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Can someone give me a basic calculus problem then break it down and solve it?

  1. Oct 19, 2012 #1
    I will be taking calculus soon and want to know what yo expect from the course. please give as many details as possible.

    thank you
  2. jcsd
  3. Oct 19, 2012 #2


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    Hey Niaboc67.

    What kind of course are you taking? Standard Calc I? Honors Calculus? Multi-variable and Vector Calculus? High school calculus?
  4. Oct 19, 2012 #3
    I assume this would be your first calculus class and that it's not proof based. In which case here's a website that covers 3 semesters of that mostly through worked problems broken down in a clear manner. http://www.khanacademy.org/math/calculus
  5. Oct 19, 2012 #4


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    A calculus course involves so many different types of problems that we can't possibly give you a good idea what the course is about this way. There are also many different types of calculus courses. I will however give you an example. To understand it, you need to understand functions, derivatives, the product rule and the chain rule.

    Problem: Find the derivative of the function ##f:\mathbb R\to\mathbb R## defined by ##f(x)=xe^{-x^2}## for all real numbers x.

    \begin{align}\frac{d}{dx}\left(xe^{-x^2}\right) &=\left(\frac{d}{dx}x\right)e^{-x^2} +x\frac{d}{dx}\left(e^{-x^2}\right)\\ &=e^{-x^2} +xe^{-x^2}\frac{d}{dx}\left(-x^2\right)\\
    &=e^{-x^2} -2x^2e^{-x^2}\\
    Last edited: Oct 20, 2012
  6. Oct 19, 2012 #5
    Given the function f(x) = x^2
    Find the slope of the line tangent to the function at the point x=2

    First find the instantaneous rate of change at x=2, which will be the slope of f(x) at the point. The instantaneous rate of change/instantaneous slope is the derivative which is 4. The tangent line will pass through the point f(2) so one of the y values of the line will be 4.

    Then use point slope.




    so y=4x-4 is the slope of the tangent of equation f(x)=x^2 at point = 2.
  7. Oct 20, 2012 #6


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    Given a unit square of paper, you are to create a box from it with the largest volume.

    Letting the new height be "x", the sides of the box becomes (1-2x).
    (By cutting away squares at the corners with sides "x")
    Thus, the volume of the box, as a function of "x" becomes V(x)=(1-2x)^2*x.

    We will find the extrema of the volume by differentiating the volume function, and determining the values of "x" giving V'(x)=0.

    We get: V'(x)=(1-2x)^2-4x(1-2x)=(1-2x)*(1-6x).
    Thus, maximum volume occurs for x=1/6, giving maximum volume V(1/6)=2/27
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