Hello there! I have this questuion for my exam. I'm pretty much unknowing in maths. Could maybe someone explain to me how you get from the second last equation from the top to he last one... I really have no Idea wh the result is supposed to be ~1/x... How would one calculate this? Help!
If this is typical for this course, you are likely going to be in for a real struggle. Did you check the prerequisites for the course?
Do you understand function notation? It makes things a little less messy.
\(\displaystyle f(x) = -\ \dfrac{x^2}{4Dt}\ and\ t_{max} = \dfrac{x^2}{2D} \implies\)
\(\displaystyle f(t_{max}) = -\ \dfrac{\dfrac{x^2}{4D}}{\dfrac{x^2}{2D}} = -\ \dfrac{x^2}{4D} * \dfrac{2D}{x^2} = -\ \dfrac{1}{2}.\)
Are you ok so far?
\(\displaystyle g(x,\ t) = 2 \sqrt{\pi D t} \implies g(x,\ t_{max}) = 2 \sqrt{ \pi D * \dfrac{x^2}{2D}} = 2 \sqrt{ x^2 * \dfrac{\pi}{2}} = 2x \sqrt{\dfrac{\pi}{2}}.\)
Still with me?
So you have \(\displaystyle c(x,\ t) = \dfrac{M^*}{g(x,\ t)} * e^{f(t)}.\) Any problem so far?
\(\displaystyle c(x,\ t_{max}) = \dfrac{M^*}{g(x,\ t_{max})} * e^{f(t_{max})} = \dfrac{M^*}{2\sqrt{\dfrac{\pi e}{2}}} * \dfrac{1}{x}.\)
But \(\displaystyle \dfrac{M^*}{2\sqrt{\dfrac{\pi e}{2}}}\) involves only constants so
\(\displaystyle K = \dfrac{M^*}{2\sqrt{\dfrac{\pi e}{2}}} \implies c(x,\ t_{max}) = K * \dfrac{1}{x} \implies c(x,\ t_{max}) \text{ ~ } \dfrac{1}{x}.\)
