MHB Calculating Area from Differentials

[ardentmed]

Hey guys,

I need some more help for this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

This is only for question 2. Ignore 1.
Question:

Alright, so from drawing a diagram, we know that width is "L" and length is "3L." Moreover, the area of the circle uses 1/2 * L for the radius.

Thus,
A = 3L^2 + $\pi$.5L^2 / 2

Thus, using differentials, we know that:
f(a+$\Delta$x) = f(a) + $\Delta$y
And
$\Delta$y = f'(x) $\Delta$x

From substituting $\Delta$y into the other equation, we get:

f(8.05) = 200.99734.
Thus,
A(8.05) = 200.99734 cm^2

For the percent error, it should be $/delta$A / A * 100 = 1.37%.

I'm not too sure about my answer though, particularly the area I calculated via differentials. What do you guys think? Thanks in advance.

 

[lfdahl]

Hi, ardentmed

Thankyou so much for sharing your problem with us at the MHB:)

First, the way you use the differentials is in principle OK, but I cannot
from the explicit area function see if you have stated it right. Parentheses are needed:

\[A(L) = 3L^2+\frac{1}{2}\pi \left ( \frac{L}{2}\right )^2=3L^2+\frac{1}{8}\pi L^2= \left ( 3+\frac{\pi }{8} \right )L^2\]

Do you agree so far?

 

[ardentmed]

Hi, ardentmed

Thankyou so much for sharing your problem with us at the MHB:)

First, the way you use the differentials is in principle OK, but I cannot
from the explicit area function see if you have stated it right. Parentheses are needed:

\[A(L) = 3L^2+\frac{1}{2}\pi \left ( \frac{L}{2}\right )^2=3L^2+\frac{1}{8}\pi L^2= \left ( 3+\frac{\pi }{8} \right )L^2\]

Do you agree so far?

Thank you for being so helpful.

Yes, that seems correct. Then how would I proceed with this question?

 

[lfdahl]

Next step would be to use the formula you have derived by means of differentials:

$\frac{\Delta A}{A(L)}\approx\frac{A'(L) \cdot \Delta L}{A(L)}\cdot 100 \%$

So you need to determine the derivative of $A$.

 

[ardentmed]

Next step would be to use the formula you have derived by means of differentials:

$\frac{\Delta A}{A(L)}\approx\frac{A'(L) \cdot \Delta L}{A(L)}\cdot 100 \%$

So you need to determine the derivative of $A$.

Alright, so I used the following formula:

$\frac{\Delta A}{A(L)}\approx\frac{A'(L) \cdot \Delta L}{A(L)}\cdot 100 \%$

And substituted 8 for L, and used A(L)= 3L^2 + (1/8)$\pi$ L^2

and

A'(L)= 6L + (1/4)$\pi$L

And ultimately computed the following:

$\frac{\Delta A}{A(L)}\approx\frac{A'(L) \cdot \Delta L}{A(L)}\cdot 100 \%$
=0.694489677 * 100

~ 6.94%.

Is that correct?

Thanks again for the help.

 

[lfdahl]

Alright, so I used the following formula:

$\frac{\Delta A}{A(L)}\approx\frac{A'(L) \cdot \Delta L}{A(L)}\cdot 100 \%$

And substituted 8 for L, and used A(L)= 3L^2 + (1/8)$\pi$ L^2

and

A'(L)= 6L + (1/4)$\pi$L

And ultimately computed the following:

$\frac{\Delta A}{A(L)}\approx\frac{A'(L) \cdot \Delta L}{A(L)}\cdot 100 \%$
=0.694489677 * 100

~ 6.94%.

Is that correct?

Thanks again for the help.

Hi, again.

Your procedure is perfectly right. Good job! But I think you´ve made a computational error. I get: $1.25 \%$

The computation can be made quite simple, if you let $k = 3+ \frac{\pi}{8}$.

Then: $A(L) = kL^2$ and $A'(L) = 2kL$

and $\frac{A'(L)\Delta L}{A(L)}=\frac{2kL \Delta L}{kL^2}=\frac{2 \Delta L}{L}=\frac{2\cdot0.05}{8}= 0.0125=1.25 \%$