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MTH 254 Calculus IV Winter 2026

Richard Zhao

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Worksheet Assigned Problems for Section 14.4

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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14.4.7.

Find an equation of the tangent plane to \(F(r,s) = r^2 s^{-\frac{1}{2}} + s^{-3}\) at the point \((2,1)\text{.}\)
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Solution.
First, we compute the function value at the point \((2,1)\text{:}\)
\begin{align*} F(2,1) \amp = (2)^2 (1)^{-1/2} + (1)^{-3} = 4 + 1 = 5 \end{align*}
Next, we find the partial derivatives:
\begin{align*} F_r(r,s) \amp = 2rs^{-1/2} \\ F_s(r,s) \amp = -\frac{1}{2}r^2s^{-3/2} - 3s^{-4} \end{align*}
Evaluate the derivatives at \((2,1)\text{:}\)
\begin{align*} F_r(2,1) \amp = 2(2)(1) = 4 \\ F_s(2,1) \amp = -\frac{1}{2}(4)(1) - 3(1) = -2 - 3 = -5 \end{align*}
The equation of the tangent plane is:
\begin{align*} z - z_0 \amp = F_r(2,1)(r - 2) + F_s(2,1)(s - 1) \\ z - 5 \amp = 4(r - 2) - 5(s - 1) \\ z \amp = 4r - 5s - 8 + 5 + 5 \\ z \amp = 4r - 5s + 2 \end{align*}
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14.4.11.

Find the points on the graph of \(z = 3x^2 - 4y^2\) at which the vector \(\v{n} = \la 3,2,2 \ra\) is normal to the tangent plane.
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Solution.
Let \(f(x,y) = 3x^2 - 4y^2\text{.}\) We know that the tangent plane is determined by the vectors \(\v{v} = \la 1, 0, f_x \ra\) and \(\v{w} = \la 0, 1, f_y \ra\text{.}\) The normal vector to the tangent plane is given by their cross product:
\begin{align*} \v{n}\amp = \v{w} \times \v{v} = \la f_x, f_y, -1 \ra \end{align*}
First, we compute the partial derivatives:
\begin{align*} f_x(x,y) \amp = 6x \\ f_y(x,y) \amp = -8y \end{align*}
So, the normal vector to the tangent plane at any point \((x,y)\) is \(\la 6x, -8y, -1 \ra\text{.}\) We are given that \(\v{n} = \la 3,2,2 \ra\) is normal to the tangent plane, so our computed vector must be parallel to \(\v{n}\text{.}\) This means there exists a constant \(k\) such that:
\begin{align*} \la 6x, -8y, -1 \ra \amp = k \la 3,2,2 \ra \end{align*}
Equating the components:
\begin{align*} -1 \amp = 2k \implies k = -\frac{1}{2} \\ 6x \amp = 3k = 3\lp -\frac{1}{2} \rp = -\frac{3}{2} \implies x = -\frac{1}{4} \\ -8y \amp = 2k = 2\lp -\frac{1}{2} \rp = -1 \implies y = \frac{1}{8} \end{align*}
Finally, we find the \(z\)-coordinate:
\begin{align*} z \amp = 3\lp -\frac{1}{4} \rp^2 - 4\lp \frac{1}{8} \rp^2 \\ \amp = \frac{3}{16} - \frac{4}{64} = \frac{12}{64} - \frac{4}{64} = \frac{8}{64} = \frac{1}{8} \end{align*}
The point is \(\lp -\frac{1}{4}, \frac{1}{8}, \frac{1}{8} \rp\text{.}\)
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14.4.13.

Find the points on the graph of \(f(x,y) = 3x^2 - xy - y^2\) at which the tangent plane is horizontal.
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Solution.
A horizontal tangent plane occurs where the partial derivatives are both zero.
\begin{align*} f_x(x,y) \amp = 6x - y = 0 \\ f_y(x,y) \amp = -x - 2y = 0 \end{align*}
From the first equation, \(y = 6x\text{.}\) Substituting this into the second equation:
\begin{align*} -x - 2(6x) \amp = 0 \\ -13x \amp = 0 \implies x = 0 \end{align*}
If \(x=0\text{,}\) then \(y=0\text{.}\) The point on the graph is \((0,0, f(0,0)) = (0,0,0)\text{.}\)
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14.4.15.

