Zol Tun Kul
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 Nov 9 accepted Applying L'Hopital on $\lim_{x\to1^+}\left( \frac{1}{\ln(x)} - \frac{1}{x - 1} \right)$ Nov 9 comment Applying L'Hopital on $\lim_{x\to1^+}\left( \frac{1}{\ln(x)} - \frac{1}{x - 1} \right)$ So $0^+ - 0^+$ is indeed $0$? And $0^+ \cdot 0^+$ is also $0$? I always thought that $0^+$ meant a number greater than $0$. Nov 9 comment Applying L'Hopital on $\lim_{x\to1^+}\left( \frac{1}{\ln(x)} - \frac{1}{x - 1} \right)$ Well, it's good to know I can do this, but I suspect I'm not supposed to know / use this yet. Do you know if $0^+ - 0^+ = 0$? Or if $0^+ \cdot 0^+ = 0$? Nov 9 asked Applying L'Hopital on $\lim_{x\to1^+}\left( \frac{1}{\ln(x)} - \frac{1}{x - 1} \right)$ Nov 9 accepted Related rates: using angle to find rate of change of opposite side Nov 9 comment Getting the rate of change of a rectangle's base if its top-right corner is defined as $y = 2^x$ You... you should've said that instead! Nov 9 accepted Getting the rate of change of a rectangle's base if its top-right corner is defined as $y = 2^x$ Nov 9 comment Getting the rate of change of a rectangle's base if its top-right corner is defined as $y = 2^x$ Hmm... So $$y = e^{x \cdot \ln(2)}$$ If I want to know the rate of change of $x$ I would need... the derivative, I guess. Which is $$\frac{dy}{dt} = e^{x \cdot \ln(2)}\cdot \ln(2) \cdot \frac{dx}{dt}$$ Since $\frac{dy}{dt} = 1$ I can have that $$1 = e^{x\cdot \ln(2)}\cdot \ln(2) \cdot \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{\frac{1}{\ln(2)}}{e^{x\cdot \ln(2)}}$$ Then replace $x = 2$ and we get that $$\frac{dx}{dt} = 0.36$$This seems to lead me to the correct answer according to the book. Thank you. Nov 9 asked Getting the rate of change of a rectangle's base if its top-right corner is defined as $y = 2^x$ Nov 9 comment Related rates: using angle to find rate of change of opposite side Ah, I see my problem. Thanks. However the answer is $$\frac{13\pi}{60}$$ Nov 9 asked Related rates: using angle to find rate of change of opposite side Nov 9 comment Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ Indeed, that's exactly it. I had interpreted $$\frac{x\cdot y}{2} = \frac{x}{2}\cdot\frac{y}{2}$$ when it should've been $$\frac{x\cdot y}{2} = \frac{x}{2} \cdot y$$ Thank you. Nov 9 comment Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ Hm, I just confirmed that I have an extra $\frac{1}{2}$ there. But I don't really get it because the derivative of $\frac{x}{2}$ is definitely $\frac{1}{2}\cdot \frac{dx}{dt}$ and then multiplied by the second part of the product, which is $\frac{y}{2}$. Nov 9 accepted Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ Nov 9 comment Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ Hmm... But if I have $\frac{x}{2}$, whose derivative is $\frac{1}{2} \cdot \frac{dx}{dt}$, by the product rule I have to multiply this by the $\frac{y}{2}$ (it's the other side of the product) which results in $$\frac{1}{2}\cdot\frac{dx}{dt}\cdot\frac{y}{2}$$. Nov 9 comment Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ What's the problem with my attempt at the product rule though? Nov 9 comment Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ I don't quite see why are you multiplying $\frac{dx}{dt}$ by $\frac{dA}{dx}$. Why $\frac{dA}{dx}$? Nov 9 asked Related rates: derivative of the function $A = \frac{x\cdot y}{2}$ Nov 9 accepted Help understand related rates problem: calculating the derivative of the distance function Nov 9 comment Help understand related rates problem: calculating the derivative of the distance function So let's see if I get this straight: $z$, $x$, $y$ are all functions of $t$ (and they are actually $z(t)$, $x(t)$ and $y(t)$), so $$[z(t)]^2$$ would have as derivative $$2\cdot z(t)$$ and by the chain rule, I have to multiply this by the derivative of $z(t)$ (which is $\frac{dz}{dt}$) and therefore the full derivative of $[z(t)]^2$ is $$2\cdot z(t) \cdot \frac{dz}{dt}$$?