Percentage change using differentials.

Question

We're given the above triangle with sides $a$ and $b$ , and area $A$. $a$ is increased by $4$% and $b$ is decreased by $3$% , we need to approximate the percentage change in the area using differentials.

What I tried :

The percentage change for a function $f(x,y)$ is given by :

$\nabla f \approx f_x \nabla x + f_y \nabla y$.

So we write , $A = \dfrac{1}{2}absin\theta$ , $\theta = \dfrac{\pi}{6}$ => $A = \dfrac{1}{4}ab$ , $A_a = \dfrac{b}{4}$ and $A_b = \dfrac{a}{4}$ ,

$\nabla A \approx A_a \nabla a + A_b \nabla b$

=> $\nabla A \approx \dfrac{b}{4}0.04 - \dfrac{a}{4}0.03$ ,

=> $\nabla A \approx \dfrac{4b-3a}{400}$ ,

But the solution says " About 1% change in area " , am I doing something wrong ?

Help!

Answer 1

Dropping a height onto side $b$ you get a 30-60-90 triangle, so the height is $a \sin \theta = a/2$ and the area of the triangle then must be

$$\frac{\text{base} \times \text{height}}{2} = \frac{b \times (a/2)}{2} = \frac{ab}{4}. $$

$a$ is increased by $4\%$ and $b$ - decreased by $3\%$, so replace $a$ with $1.04a$ and $b$ with $0.97b$, to get $$ A' = \frac{1.04 a \times 0.97 b}{4} = 1.04 \times 0.97 A = 1.0088 A,$$

so the net increase in the area is $0.88\%$.

Answer 2

You get 1% if you do this:

=> $\nabla A \approx \dfrac{ab}{4}0.04 - \dfrac{ab}{4}0.03$ ,

=> $\nabla A \approx \dfrac{ab}{4}(0.04 - 0.03)$ ,

=> $\nabla A \approx \dfrac{ab}{4}(0.01)$ ,

Answer 3

You have $\log A=\log a + \log b - \log 4, \frac {dA}A=\frac {da}a+\frac {db}b, \frac {dA}A=4\%-3\%=1\%$

Answer 4

If we take a and b to be functions of time, and take area as A(a(t), b(t)) we arrive at the following equations: $$ a(t) = a_0 * 1.04^t \\ b(t) = b_0 * 0.96^t \\ A(t) = \frac{1}{4}a(t)b(t) $$ If we now consider the derivatives with respect to time: $$ a'(t) = a_0(1.04^t * \ln(1.04)) \\ b'(t) = b_0(0.97^t * \ln(0.97)) \\ A'(t) = \frac{1}{4}(a'(t)b(t) + a(t)b'(t)) $$ If we take $A'(t)$ a bit further we arrive at the expression: $$ A'(t) = \frac{a_0b_01.0088^t}{4}\ln(1.0088) $$ And then integrating with respect to $t$, and using the initial condition $A(0) = \frac{a_0b_0}{4}$ we arrive at: $$ A(t) = \frac{a_0b_0}{4}1.0088^t $$ It doesn't take too much to see that this is a percentage change of about 1%, but if you do want to be slightly more rigorous, you could proceed as: $$ \text{Percentage change} = \frac{\text{difference}}{\text{original}} \\ \frac{A(t + \delta t) - A(t)}{A(t)} = \frac{1.0088^{t + \delta t}}{1.0088^t} - 1 = 1.0088^{\delta t} - 1 $$ Plugging in $\delta t = 1$ gives you $\text{Percentage change} = 0.0088 \approx 1%$