Derivative of a log function

[Adi]

Hey guys I'm in a lot of strife with this question. I've solved part a and b but am really stuck on c because I don't really understand logarithms well or how to differentiate them. This is what I have so far but I think I might be tackling it the wrong way. Can someone please tell me if there is a better way of if my way is ok how to go about the next step because I have no idea. Thanks heaps

 

[HallsofIvy]

What happened to all of the responses you had just a few minutes ago?

If E is a function of p and r and p and r themselves are functions of t the "chain rule" says that
$\displaystyle \frac{dE}{dt}= \frac{\partial E}{\partial p}\frac{dp}{dt}+ \frac{\partial E}{\partial r}\frac{dr}{dt}$.

In this problem $\displaystyle E(p, r)= Ap^{-a}r^b$ so $\displaystyle \frac{\partial E}{\partial p}= -aAp^{-a-1}r^b$ and $\displaystyle \frac{\partial E}{\partial r}= bAp^{-a}t=r^{b-1}$.

Further $\displaystyle p(t)= p_0(1.06)^t$. Since this problem is about "logarithmic differentiation" we can write this as $\displaystyle ln(p_0)+ tln(1.06)$ and then, differentiating both sides, $\displaystyle \frac{1}{p}\frac{dp}{dt}= ln(1.06)$ so that $\displaystyle \frac{dp}{dt}= ln(1.06)p= p_0ln(1.06)(1.06)^t$.

Similarly $\displaystyle r(t)= r_0(1.08)^t$ so that $\displaystyle ln(r)= ln(r_0)+ t ln(1.08)$, $\displaystyle \frac{1}{r}\frac{dr}{dt}= ln(1.08)$, and $\displaystyle \frac{dr}{dt}= ln(1.08)r= r_0ln(1.08)(1.08)^t$.

I'll leave it to you to put those together in

If E is a function of p and r and p and r them selves are functions of t the "chain rule" says that
$\displaystyle \frac{dE}{dt}= \frac{\partial E}{\partial p}\frac{dp}{dt}+ \frac{\partial E}{\partial r}\frac{dr}{dt}$.
$\displaystyle \frac{dE}{dt}= \frac{\partial E}{\partial p}\frac{dp}{dt}+ \frac{\partial E}{\partial r}\frac{dr}{dt}$.

In this problem $\displaystyle E(p, r)= Ap^{-a}r^b$ so $\displaystyle \frac{\partial E}{\partial p}= -aAp^{-a-1}r^b$ an $\displaystyle \frac{\partial E}{\partial r}= bAp^{-a}t=r^{b-1}$.

Further $\displaystyle p(t)= p_0(1.06)^t$. Since this problem is about "logarithmic differentiation" we can write this as $\displaystyle ln(p_0)+ tln(1.06)$ and then, differentiating both sides, $\displaystyle \frac{1}{p}\frac{dp}{dt}= ln(1.06)$ so that $\displaystyle \frac{dp}{dt}= ln(1.06)p= p_0ln(1.06)(1.06)^t$.

Similarly $\displaystyle r(t)= r_0(1.08)^t$ so that $\displaystyle ln(r)= ln(r_0)+ t ln(1.08)$, $\displaystyle \frac{1}{r}\frac{dr}{dt}= ln(1.08)$, and $\displaystyle \frac{dr}{dt}= ln(1.08)r= r_0ln(1.08)(1.08)^t$.

 

[JeffM]

I told you that trying to work from multiple attachments is very difficult. I shall not look at any future posts from you that do not permit viewing all relevant information in a single window.

 

[MarkFL]

I told you that trying to work from multiple attachments is very difficult. I shall not look at any future posts from you that do not permit viewing all relevant information in a single window.

I've edited the OP to attach the images as full images inline. I don't know why so many people choose to attach as thumbnails, and I do a lot of editing to change that (because like you I find the thumbnailed attachments difficult to use), but I missed this one.