\partial y} \amp = \frac{\partial}{\partial y}\lp x^4y \rp + \frac{\partial}{\partial y}\lp xy^{-2} \rp \\ \amp = x^4 - 2xy^{-3} \end{align*}
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14.3.23.
\(z = \lp \sin(x) \rp\lp \cos(y) \rp\)
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Solution.
\begin{align*} \frac{\partial z}{\partial x} \amp = \cos(x)\cos(y) \end{align*}
\begin{align*} \frac{\partial z}{\partial y} \amp = -\sin(x)\sin(y) \end{align*}
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14.3.33.
\(z = e^{-x^2 - y^2}\)
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Solution.
\begin{align*} \frac{\partial z}{\partial x} \amp = e^{-x^2 - y^2} \cdot \frac{\partial}{\partial x} \lp -x^2 - y^2 \rp \\ \amp = -2xe^{-x^2 - y^2} \end{align*}
\begin{align*} \frac{\partial z}{\partial y} \amp = e^{-x^2 - y^2} \cdot \frac{\partial}{\partial y} \lp -x^2 - y^2 \rp \\ \amp = -2ye^{-x^2 - y^2} \end{align*}
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14.3.39.
\(w = xy^2z^3\)
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Solution.
\begin{align*} \frac{\partial w}{\partial x} \amp = y^2z^3 \end{align*}
\begin{align*} \frac{\partial w}{\partial y} \amp = 2xyz^3 \end{align*}
\begin{align*} \frac{\partial w}{\partial z} \amp = 3xy^2z^2 \end{align*}
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14.3.45.

Given that \(g(u,v) = u \ln\lp u + v \rp\text{,}\) compute the partial derivative \(g_u(1,2)\text{.}\)
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Solution.
First, we find the partial derivative function \(g_u(u,v)\) using the Product Rule.
\begin{align*} g_u(u,v) \amp = 1 \cdot \ln\lp u + v \rp + u \cdot \frac{1}{u+v} \cdot 1 \\ \amp = \ln\lp u + v \rp + \frac{u}{u+v} \end{align*}
Now we substitute \((1,2)\) into the derivative.
\begin{align*} g_u(1,2) \amp = \ln\lp 1 + 2 \rp + \frac{1}{1+2} \\ \amp = \ln(3) + \frac{1}{3} \end{align*}
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Exercise Group.

For the following exercises, compute the derivatives indicated.
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14.3.53.
\(f(x,y) = 3x^2y - 6xy^4\, , \quad \dfrac{\partial^2 f}{\partial x^2} \text{ and } \dfrac{\partial^2 f}{\partial y^2}\)
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Solution.
\begin{align*} \frac{\partial f}{\partial x} \amp = 6xy - 6y^4 \\ \frac{\partial^2 f}{\partial x^2} \amp = 6y \end{align*}
\begin{align*} \frac{\partial f}{\partial y} \amp = 3x^2 - 24xy^3 \\ \frac{\partial^2 f}{\partial y^2} \amp = -72xy^2 \end{align*}
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14.3.55.
\(h(u,v) = \dfrac{u}{u + 4v}\, , \quad h_{vv}(u,v)\)
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Solution.
We can rewrite \(h(u,v) = u(u+4v)^{-1}\) to make differentiation easier.
\begin{align*} h_v(u,v) \amp = u(-1)(u+4v)^{-2}(4) = -4u(u+4v)^{-2} \\ h_{vv}(u,v) \amp = -4u(-2)(u+4v)^{-3}(4) \\ \amp = 32u(u+4v)^{-3} = \frac{32u}{\lp u + 4v \rp^3} \end{align*}
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Exercise Group.

In the following exercises, compute the derivative indicated.
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14.3.61.
\(f(u,v) = \cos\lp u + v^2 \rp \, , \quad f_{uuv}\)
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Solution.
\begin{align*} f_u \amp = -\sin\lp u + v^2 \rp \\ f_{uu} \amp = -\cos\lp u + v^2 \rp \\ f_{uuv} \amp = \frac{\partial}{\partial v}\lp -\cos\lp u + v^2 \rp \rp = -\lp -\sin\lp u + v^2 \rp \rp \cdot 2v \\ \amp = 2v\sin\lp u + v^2 \rp \end{align*}
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14.3.67.
\(g(x,y,z) = \sqrt{x^2 + y^2 + z^2} \, , \quad g_{xyz}\)
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Solution.
\begin{align*} g_x \amp = \frac{1}{2}\lp x^2 + y^2 + z^2 \rp^{-1/2} \cdot 2x = x\lp x^2 + y^2 + z^2 \rp^{-1/2} \\ g_{xy} \amp = x \cdot \lp -\frac{1}{2} \rp \lp x^2 + y^2 + z^2 \rp^{-3/2} \cdot 2y = -xy\lp x^2 + y^2 + z^2 \rp^{-3/2} \\ g_{xyz} \amp = -xy \cdot \lp -\frac{3}{2} \rp \lp x^2 + y^2 + z^2 \rp^{-5/2} \cdot 2z \\ \amp = 3xyz\lp x^2 + y^2 + z^2 \rp^{-5/2} \end{align*}
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14.3.75.

