University of Technology Sydney
 

MSSC supporting maths @ UTS

Health Sciences

Health Sciences

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If you are interested in health areas why not try out one of our Assess yourself activities?

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Equity & Diversity Unit

Equity & Diversity Unit

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2012 - 2014 saw a partnership in the form of the Federally funded Pathways to Mathematics Project.

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Learning Maths

Learning Maths Learning maths is a process that takes time and perseverance.

Consider this....




Mathematics Study Support Centre

MSSC Welcome. Pathways to UTS Maths.

How do I use this site? (We highly recommend bookmarking this page as UTS is a large Australian University)


Explore: Browse the Resources section and discover mathematics and tools using technology.


Assess yourself: follow these steps

 User guide

There are three steps to use this site for your learning.

  1. We recommend that you use the "More..." link on the images carousel above to attempt a variety of tests to find out what areas need your attention.
  2. Guided exercises use Resources from the Resources/Library section and resources found on the WWW. The UTS resources are of a wide variety and are designed to help you. These are written to provide you with the requisite skills. You may also follow the recommended revision from your selected area after testing.
  3. Use the "More..." link on the images carousel again to re-test your skill in the area you have chosen. Repeat until mastery has taken hold.


Register:

    • Current Staff and Students can log-in at the MSSC LOGIN module on the left, using the usual UTS, credentials. After login new menu items with more activities appear on the left sub-menus.
    • If you are not a current student or staff-member you can use the public items.


Everyone is welcome to view the online UTS materials in the Resources repository, which is constantly being updated.


The next step is up to you!

Registration offers the following benefits:

  • you can use the Mathematics Study Support Centre (Drop-in Room), CB04.03.331. Tutors will assist you with your mathematical learning.
  • on further investigation inside this website you will find activities which are games based. Innovative technology provides you with a range of challenges to help you develop your mathematical skills and vocabulary, why not try it out?
  • there are materials which embrace traditional modes of learning. Also there are resources which use blended modes of learning, which utilise online technology to augment the teaching and learning process.



MSSC strongly supports your ambitions in the mathematical sciences.

Last Updated on Wednesday, 29 March 2017 11:05
 

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More maths support from MSSC.....

Short courses review
2014-08-06

I got into the course of my choice and the extended maths short course gave me the grounding I needed to succeed. Thank you so much!


2014-02-11

I did 3unit maths in the past and it just didn't click. The MSSC approach has made it different. Now i am familiar and confident with requisite tasks. Thank you.


Tips and Tricks
2014-09-26

Adding fractions requires deliberate steps and decisions. Notice the operator on the expression, is it addition/subtraction or multiplication/division. They require different strategies. $\frac{1}{3}+\frac{1}{2}$
Are the denominators the same? If not as in this example we need to form a new fractional sum of an appropriate 'family' by multiplying across the denominator only.
$\frac{\fbox{}}{6}+\frac{\fbox{}}{6}$
Now to fill in the blanks
Multiplying each term by the appropriate form of one!
$\frac{2}{2}\times\frac{1}{3}+\frac{3}{3}\times\frac{1}{2}=\frac{2}{6}+\frac{3}{6}$
Simplifys to:
$\frac{2+3}{6}=\frac{5}{6}$


JIntegrationTree
php tree script
Expand/Collapse all


  • 1: up and under
    • 2: \(\int{} \,d{x}=x+C\)
      • \(\int{} \,d{t}=t+C\)
      • \(\int{} \,d{v}=v+C\)
    • 2: \(\int{k} \,d{x}=kx+C\)
      • \(\int{5} \,d{x}=5x+C\)
      • \(\int{10} \,d{x}=10x+C\)
      • \(\int{22a} \,d{x}=22ax+C\)
    • 2: \(\int{{x}^{{n}}} \,d{x}=\frac{x^{n+1}}{n+1}+C,\left(n\neq-1\right)\)
      • \(\int{{x}^{{2}}} \,d{x}=\frac{x^{2+1}}{2+1}+C=\frac{x^{3}}{3}+C\)
      • \(\int{{x}^{{7}}} \,d{x}=\frac{x^{8}}{8}+C\)
  • 1: quotient - derivative over function
    • 2: \(\int{\frac{f^`\left(x\right)}{f\left(x\right)}} \,d{x}=ln\left|{f\left(x\right)}\right|+C\)
      • \(\int{\frac{1}{x}} \,d{x}=ln\left|{x}\right|+C\)
      • \(\int{\frac{2}{2x+1}} \,d{x}=ln\left|{2x+1}\right|+C\)
  • 1: u substitution
    • 2: \(\int{\sqrt{{2x+1}}} \,d{x}=\frac{{1}}{{2}}\int{\sqrt{{u}}} \,d{u}\)
    • 2: \(\int{{x}^{{3}}} \cos \left(x^4+2\right)\,d{x}\)
      • let \(u=x^4+2\)
      • \(\frac{d{u}}{d{x}}=4x^3\), now re-arrange
      • therefore\(\frac{1}{4}d{u}=x^{3}d{x}\)
      • this is the point of substitution!
      • so substitute........BIG STEP
      • \(=\frac{1}{4}\int{\cos \left({u}\right)} \,d{u}\)
      • this is the NEW easier
      • integral to be done
      • \(=\frac{{1}}{{4}}\sin \left({u}\right)+C\)
      • return to original variable, in x
      • \(=\frac{{1}}{{4}}\sin \left({x^4+2}\right)+C\)
      • complete....well done!
    • 2: \(\int{{e}^{{5x}}}\,d{x}\)
      • let \(u=5x\)
      • \(\frac{d{u}}{d{x}}=5\), now re-arrange
      • therefore\(\frac{1}{5}d{u}=d{x}\)
      • this is the point of substitution!
      • so substitute........BIG STEP
      • \(=\frac{1}{5}\int{e^{\left({u}\right)}} \,d{u}\)
      • this is the NEW easier
      • integral to be done
      • \(=\frac{{1}}{{5}}e^{\left({u}\right)}+C\)
      • return to original variable, in x
      • \(=\frac{{1}}{{5}}e^{\left({5x}\right)}+C\)
      • complete....well done!
  • 1: integration by parts
    • 2: \(\int{u} \,d{v}=uv-\int{v}\,d{u}\)