Delta - process, word problem

[Kamhogo]

Hello. I have trouble working a calculus problem and I would appreciate if someone could help please. Here's the problem:

"A holograph of concentric circles is formed. The radius r of each circle varies directly as the square root of the wavelength λ of the light used. If r= 3.72 cm for λ= 592 nm, find the expression for the instantaneous rate of change or r with respect to λ.

The attempt at a solution

A) r = k * sqrt( λ) ==> 3.72 = k * sqrt ( 0.0000592) ==> k = 483.48 ==> r = 483.48 * sqrt ( λ )

B) Delta - process
(r + delta r) ^2 = 483.48 * (sqrt ( λ + delta λ))^2

r^2 + 2*r*delta r + ( delta r)^2 = 483.48 λ + 483.48 delta λ

r^2 +2*r*delta r - r^2 = 483.48 λ + 483.48 delta λ - 483.48 (sqrt λ)^2

(2*r + delta r)* (delta r /delta λ ) = (483.48* delta λ )/ delta λ

delta r/ delta λ = 483.48 / (2*r + delta r)

As delta λ approaches 0, so does delta r

Limit (delta r/delta λ) as delta λ approaches 0 = 483.48/2*r
= 483.48/2*483.48*sqrt λ
= 1/sqrt λ ==> answer.

The answer given by the textbook is 24.2/sqrt λ. I've been working on this problem for a week and I can't see what I'm doing wrong. Please someone help. Thanks!

 

[Ishuda]

Hello. I have trouble working a calculus problem and I would appreciate if someone could help please. Here's the problem:

"A holograph of concentric circles is formed. The radius r of each circle varies directly as the square root of the wavelength λ of the light used. If r= 3.72 cm for λ= 592 nm, find the expression for the instantaneous rate of change or r with respect to λ.

...
The answer given by the textbook is 24.2/sqrt λ. I've been working on this problem for a week and I can't see what I'm doing wrong. Please someone help. Thanks!

You have a couple of mistakes: First, as you do, start with
r = k \(\displaystyle \sqrt{\lambda}\)
and go through the delta process:
\(\displaystyle \frac{\Delta r}{\Delta \lambda}\, =\, \frac{(r+\Delta r)\, -\, r}{\Delta \lambda}\, =\, \frac{k\, [\, \sqrt{\lambda+\Delta\lambda}\, -\, \sqrt{\lambda}\, ]}{\Delta \lambda}\)
Now multiply numerator and denominator by \(\displaystyle \sqrt{\lambda+\Delta\lambda}\, +\, \sqrt{\lambda}\)
\(\displaystyle \frac{\Delta r}{\Delta \lambda}\, =\, \frac{k\, [\, \sqrt{\lambda+\Delta\lambda}\, -\, \sqrt{\lambda}\, ]}{\Delta \lambda}\, \frac{\sqrt{\lambda+\Delta\lambda}\, +\, \sqrt{\lambda}}{\sqrt{\lambda+\Delta\lambda}\, +\, \sqrt{\lambda}}\)
\(\displaystyle =\, \frac{k}{\sqrt{\lambda+\Delta\lambda}\, +\, \sqrt{\lambda}}\)\(\displaystyle \, \frac{[\, \sqrt{\lambda+\Delta\lambda}\, -\, \sqrt{\lambda}\, ]\, [\, \sqrt{\lambda+\Delta\lambda}\, +\, \sqrt{\lambda}\, ]}{\Delta \lambda}\)
and continue.

For determining k, your answer is correct for units of k of cm^1/2. However, the units for the textbook solution appears to be m^1/2 so there would be a factor of 10 difference between your answer and the textbooks answer.

 

[Kamhogo]

Resolved!

Thank you so very much! Now I can finally move on to another problem!