Calculus 3 Homework 1

Question

梯形之下底、上底、高各為100，40，30。計算梯形面積的相對誤差。若測量這些量時之相對誤差控制在 1% 內，求面積相對誤差之最大可能。

- (Method 1) The formula or the area $A(a,b,h)=(a+b)h/2$. By the question, we know $\frac{\Delta a}{a}, \frac{\Delta b}{b}, and \frac{\Delta h}{h}< 1\\%.$ In particular, we know the error of $a+b$ is at most $2\%$, i.e. $\frac{\Delta a+\Delta b}{a+b}<2\\%$ We want to find a bound for $\frac{\Delta A}{A}$, and $A+\Delta A = A(a+\Delta a, b+\Delta b, h+\Delta h) = (a+b+\Delta a+\Delta b)(h+\Delta h)/2.$ After ignoring all nonlinear term of $\Delta a$, $\Delta b$, and $\Delta h$, we arrive $A+\Delta A = \frac{(a+b)h+(\Delta a+\Delta b)h + (a+b)\Delta h}{2},$ which implies $\left|\frac{\Delta A}{A}\right| = \left |\frac{(\Delta a +\Delta b) h + (a+b)\Delta h}{(a+b)h}\right| = \left|\frac{\Delta a+\Delta b}{a+b}+\frac{\Delta h}{h}\right|\leq 2\\% + 1\\%$
- (Method 2) We can find the linear approximation $L(x,y,z)$ of $A(x,y,z)$ at $(a,b,h)$ which is $L(a,b,h) = A(a,b,h)+\frac{h}{2}(x-a)+\frac{h}{2}(y-b)+\frac{(a+b)}{2}(z-h),$ so $\Delta A = L(a+\Delta a,b+\Delta b,h+\Delta h)-A(a,b,h) = \frac{h\Delta a}{2}+\frac{h\Delta b}{2} +\frac{(a+b)\Delta h}{2}.$ Then, the following are the same as Method 1.
We need $\frac{\partial z}{\partial x} = \frac{y^3-x^2y}{(x^2+y^2)^2}\ \text{and}\ \frac{\partial z}{\partial y} = \frac{x^3-xy^2}{(x^2+y^2)^2}.$ Thus, $z = \frac{1}{2} + 0 \cdot (x-1) + 0\cdot (y-1)$ is the tangent plain.