iiomqjessica

07.12.2021 •

Mathematics

Matthhew's school has 32 classrooms.Each classroom has 16 desks

Solved

Show answers

Ответ:

werty775581

07.09.2020

Mathematics

$\displaystyle \frac{d}{dt}[c^2=a^2+b^2-2ab\cos(\theta)]\\\\\\\Rightarrow2c\frac{dc}{dt}=\frac{da}{dt}(2a-2b\cos(\theta))+\frac{db}{dt}(2b-2a\cos(\theta))+2ab(\sin(\theta)\frac{d\theta}{dt})$

Step-by-step explanation:

The Law of Cosines is:

$c^2=a^2+b^2-2ab\cos(\theta)$

And we want to differentiate this with respect to time t.

So, let's take the derivative of both sides with respect to t:

$\displaystyle\frac{d}{dt}[c^2]=\frac{d}{dt}[a^2+b^2-2ab\cos(\theta)]$

First, let's expand our differentiation:

$\displaystyle \frac{d}{dt}[c^2]=\frac{d}{dt}[a^2]+\frac{d}{dt}[b^2]-\frac{d}{dt}[2ab\cos(\theta)]$

Implicitly differentiate. Let's ignore the third term for now aside from moving the two outside. Thus:

$\displaystyle 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}-2\frac{d}{dt}[ab\cos(\theta)]$

For the third term, this will require the product rule. We have:

$\displaystyle \frac{d}{dt}[ab\cos(\theta)]$

The product rule states:

$\displaystyle(uv)^\prime=u^\prime v+uv^\prime$

Therefore, we will let:

$u=ab\text{ and } v=\cos(\theta)$

Then, we will take their derivatives with respect to t. Hence:

$\displaystyle u'=\frac{d}{dt}[ab]\text{ and } v'=\frac{d}{dt}[\cos(\theta)]$

The left requires another application of the product rule. Let's ignore it for now. Implicitly differentiate the right:

$\displaystyle v^\prime=-\sin(\theta)\frac{d\theta}{dt}$

On the left, we have:

$\displaystyle u^\prime=\frac{d}{dt}[ab]$

Again, let's let: (The subscript is used to differentiate between variables)

$\displaystyle u_2=a \text{ and } v_2=b$

Differentiate with respect to t:

$\displaystyle u_2^\prime=\frac{d}{dt}[a]\text{ and }v_2^\prime=\frac{d}{dt}[b]$

Implicitly differentiate:

$\displaystyle u_2^\prime=\frac{da}{dt}\text{ and } v_2^\prime=\frac{db}{dt}$

Hence, it follows that the above is:

$\displaystyle u{^\prime}=\frac{d}{dt}[ab]=u_2^\prime v_2+u_2v_2^\prime=\frac{da}{dt}b+a\frac{db}{dt}$

Therefore:

$\displaystyle \frac{d}{dt}[ab\cos(\theta)]=u^\prime v+uv^\prime\\\\ =(\frac{da}{dt}b+a\frac{db}{dt})\cos(\theta)-(ab)(\sin(\theta)\frac{d\theta}{dt})$

Hence:

$\displaystyle 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}-2\frac{d}{dt}[ab\cos(\theta)]$

By substitution:

$\displaystyle 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}-2\Big[(\frac{da}{dt}b+a\frac{db}{dt})\cos(\theta)-(ab)(\sin(\theta)\frac{d\theta}{dt})\Big]$

Distributive property:

$\displaystyle 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}-2\frac{da}{dt}b\cos(\theta)-2a\frac{db}{dt}\cos(\theta)+2ab(\sin(\theta)\frac{d\theta}{dt})$

Rearranging gives:

$\displaystyle 2c\frac{dc}{dt}=\big[2a\frac{da}{dt}-2\frac{da}{dt}b\cos(\theta)\big]+\big[2b\frac{db}{dt}-2a\frac{db}{dt}\cos(\theta)\big]+2ab(\sin(\theta)\frac{d\theta}{dt})$

Factor out the derivative from the first and second brackets:

$\displaystyle 2c\frac{dc}{dt}=\frac{da}{dt}(2a-2b\cos(\theta))+\frac{db}{dt}(2b-2a\cos(\theta))+2ab(\sin(\theta)\frac{d\theta}{dt})$

This cannot be simplified further.

Hence, the derivative of the Law of Cosines with respect to time t is:

$\displaystyle 2c\frac{dc}{dt}=\frac{da}{dt}(2a-2b\cos(\theta))+\frac{db}{dt}(2b-2a\cos(\theta))+2ab(\sin(\theta)\frac{d\theta}{dt})$

0,0(0 оценок)

Ask your question to the AI advisor

Ai-bot is an expert in any field and is the perfect companion for reliable and useful answers and advice on a variety of topics, including science, history, technology, art, sports, health, culture and more.

Ask your question

More tips

Answers on questions: Mathematics

Ask an AI advisor a question

I want advice