Recent content by Nemo1

Hmm, your explanation is great! I am yet to confidently say I get it tho! My teacher has given me some sample problems from Schaum's Outlines to go through which start easy and get harder. I am hoping to get through them before the exam and also some practice at $related rates$ word problems...

Thanks Mark for the extra explanation. What happens to the $"+c"$ after the anti-derivative? Is it a case of $+c$ on the first integral and $-c$ on the other? Cheers Nemo

Hi Mark, So after applying the FTOC: \d{}{x}\int_{a}^{x} f(t) \,dt = f(x)h(t)=-k\cdot t+c Which set to zero: h(0)=-k\cdot 0+c We get: c=0Which makes sense as when the height is zero the surface area is also zero. Does my explanation make sense? Cheers Nemo

Hi Mark, Ok, so I have been working on this one for a while now, and think I have solved part a from your help. I feel like it is simple algebraic manipulation but wish I fully understood the concepts. So if we have: V=\int_{0}^{h}A(x) \,dx Then setup the derivative with respect to $t$...

For Part C: Using the formula for volume around an axis: \pi\int_{a}^{b}r^2 \,dx I can set up my integral as: \pi\int_{0}^{1}(\frac{e^{x}+e^{-x}}{2})^2 \,dx Simplifying and taking the indefinite integral (Take too long to type in my working out): \pi \int \frac{1}{4}(e^{x}+e^{-x})^2 \,dx =...

Hi Mark, to find the length, I think its the following, Using the formula: Length=\int_{a}^{b}\sqrt{1+f'(x)^2} \,dx Knowing that the derivative of \frac{e^{x}+e^{-x}}{2} = \frac{1}{2}(e^{-x}+e^{x}) We can plug this into get: Length=\int_{a}^{b}\sqrt{1+(\frac{1}{2}(e^{-x}+e^{x}))^2} \,dx...

I am still confused even with your explanation, sorry. How do you differentiate a function or apply the chain rule when it is only $A(x)$. Totally lost. Cheers Nemo

\frac{e^{2x}+2+e^{-2x}}{4} Can become: \frac{e^{2x}+2(e^xe^{-x})+e^{-2x}}{4} Which is in the format of: (a+b)^2=A^2+2ab+b^2 Which then gives us: \frac{(e^{x}+e^{-x})^2}{4} Which is finally equal to: (\frac{e^{x}+e^{-x}}{2})^2 which is the function we are trying to get. Wow! I would not...

Slight light bulb moment: 1+\frac{(e^{x})^2+(-e^{-x})^2-2}{4} Can become: \frac{4}{4}+\frac{(e^{x})^2+(-e^{-x})^2-2}{4} Then can become: \frac{4+(e^{x})^2+(-e^{-x})^2-2}{4} Then: \frac{(e^{x})^2+(-e^{-x})^2+2}{4} Then: \frac{(e^{x})(e^{x})+(-e^{-x})(-e^{-x})+2}{4} Then...

Hi Mark, I am completely lost now, When I read: \d{V}{t}=\frac{d}{dt}\int_0^h A(x)\,dx=A(h)\d{h}{t} I am thinking that the derivative of the volume with respect to time is equal to the derivative of time from $0$ to $h$ of $A(x)dx$ If $A(x)dx$ is the area of the surface, the volume...

So I have been working on this and must admit I am overwhelmed! I have found that we are missing a $1+$ to the front of our formula to be equal to the original 1+({\frac{e^{x}-e^{-x}}{2}})^2 By using the exponent rule a^b\cdot a^c=a^{b+c} I can simplify the centre to...

Hi Mark, When I work it out I get: V=\int_0^h A(x)\,dx = \frac{1}{2}Ah^2 By using the Power rule and the F.T.O.C, this seems very abstract for me as I am struggling to understand what the dx truly means in this context. At a basic level, I know that dx is the same as \Delta x and visually I...

Would it be? \frac{1}{4}(e^x)^2+2(e^x)(-e^{-x})+(-e^{-x})^2

Hi Community, I have the following problem and I am completely stuck. I really struggle to get my head around how to break down these questions into chunks that I can then apply the math to. From what I can see so far, I have a to be able calculate the surface area at any height to get the...

Hi Community, I have the following problem and I would like some help in understanding part a. So far I far I have been able to show that: 1+\frac{(e^x-e^{-x})^2}{4} = \frac{(e^x)^2-2(e^x-e^{-x})+(e^{-x})2}{4}+1 But I am unsure of how to proceed. Also any pointers on how to look at the...