How to find the Tafel slope on the Tafel plot?
370091
"a" and "b" are characteristic constants of the electrode system. A plot of electrode potential versus the logarithm of current density is called the "Tafel plot" and the resulting straight line the "Tafel line." "b" is the "Tafel slope" that provides information about the mechanism of the reaction, and "a" provides information about the rate constant of the reaction. \nRigorously, the equation should be written using the absolute value of the current density and the ± sign for anodic and cathodic overpotentials, respectively. Also, the equation holds only for relatively high overpotentials, approximately 0.1 volt or higher. \nFor example:a plot of  h (greek alphabet not available,its "eta") vs. log of anodic current with a=0.5 (a is the cathodic electron transfer coefficient). The linear plot is of the Tafel equation. The slope b is 2.3RT/anF. The intercept at h=0 gives the exchange current density i0. This is known as a Tafel plot. With this we can calculate a and i0.
751258
y' = 2x + 4\ny'(1) = 6 this is the slope.\ntangent: yt = 6x + b\n\ny(1) = 1 + 4 + 10 = 15\nyt(1) = y(1) = 15 = 6 + b ===> b = 9\n\nyt = 6x + 9
766685
You need to differentiate the equation y=2x^2. \n\ndy/dx = 4x therfore the gradient at x=3 = 4*x = 12
926663
the slope of the graph. the formula is rise/run. take 2 points X1,Y1 and X2,Y2  plug them in the formula X2-X1/Y2-Y1. youll get the rate of change! Best of Luck!
872809
If the equation is lineear, or a line like it sounds like, the equation is y=mx+b. m is the slope or ten. and b is the y-int. or -130, and you are done
354280
Take a look at this webpage:\n\nhttp://www.walterzorn.com/grapher/grapher_e.htm\n\nType: \nx min = 0\nx max = 20\ny min = 0\ny max = 10000\n\nFunction f(x):\n1000*exp(0.1x)\n\nThen click plot graph.\n\nVoila!
307198
It is linear on intervals, on (- infinite, 0) and on (0, +infinite) but not on the whole R. \n\nTry plotting it, you get a reversed V, not a straight line.
450157
slope = (y2-y1)/(x2-x1)\n5= (k-4)/(2-3)\n5= (k-4)/(-1)\nk-4= (-1)*5\nk-4= -5\nk= 4-5\nso k= -1.\ni hope you understand they way I worked it out. If you have any more questions, pls feel free to ask me on: pinkieflava@yahoo.co.uk
877049
use the slope formula.\n\ny sub 2 minus y sub 1 divided by x sub 2 minus y sub 1\n\nthen you get 1-5 divided by 0-2\n\nthen you get -4 divided by -2\n\nthe answer is 2
850129
Firstly, determine the slope or gradient of the lines;\n\nLine 1:\n\nm1 = (y2-y1)/(x2-x1)\n     = (3 - -7)/(1 - -4)\n     = 10/5\n     = 2\n\nLine 2:\n\nm2 = (y2-y1)/(x2-x1)\n    = (10 - 6)/(4 - 2)\n    = 4/2\n    = 2\n\nBoth gradients are the same, therefore the lines are parallel. if the product of the gradients (m1m2) was -1 then they would have been perpendicular.\n\nWe can take this further by checking to see if they are the same line.\n\nThe equation y - y1 = m(x - x1) will be the same for both lines if this is the case.\n\nLine 1:\n\ny - 3 = 2(x - 2)\ny = 2x + 1\n\nLine 2:\n\ny - 10 = 2(x - 4)\ny = 2x + 2\n\nThese are not equal, however we now know that line 2 sits 1 unit higher on a graph than line 1, and both lines are parallel to each other.
1154399
First you need to find the slope of the line, by finding two points on the line and making a fraction found by subtracting the two y-coordinates and putting this over what you get by subtracting the two x-coordinates.  Then if you can, find the y-intercept.  The equation is y = mx + b where m = slope and b = y-intercept.\n\nSuppose your trend line contains the point (4, 20) and hits the y-axis at (0,8)\n\nThe slope is (20-8)/(4-0) = 12/4 = 3\n\nThe y-intercept is 8\n\nSo the equation is y = 3x + 8
729746
you solved it right, since the line BC never goes up/down(the y-coordinate never changed, graph it and you'll see) the slope is 0. and 0/7=0, so you're right.\nit would be impossible if you got 7/0, because you can't divide by 0.
885688
slope m=(y1-y2)/(x1-x2)\n          m=(1--1)/(3-3)\n          m=2/0    which gives us a singularity,\nthat is, a vertical line.
176854
You can graph it but probably that is not the answer your teacher is looking for. If you wanna do this by the good old method:\n\nf'(x)=3^x*ln(3)\n\nYou are looking for a point where the slope equals to 1 meaning 1 equals to your derivative.\n\n3^x*ln(3)=1\n\nNow use the calculator: x=-0.0856
712712
[on] [y=] [(] [x,t,o,n] [+] [2] [)] [/] [x,t,o,n] [enter] [5] [-] [2] [x,t,o,n] [enter] [(-)] [2] [x,t,o,n] [-] [3] [enter] [window] [(-)] [5] [enter] [5] [enter] [1] [enter] [(-)] [5] [enter] [5] [enter] [graph]\n\nYou should have on screen a vertical asymptote at x=0, horizontal asymptote at y=1, a curve going to negative infinity towards 0 on the left, a curve going to positive infinity towards 0 on the right, on the left tangent line y=-2x-3 touches point(-1,-1), and on the right tangent line y=5-2x touches point(1,3).
764932
slope = f'(0),\nf'(x)= e^x 1/(x^2 +3) 2x + e^x ln( x^2 +e)\nf'(0)= e^0 1/3 (2(0)) + e^0 ln(e)\n= 1\n\nnow y=f(0)=e^0 ln( e) =1\nso the eq. of the tangent line is:\n\ny-1= x\ny=x+1\ns
851755
the slope of the line can be figured out with a simple formula.  If the question gives you two points all you have to do is plug into the formula delta y over delta x.  this means take the two y coordinates and subtract them from each other and divide that by the same thingy with the x coordinates. \nFor example. ( 0, 0) and ( 1, 2) What is the slope between those two points?\njust do (2 - 0 ) over ( 1 - 0 ) = 2/1 = slope of 2\n\nThe other method is to draw a line between the two points and count the boxes (on graph paper)  you first count to the left then up. so it would be starting from (0,0), you count 2 boxes to the left, then 1 up.  that would be 2/1 which is slope of 2.