Differentials & Error Estimation - Calculus 3

1. The function $z = x^2y - xy^2$ is evaluated near the point $(3,2)$. If the maximum errors in $x$ and $y$ are both $0.05$, what is the maximum possible error in $z$ using linear approximation?
   Answer: $|dz| \leq 0.55$ $\frac{\partial z}{\partial x} = 2xy - y^2$ and $\frac{\partial z}{\partial y} = x^2 - 2xy$. At $(3,2)$: $\frac{\partial z}{\partial x} = 2(3)(2) - 2^2 = 12 - 4 = 8$ and $\frac{\partial z}{\partial y} = 3^2 - 2(3)(2) = 9 - 12 = -3$. The maximum error is $|dz| \leq |\frac{\partial z}{\partial x}||dx| + |\frac{\partial z}{\partial y}||dy| = 8(0.05) + |-3|(0.05) = 0.4 + 0.15 = 0.55$.
2. A cylindrical tank has radius $r = 5.0$ m and height $h = 12.0$ m. Due to thermal expansion, the radius increases by $0.02$ m and the height increases by $0.05$ m. Using differentials, what is the approximate change in the volume of the tank?
   Answer: $\Delta V \approx 11.47$ m³ For a cylinder, $V = \pi r^2 h$. The differential is $dV = \frac{\partial V}{\partial r}dr + \frac{\partial V}{\partial h}dh = 2\pi rh \, dr + \pi r^2 \, dh$. With $r = 5.0$, $h = 12.0$, $dr = 0.02$, and $dh = 0.05$: $dV = 2\pi(5.0)(12.0)(0.02) + \pi(5.0)^2(0.05) = 2.4\pi + 1.25\pi = 3.65\pi \approx 11.47$ m³.
3. The electric field magnitude at distance $r$ from a point charge is $E = \frac{kq}{r^2}$. If $k = 9.0 \times 10^9$ N⋅m²/C², $q = 2.0 \times 10^{-6}$ C, and $r = 0.50$ m, and $r$ has an uncertainty of $\pm 0.01$ m, what is the approximate relative error in $E$?
   Answer: $\frac{|dE|}{E} \leq 0.04$ When you encounter error propagation problems in calculus, you're applying differential approximation to estimate how uncertainties in measurements affect calculated quantities. To find the relative error in $E = \frac{kq}{r^2}$, you need to determine how changes in $r$ affect $E$. Start by taking the differential: $\frac{dE}{dr} = \frac{d}{dr}\left(\frac{kq}{r^2}\right) = kq \cdot (-2r^{-3}) = -\frac{2kq}{r^3}$ The absolute error is $|dE| = \left|\frac{dE}{dr}\right| \cdot |dr| = \frac{2kq}{r^3} \cdot |dr|$ For relative error, divide by the original function: $\frac{|dE|}{E} = \frac{\frac{2kq}{r^3} \cdot |dr|}{\frac{kq}{r^2}} = \frac{2|dr|}{r}$ Substituting the values $r = 0.50$ m and $|dr| = 0.01$ m: $\frac{|dE|}{E} = \frac{2(0.01)}{0.50} = \frac{0.02}{0.50} = 0.04$ Therefore, answer A ($\frac{|dE|}{E} \leq 0.04$) is correct. Answer B gives 0.02, which would result from forgetting the factor of 2 when differentiating $r^{-2}$. Answer C gives 0.06, which might come from incorrectly using $\frac{3|dr|}{r}$ instead of $\frac{2|dr|}{r}$. Answer D gives 0.08, which could result from doubling the correct answer or other calculation errors. Study tip: For relative error in functions with power laws like $r^n$, the relative error is always $|n| \times \frac{|dr|}{r}$. This pattern saves time on exams.
4. The period $T$ of a simple pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$, where $L$ is length and $g$ is gravitational acceleration. If $L = 1.00$ m with uncertainty $\pm 0.02$ m and $g = 9.80$ m/s² with uncertainty $\pm 0.05$ m/s², what is the approximate relative error $\frac{|dT|}{T}$ in the period?
   Answer: $\frac{|dT|}{T} \leq 0.013$ When you encounter error propagation problems in calculus, you're applying partial derivatives to estimate how uncertainties in input variables affect the final result. For functions of multiple variables, the relative error formula is your key tool. Given $T = 2\pi\sqrt{\frac{L}{g}}$, you need to find how uncertainties in $L$ and $g$ propagate to $T$. The relative error formula states: $\frac{|dT|}{T} \leq \left|\frac{\partial T}{\partial L} \cdot \frac{dL}{T}\right| + \left|\frac{\partial T}{\partial g} \cdot \frac{dg}{T}\right|$ First, find the partial derivatives. Since $T = 2\pi L^{1/2} g^{-1/2}$: - $\frac{\partial T}{\partial L} = \pi L^{-1/2} g^{-1/2} = \frac{T}{2L}$ - $\frac{\partial T}{\partial g} = -\pi L^{1/2} g^{-3/2} = -\frac{T}{2g}$ Substituting into the relative error formula: $\frac{|dT|}{T} \leq \frac{1}{2}\left|\frac{dL}{L}\right| + \frac{1}{2}\left|\frac{dg}{g}\right|$ Calculate the relative uncertainties: - $\frac{|dL|}{L} = \frac{0.02}{1.00} = 0.02$ - $\frac{|dg|}{g} = \frac{0.05}{9.80} \approx 0.0051$ Therefore: $\frac{|dT|}{T} \leq \frac{1}{2}(0.02 + 0.0051) = \frac{1}{2}(0.0251) \approx 0.013$ Answer D is correct. Answer A (0.008) underestimates by ignoring one variable's contribution. Answer B (0.023) incorrectly adds the full relative errors without the $\frac{1}{2}$ factor. Answer C (0.018) represents a computational error in the partial derivative coefficients. Remember: for power-law functions, the coefficient in front of each relative uncertainty equals the exponent of that variable in the function.
