User Portlet
| Discussions |
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| Hi; Please see attached notebook detailing confusion taking the partial derivative of y' with respect to y. In my way of thinking (which may be wrong) y' and y are totally different and therefore y' should be considered a constant when taking the... |
| Hi Gianluca; I have done quite a bit of research today and believe that you are exactly right. Thanks so much for your help, Mitchell Sandlin |
| Hi; In working with StreamPlot and Vector Plot, I am getting two different representations - see attached. It appears that Vector Plot is representing expected results whereas StreamPlot seems to be having problems, which I cannot figure out.... |
| I am creating a user defined function f[t_] := where I am evaluating an integral in the body of the function without any issues. I can subsequently evaluate the user function f[t] and obtain the correct answer. However, when I plot the expression... |
| My goal is to display the inverse trigonometric function of ArcCos and Cos through a specific x range of 0 to Pi for Cos and -1 to 1 for the ArcCos since those domain ranges will pass the horizontal line test as being one-to-one and thus inverses... |
| Hi; In calculating an integral (using both the icon from the Classroom Pallett and the Integrate[] function in Mathematica) , it appears that Mathematica is placing an additional value in the results that is both unexpected and unwanted (see... |
| Hi; I am having difficulty creating a plottable user function using the Integrate[] function - please see attached. Given, I am using the Integrate[] function to create an antiderivative with one variable and assuming that the constant is 0 in my... |
| Hi; When I solve the function f[x] = 3x^(2/3)-2x at the x values of [-1,0,1], the independent value of -1 gives a complex number and I cannot understand why - see attached. When I treat the statement of x^(2/3) as the cubed root of x squared... |
| Hi; I am having difficulty displaying my two entering functions in the ContourPlot[] function (Please see attached). As my attachment shows, I can enter my two functions individually in two separate ContourPlot[] functions and combine them using... |
| Hi; In the function that I am using x is undefined at value 3 (see attached). To circumvent the problem of x being undefined, I created a piecewise function defining x at value 3 to be zero and subsequently created a table using the piecewise... |