Approximation used in physics explanation

[tellmesomething]

Homework Statement

Find percent change in KE if speed increases by 1%

Relevant Equations

KE= 0.5 m v^2

I dont know if this question is more fit for the physics forums but regardless i have a doubt in a suggested approach to these questions.

So this is really easy right, you can do this simply by taking out the change in KE which comes out to be 0.0201 K (K being the initial KE)
and then find the percentage which comes out to be 2.01%

Now while this question is easy on the calculation, a suggested approach i saw for questions where the change is very small is:
K=0.5 mv^2
Taking natural log on both sides and then differentiating.
Maybe its because of my poor math skills but i cannot figure out why this works,
we get a close enough answer of 2% by this method

Again i know this sum's calculation can easly be done by hand but for changes small than 1 say 0.001, it must be difficult.
Can someone explain why this works? Why do we take log and then differentiate?

 

[PeroK]

What's your alternative calculation precisely.

 

[tellmesomething]

What's your alternative calculation precisely.

We start with K=0.5mv² (initial KE)
Taking natural log on both sides
ln K=ln (0.5) + ln (m) + 2 ln(v)
Differentiating
dk/k = 0 + dm/m + 2 dv/v

For percentage change

dk/k × 100 = (dm/m +2dv/v) ×100
% change in KE = 0 + 2(1%)
=2%

 

[gleem]

You could find the differential of E first. You get dE = mvdv. Note that it contains m explicitly. If we multiply and divide the right side ( love this trick) by v/2 we get dE =2mv²/2v =2Edv/v which yields dE/E =2dv/v which is what you get by taking the logarithm of the original equation and then finding the differential and it hides the mass. It also directly relates the fractional change of E due to v.

 

[tellmesomething]

You could find the differential of E first. You get dE = mvdv. Note that it contains m explicitly. If we multiply and divide the right side ( love this trick) by v/2 we get dE =2mv²/2v =2Edv/v which yields dE/E =2dv/v which is what you get by taking the logarithm of the original equation and then finding the differential and it hides the mass. It also directly relates the fractional change of E due to v.

I see...that does make some sense by writing it down in statements..i ll think on it Thankyou

 

[tellmesomething]

What's your alternative calculation precisely.

is there something wrong with the approach?

 

[PeroK]

is there something wrong with the approach?

Your approach is unsound. You appear to have differentiated each term with respect to a different variable. There was no reason to take the log or differentiate. If we increase ##v## by ##\Delta v##, then ##K## increases by ##\Delta K##, where:
$$K = \frac 1 2 mv^2$$$$ \implies K + \Delta K = \frac 1 2 m(v + \Delta v)^2 = \frac 1 2 m \big (v^2 + 2v\Delta V + (\Delta v)^2 \big )$$$$= K + \frac 1 2 m(2v\Delta v) + \frac 1 2 m (\Delta v)^2$$If ##\Delta v## is small, we can drop the last term to get the approximation:$$\Delta K \approx \frac 1 2 m(2v\Delta v) = mv\Delta v$$And, finally:$$\frac{\Delta K}{K} \approx \frac{mv\Delta v}{\frac 1 2 mv^2} = 2\frac{\Delta v}{v}$$Note that this holds in general: if ##A = kB^2## for some constant ##k## and ##\Delta B## is a small change in ##B##, then the change in ##A## satisfies:
$$\frac{\Delta A}A = 2\frac{\Delta B}{B}$$

 

[tellmesomething]

Your approach is unsound. You appear to have differentiated each term with respect to a different variable. There was no reason to take the log or differentiate. If we increase ##v## by ##\Delta v##, then ##K## increases by ##\Delta K##, where:
$$K = \frac 1 2 mv^2$$$$ \implies K + \Delta K = \frac 1 2 m(v + \Delta v)^2 = \frac 1 2 m \big (v^2 + 2v\Delta V + (\Delta v)^2 \big )$$$$= K + \frac 1 2 m(2v\Delta v) + \frac 1 2 m (\Delta v)^2$$If ##\Delta v## is small, we can drop the last term to get the approximation:$$\Delta K \approx \frac 1 2 m(2v\Delta v) = mv\Delta v$$And, finally:$$\frac{\Delta K}{K} \approx \frac{mv\Delta v}{\frac 1 2 mv^2} = 2\frac{\Delta v}{v}$$Note that this holds in general: if ##A = kB^2## for some constant ##k## and ##\Delta B## is a small change in ##B##, then the change in ##A## satisfies:
$$\frac{\Delta A}A = 2\frac{\Delta B}{B}$$

I see okay.... thanks for the clarifications :)

 

[Steve4Physics]

Hi @tellmesomething. There are various ways to explain. Here’s a fairly basic approach. I'm adding it out of nostalgia, as it's based on what I was taught (before doing any calculus) many##^2## years ago. It’s a bit long though.

1. Note that ##(1+x)^2=1+2x+x^2##.

2. If ##x## is much smaller than ##1##, then ##x^2## is very much smaller than 1 and can be ignored.

3. Look at an example. Suppose ##x = 0.01##. Then
##(1+x)^2 = (1+0.01)^2 = 1^2 + 2*0.01 + 0.01^2 = 1 + 0.02 + 0.0001 = 1.0201##.
The ##x^2## term ##(0.0001)## is so small that to a good approximation we can ignore it and write ##(1+0.01)^2 \approx 1 + 0.02 \approx 1.02##.

4. In general, when ##x## is much smaller than 1, we can write ##(1+x)^2 \approx 1 + 2x##.

5. Increasing something by p% is the same as multiplying it by a factor ##1+\frac p{100}##.

Decreasing something by p% is the same as multiplying it by a factor ##1-\frac p{100}##.

E.g. increasing ##v## by 1% is the same as multiplying it by a factor ##1+\frac 1{100} = 1+0.01 = 1.01##.

6. If we increase ##v## by 1% then its value is multiplied by a factor ##(1+0.01)##, That means ##v^2## increases by a factor ##(1+0.01)^2##. We now know this is approximately 1.02 (no need to find the exact value 1.0201).

So ##v^2## has increased by a factor of approximately 1.02, which is the same as saying it has increased by 2% (approx.).

7. We know ##KE = \frac 12 mv^2##. Since ##\frac 12 m## is a fixed constant, ##KE## has increased by same factor as ##v^2## - which is 1.02. I.e. KE has increased by 2% (approx.).

This means a 1% increase in ##v## produces (to a good approximation) a 2% increase in KE.

8. Another example: ##v## decreases by 0.07%. What happens to KE?

Answer: KE decreases by 2x0.07 = 0.14%.

9.. This works for other powers too. For example take ##y = ax^5##. Note the power ‘5’. If ##x## increases by 0.3%, ##y## increases by 5x0.3 = 1.5% approx. I.e. ##y## is multiplied by a factor ##1+5*\frac {0.3}{100} =1. 015## approx.