5. The resistance $R$ of a wire is given by $R = \frac{\rho L}{A}$, where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area. If $\rho = 1.7 \times 10^{-8}$ Ω⋅m, $L = 100$ m, and $A = 2.0 \times 10^{-6}$ m², and measurements have uncertainties $d\rho = \pm 0.1 \times 10^{-8}$, $dL = \pm 0.5$, and $dA = \pm 0.1 \times 10^{-6}$, what is the maximum possible error in $R$?
   Answer: $|dR| \leq 0.047$ Ω $dR = \frac{\partial R}{\partial \rho}d\rho + \frac{\partial R}{\partial L}dL + \frac{\partial R}{\partial A}dA = \frac{L}{A}d\rho + \frac{\rho}{A}dL - \frac{\rho L}{A^2}dA$. Substituting values: $\frac{L}{A} = \frac{100}{2.0 \times 10^{-6}} = 5 \times 10^7$, $\frac{\rho}{A} = \frac{1.7 \times 10^{-8}}{2.0 \times 10^{-6}} = 8.5 \times 10^{-3}$, $\frac{\rho L}{A^2} = \frac{1.7 \times 10^{-8} \times 100}{(2.0 \times 10^{-6})^2} = 4.25 \times 10^{2}$. Maximum error: $|dR| \leq (5 \times 10^7)(0.1 \times 10^{-8}) + (8.5 \times 10^{-3})(0.5) + (4.25 \times 10^{2})(0.1 \times 10^{-6}) = 0.005 + 0.0425 + 0.0000425 \approx 0.047$ Ω.
6. The temperature at point $(x,y)$ in a metal plate is given by $T(x,y) = x^2 + 3xy - y^2$. A thermometer is moved from point $(2,1)$ to point $(2.1, 0.9)$. Using differentials, what is the approximate change in temperature?
   Answer: $\Delta T \approx 0.3$ $\frac{\partial T}{\partial x} = 2x + 3y$ and $\frac{\partial T}{\partial y} = 3x - 2y$. At $(2,1)$: $\frac{\partial T}{\partial x} = 2(2) + 3(1) = 7$ and $\frac{\partial T}{\partial y} = 3(2) - 2(1) = 4$. The changes are $dx = 0.1$ and $dy = -0.1$. Therefore, $dT = 7(0.1) + 4(-0.1) = 0.7 - 0.4 = 0.3$.
7. The function $f(x,y) = xe^{xy}$ is evaluated at $(1,0)$. If $x$ has uncertainty $\pm 0.05$ and $y$ has uncertainty $\pm 0.03$, what is the maximum absolute error in $f$ using linear approximation?
   Answer: $|df| \leq 0.08$ $\frac{\partial f}{\partial x} = e^{xy} + xye^{xy} = e^{xy}(1 + xy)$ and $\frac{\partial f}{\partial y} = x^2e^{xy}$. At $(1,0)$: $\frac{\partial f}{\partial x} = e^0(1 + 0) = 1$ and $\frac{\partial f}{\partial y} = 1^2 e^0 = 1$. The maximum error is $|df| \leq |\frac{\partial f}{\partial x}||dx| + |\frac{\partial f}{\partial y}||dy| = 1(0.05) + 1(0.03) = 0.08$.
8. The volume of an ellipsoid is $V = \frac{4}{3}\pi abc$, where $a$, $b$, and $c$ are the semi-axes. If $a = 3.0 \pm 0.1$, $b = 2.0 \pm 0.15$, and $c = 1.5 \pm 0.05$, which contributes most to the uncertainty in volume?
   Answer: The uncertainty in $b$ contributes most to $|dV|$ When you encounter uncertainty propagation problems, you need to use differential calculus to determine how small changes in each variable affect the overall result. This is a classic application of partial derivatives and the differential. For the ellipsoid volume $V = \frac{4}{3}\pi abc$, the total differential is: $dV = \frac{\partial V}{\partial a}da + \frac{\partial V}{\partial b}db + \frac{\partial V}{\partial c}dc$ Computing the partial derivatives: - $\frac{\partial V}{\partial a} = \frac{4}{3}\pi bc$ - $\frac{\partial V}{\partial b} = \frac{4}{3}\pi ac$ - $\frac{\partial V}{\partial c} = \frac{4}{3}\pi ab$ Now substitute the given values and uncertainties: - Contribution from $a$: $\frac{4}{3}\pi(2.0)(1.5)(0.1) = 1.257$ - Contribution from $b$: $\frac{4}{3}\pi(3.0)(1.5)(0.15) = 2.827$ - Contribution from $c$: $\frac{4}{3}\pi(3.0)(2.0)(0.05) = 1.257$ Choice A is correct because the uncertainty in $b$ produces the largest contribution to $|dV|$. Choice B is wrong because $a$'s contribution (1.257) is smaller than $b$'s contribution. Choice C is incorrect since $c$'s contribution equals $a$'s but is still less than $b$'s. Choice D is false because the contributions are clearly unequal. Strategy tip: In uncertainty problems, don't just look at the absolute uncertainties (±0.1, ±0.15, ±0.05). You must multiply each uncertainty by its corresponding partial derivative. The variable with the largest product of (partial derivative × uncertainty) dominates the total uncertainty.

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