Find the linearization \(L(x,y)\) of \(f(x,y) = x^2y^3\) at \((a,b) = (1,2)\text{.}\) Use it to estimate \(f(2.01, 1.02)\) and \(f(1.97, 1.01)\text{,}\) and compare with values obtained using a calculator.
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Solution.
We compute the function value and derivatives at \((2,1)\text{:}\)
\begin{align*} f(2,1) \amp = 2^2(1)^3 = 4 \\ f_x(x,y) \amp = 2xy^3 \implies f_x(2,1) = 2(2)(1) = 4 \\ f_y(x,y) \amp = 3x^2y^2 \implies f_y(2,1) = 3(4)(1) = 12 \end{align*}
The linearization is:
\begin{align*} L(x,y) \amp = f(2,1) + f_x(2,1)(x-2) + f_y(2,1)(y-1) \\ \amp = 4 + 4(x-2) + 12(y-1) \end{align*}
Using \(L(x,y)\) for estimation:
\begin{align*} f(2.01, 1.02) \amp \approx 4 + 4(0.01) + 12(0.02) = 4 + 0.04 + 0.24 = 4.28 \\ \text{Calculator: } \amp 2.01^2 \cdot 1.02^3 \approx 4.286 \end{align*}
\begin{align*} f(1.97, 1.01) \amp \approx 4 + 4(-0.03) + 12(0.01) = 4 - 0.12 + 0.12 = 4 \\ \text{Calculator: } \amp 1.97^2 \cdot 1.01^3 \approx 3.998 \end{align*}
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14.4.17.

Let \(f(x,y) = x^3 y^{-4}\text{.}\) Use the formula for linear approximation using differentials to estimate the change
\begin{equation*} \Delta f = f(2.03, 0.9) - f(2,1) \end{equation*}
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Solution.
The differential is \(df = f_x \, dx + f_y \, dy\text{.}\)
\begin{align*} f_x \amp = 3x^2y^{-4} \\ f_y \amp = -4x^3y^{-5} \end{align*}
At \((2,1)\) with \(dx = 0.03\) and \(dy = -0.1\text{:}\)
\begin{align*} f_x(2,1) \amp = 3(4)(1) = 12 \\ f_y(2,1) \amp = -4(8)(1) = -32 \end{align*}
\begin{align*} \Delta f \approx df \amp = 12(0.03) + (-32)(-0.1) \\ \amp = 0.36 + 3.2 = 3.56 \end{align*}
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14.4.23.

Estimate \(f(2.1, 3.8)\) assuming that \(f(2,4) = 5\text{,}\) \(f_x(2,4) = 0.3\text{,}\) and \(f_y(2,4) = -0.2\text{.}\)
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Solution.
We use the linearization centered at \((2,4)\text{.}\) Here \(\Delta x = 0.1\) and \(\Delta y = -0.2\text{.}\)
\begin{align*} f(2.1, 3.8) \amp \approx f(2,4) + f_x(2,4)\Delta x + f_y(2,4)\Delta y \\ \amp = 5 + (0.3)(0.1) + (-0.2)(-0.2) \\ \amp = 5 + 0.03 + 0.04 \\ \amp = 5.07 \end{align*}
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14.4.29.