The Laplace operator \(\Delta\) is defined by
\begin{equation*} \Delta f = f_{xx} + f_{yy} \end{equation*}
A function \(u(x,y)\) satisfying the Laplace equation \(\Delta u = 0\) is called harmonic.
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Show that the following functions are harmonic.
  1. \(\displaystyle u(x,y) = x\)
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  2. \(\displaystyle u(x,y) = e^x\cos(y)\)
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  3. \(\displaystyle u(x,y) = \tan^{-1} \lp \dfrac{y}{x} \rp\)
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  4. \(\displaystyle u(x,y) = \ln\lp x^2 + y^2 \rp\)
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Solution.
To demonstrate that each function is harmonic, we must compute the second partial derivatives with respect to \(x\) and \(y\text{,}\) and show that their sum is zero.
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  1. We first compute the derivatives for \(u(x,y) = x\text{.}\)
    \begin{align*} u_x \amp = 1 \\ u_{xx} \amp = 0 \\ u_y \amp = 0 \\ u_{yy} \amp = 0 \end{align*}
    Now we calculate the Laplacian:
    \begin{align*} \Delta u \amp = u_{xx} + u_{yy} = 0 + 0 = 0 \end{align*}
    Since \(\Delta u = 0\text{,}\) the function is harmonic.
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  2. Next, we compute the derivatives for \(u(x,y) = e^x\cos(y)\text{.}\)
    \begin{align*} u_x \amp = e^x\cos(y) \\ u_{xx} \amp = e^x\cos(y) \\ u_y \amp = -e^x\sin(y) \\ u_{yy} \amp = -e^x\cos(y) \end{align*}
    Summing the second derivatives gives:
    \begin{align*} \Delta u \amp = e^x\cos(y) - e^x\cos(y) = 0 \end{align*}
    Therefore, the function is harmonic.
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  3. For \(u(x,y) = \tan^{-1} \lp \dfrac{y}{x} \rp\text{,}\) we use the Chain Rule to find the derivatives.
    \begin{align*} u_x \amp = \frac{1}{1 + (y/x)^2} \cdot \lp -yx^{-2} \rp = \frac{-y}{x^2 + y^2} \\ u_{xx} \amp = \frac{\partial}{\partial x}\lp -y\lp x^2 + y^2 \rp^{-1} \rp = -y(-1)\lp x^2 + y^2 \rp^{-2}(2x) = \frac{2xy}{(x^2 + y^2)^2} \\ u_y \amp = \frac{1}{1 + (y/x)^2} \cdot \lp x^{-1} \rp = \frac{x}{x^2 + y^2} \\ u_{yy} \amp = \frac{\partial}{\partial y}\lp x\lp x^2 + y^2 \rp^{-1} \rp = x(-1)\lp x^2 + y^2 \rp^{-2}(2y) = \frac{-2xy}{(x^2 + y^2)^2} \end{align*}
    Adding the second partials:
    \begin{align*} \Delta u \amp = \frac{2xy}{(x^2 + y^2)^2} + \frac{-2xy}{(x^2 + y^2)^2} = 0 \end{align*}
    Thus, the function is harmonic.
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  4. Finally, for \(u(x,y) = \ln\lp x^2 + y^2 \rp\text{,}\) we compute the derivatives using the Chain Rule and Quotient Rule.
    \begin{align*} u_x \amp = \frac{1}{x^2 + y^2} \cdot 2x = \frac{2x}{x^2 + y^2} \\ u_{xx} \amp = \frac{2(x^2 + y^2) - 2x(2x)}{(x^2 + y^2)^2} = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2} \\ u_y \amp = \frac{1}{x^2 + y^2} \cdot 2y = \frac{2y}{x^2 + y^2} \\ u_{yy} \amp = \frac{2(x^2 + y^2) - 2y(2y)}{(x^2 + y^2)^2} = \frac{2x^2 - 2y^2}{(x^2 + y^2)^2} \end{align*}
    We sum \(u_{xx}\) and \(u_{yy}\text{:}\)
    \begin{align*} \Delta u \amp = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2} + \frac{2x^2 - 2y^2}{(x^2 + y^2)^2} \\ \amp = \frac{2y^2 - 2x^2 + 2x^2 - 2y^2}{(x^2 + y^2)^2} = 0 \end{align*}
    Since the sum is zero, we conclude that the function is harmonic.
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