How do you know you understand calculus?

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I'm studying concavity and the second derivative test and it is starting to become more confusing because the second derivate is the slope of the slope of the original function. Do you need to become so familiar with this that you have a "picture" of the graphs in your head? I mean, one can just plug in numbers and test away without understanding the behavior of the graphs right? :rolleyes:
 
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i think is easy visualizing it, some of my friends prefers numbers and formulas, (and sometimes i simply have to plug in then),

sometimes it not useful, but i prefer understanding that solving
 
Yes you can plug in numbers but it is better to understand the process that occurs and the significance of that value of the 2nd derivative. Thinking of the derivative of the slope is extremely limiting when it comes to understanding, as you have just mentioned. It is better to think of the derivative as the rate of change of the original function, which for the first derivative is easy to think of as slope, but for the second and so on derivatives, becomes harder.
 
the slope of the slope
:smile:
so you have a collection of slopes; each is slightly different, if it is a "smooth" curve the change from one to another is infinitestimal and so the collection of slopes really is a "curve" itself (for these collection of slopes sketches out the change of the slopes inasmuch as the original curve sketches out each slope) and so it can have a slope of its own and so on.

things will get increasingly difficult to visualise as to go to more variables and higher derivatives, but that's when the power of mathematics comes in... it allows us to explore things that we can't visualise
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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