Use the Linear Approximation to estimate the value of \(\sqrt{(1.9)(2.02)(4.05)}\text{.}\) Compare with the value given by a calculator.
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Solution.
Let \(f(x,y,z) = \sqrt{xyz}\text{.}\) We approximate near \((2,2,4)\) where \(f(2,2,4) = \sqrt{16} = 4\text{.}\) The differentials are \(dx = -0.1\text{,}\) \(dy = 0.02\text{,}\) and \(dz = 0.05\text{.}\)
\begin{align*} f_x \amp = \frac{yz}{2\sqrt{xyz}} \implies f_x(2,2,4) = \frac{8}{8} = 1 \\ f_y \amp = \frac{xz}{2\sqrt{xyz}} \implies f_y(2,2,4) = \frac{8}{8} = 1 \\ f_z \amp = \frac{xy}{2\sqrt{xyz}} \implies f_z(2,2,4) = \frac{4}{8} = 0.5 \end{align*}
\begin{align*} L(1.9, 2.02, 4.05) \amp = 4 + 1(-0.1) + 1(0.02) + 0.5(0.05) \\ \amp = 4 - 0.1 + 0.02 + 0.025 \\ \amp = 3.945 \end{align*}
Using a calculator: \(\sqrt{15.5439} \approx 3.9426\text{.}\)
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14.4.31.

Suppose that the plane tangent to \(z = f(x,y)\) at \((-2,3,4)\) has equation \(4x + 2y + z = 2\text{.}\) Estimate \(f(-2.1, 3.1)\text{.}\)
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Solution.
The tangent plane \(z = L(x,y)\) provides the linear approximation for \(f\text{.}\) We rewrite the plane equation to solve for \(z\text{:}\)
\begin{align*} z \amp = 2 - 4x - 2y \end{align*}
Substituting the point \((-2.1, 3.1)\text{:}\)
\begin{align*} f(-2.1, 3.1) \amp \approx 2 - 4(-2.1) - 2(3.1) \\ \amp = 2 + 8.4 - 6.2 \\ \amp = 4.2 \end{align*}
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14.4.37.

A cylinder of radius \(r\) and height \(h\) has volume \(V(r,h) = \pi r^2 h\text{.}\)
  1. Use the Linear Approximation to show that
    \begin{equation*} \frac{\Delta V}{V} \approx \frac{2\Delta r}{r} + \frac{\Delta h}{h} \end{equation*}
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  2. Estimate the percentage increase in \(V\) if \(r\) and \(h\) are each increased by 2%.
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  3. The volume of a certain cylinder \(V\) is determined by measuring \(r\) and \(h\text{.}\) Which will lead to a greater error in \(V\text{:}\) a 1% error in \(r\) or a 1% error in \(h\text{?}\)
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Solution.
  1. The total differential of \(V\) is:
    \begin{align*} dV \amp = \frac{\partial V}{\partial r}dr + \frac{\partial V}{\partial h}dh \\ \amp = 2\pi rh \, dr + \pi r^2 \, dh \end{align*}
    Approximating \(\Delta V \approx dV\) and dividing by \(V = \pi r^2 h\text{:}\)
    \begin{align*} \frac{\Delta V}{V} \amp \approx \frac{2\pi rh \Delta r + \pi r^2 \Delta h}{\pi r^2 h} \\ \amp \approx \frac{2\pi rh \Delta r}{\pi r^2 h} + \frac{\pi r^2 \Delta h}{\pi r^2 h} \\ \amp \approx \frac{2\Delta r}{r} + \frac{\Delta h}{h} \end{align*}
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  2. Given \(\dfrac{\Delta r}{r} = 0.02\) and \(\dfrac{\Delta h}{h} = 0.02\text{:}\)
    \begin{align*} \frac{\Delta V}{V} \amp \approx 2(0.02) + 0.02 = 0.04 + 0.02 = 0.06 \end{align*}
    The percentage increase is approximately 6%.
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  3. Looking at the formula \(\frac{\Delta V}{V} \approx 2\frac{\Delta r}{r} + \frac{\Delta h}{h}\text{,}\) A 1% error in \(r\) contributes \(2(1\%) = 2\%\) to the error in \(V\text{.}\) A 1% error in \(h\) contributes \(1(1\%) = 1\%\) to the error in \(V\text{.}\) Therefore, a 1% error in \(r\) leads to a greater error in \(V\text{.}\)
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