How to derive? (1 Viewer)

specificagent1 · Sep 30, 2021

21 b, how do you derive it?

Screen Shot 2021-09-30 at 3.47.24 pm.png

Lith_30 · Sep 30, 2021

You use the identity $a=e^{\ln(a)}$

$\\T=25+70(1.5)^{-0.4t}\\T=25+70e^{\ln(1.5^{-0.4t})}\\T=25+70e^{-0.4t\ln(1.5)}$

Then you differentiate from there.

jimmysmith560 · Sep 30, 2021

Going by Lith_30's method, this would give:

yashbb · Sep 30, 2021

Lith_30 said:
You use the identity $a=e^{\ln(a)}$

$\\T=25+70(1.5)^{-0.4t}\\T=25+70e^{\ln(1.5^{-0.4t})}\\T=25+70e^{-0.4t\ln(1.5)}$

Then you differentiate from there.

do you have to do it in terms of e or can it be done simply through use of ln

Lith_30 · Sep 30, 2021

yashbb said:
do you have to do it in terms of e or can it be done simply through use of ln

I'm assuming you mean to take the ln of both sides and then differentiate.

$\\T=25+70(1.5)^{-0.4t}\\\ln(T)=\ln(25+70(1.5)^{-0.4t})\\\\\frac{dT}{dt}(\ln(T))=\frac{dT}{dt}(\ln(25+70(1.5)^{-0.4t}))$

On the LHS I'm not even sure how to derive $\ln(T)$ , and on the RHS you are going to have to derive the equation anyway. So I believe that the the best method would just be to do what I did in the first post.

specificagent1 · Sep 30, 2021

Lith_30 said:
I'm assuming you mean to take the ln of both sides and then differentiate.

$\\T=25+70(1.5)^{-0.4t}\\\ln(T)=\ln(25+70(1.5)^{-0.4t})\\\\\frac{dT}{dt}(\ln(T))=\frac{dT}{dt}(\ln(25+70(1.5)^{-0.4t}))$

On the LHS I'm not even sure how to derive $\ln(T)$ , and on the RHS you are going to have to derive the equation anyway. So I believe that the the best method would just be to do what I did in the first post.

derive it as 1/T x DT/Dt

Lith_30 · Sep 30, 2021

specificagent1 said:
derive it as 1/T x DT/Dt

$\\\frac{\frac{dT}{dt}}{T}=\frac{\frac{d}{dt}(25+70(1.5)^{-0.4t})}{25+70(1.5)^{-0.4t}}\\\\\\\frac{\frac{dT}{dt}}{25+70(1.5)^{-0.4t}}=\frac{\frac{d}{dt}(25+70(1.5)^{-0.4t})}{25+70(1.5)^{-0.4t}}\\\\\\\frac{dT}{dt}=\frac{d}{dt}(25+70(1.5)^{-0.4t})$

We kind of end up going back to where we started, better to just differentiate it using e.

yashbb · Sep 30, 2021

Lith_30 said:
$\\\frac{\frac{dT}{dt}}{T}=\frac{\frac{d}{dt}(25+70(1.5)^{-0.4t})}{25+70(1.5)^{-0.4t}}\\\\\\\frac{\frac{dT}{dt}}{25+70(1.5)^{-0.4t}}=\frac{\frac{d}{dt}(25+70(1.5)^{-0.4t})}{25+70(1.5)^{-0.4t}}\\\\\\\frac{dT}{dt}=\frac{d}{dt}(25+70(1.5)^{-0.4t})$

We kind of end up going back to where we started, better to just differentiate it using e.

ah i see thanks man.

CM_Tutor · Sep 30, 2021

A pair of useful results, that you can use unless asked to prove them, that apply so long as $a > 0$ :

$\cfrac{d}{dx} a^x = a^x\ln{a}$

$\int a^x\ dx = \cfrac{a^x}{\ln{a}} + C,\ \text{for some constant $C$}$

$\begin{align*} \text{Applied here, we have} \qquad T &= 25 + 70(1.5)^{-0.4t} \\ \cfrac{dT}{dt} &= 0 + 70(1.5)^{-0.4t}\ln{1.5} \times -0.4 \\ &= -28\ln{1.5} \times (1.5)^{-0.4t} \\ \\ \text{At $t = 4$ min:} \qquad \cfrac{dT}{dt} &= -28\ln{1.5} \times (1.5)^{-0.4 \times 4} \\ &= -28\ln{1.5} \times (1.5)^{-1.6} \\ &= -\cfrac{28\ln{1.5}}{(1.5)^{1.6}} \\ &= -5.934...\ ^\circ \text{C min}^{-1} \end{align*}$

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How to derive? (1 Viewer)

specificagent1

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Lith_30

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jimmysmith560

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yashbb

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Lith_30

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specificagent1

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Lith_30

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yashbb

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CM_Tutor

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