Here is a link to a group of questions and answers to volumes, obtained by rotating graphs around a given line.
http://hotmath.com/help/bookindexes2/minicalcgt/problems_6_1_1.html
Below is a set of practices for derivatives and Tanget lines
http://hotmath.com/help/bookindexes2/minicalcgt/problems_3_1_1.html
Math-Talk-Blog is created for Booker T. Washington Mathematics students, in order to ensure success knowledge and appreciation for all topics in Mathematics.
Monday, April 30, 2007
- Also, go here for thje list aof all the general Calculus formulas you need to know.
- http://www.calculus-help.com/funstuff/cheatsheet.html
Friday, April 13, 2007
Calculus Mock exam
Townview Center
Friday April 13th 4:30 - 9:00pm
Saturday April 14th 9:00 am - 1:30pm
BE THERE OR BE SQUARE...
Part A: 28 mult. choice questions in 55 min. no Calculator
Part B: 17 mult. choice questions in 50 min. w/ Calculator
Free Respond questions
Part A: 3 questions in 45 min. w/ Calculator
Part B: 3 questions in 45 min. no Calculator
Sunday, February 25, 2007
Sunday, February 11, 2007
Algebra II
Undersatnding Rational functions and their Graphs
http://www.analyzemath.com/Graphing/GraphRationalFunction.html
Saturday, February 10, 2007
Calculus
Saturday Calculus FUN!
2-24-07 Town View HS / 9 am - 2:30 pm
Check out the new link for Fundamental Theorem of Calculus.
Click here http://www.math.hmc.edu/calculus/tutorials/fundamental_thm/
The graphs of first derivative and second derivative of a given function.
Click here http://archives.math.utk.edu/visual.calculus/3/graphing.1/1.html
for self quiz.
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How do you find the derivitive of a funtion that has more than one unknown variable?
Find dy/dx using implicit differentiation.
Start by using the Product Rule on the two terms on the left.
We take the derivative of both sides to obtain
(x2) ′y + x2y′ + x′y2 + x(y2) ′ = (3x) ′ .
Then, we simplify to
2xy + x2y′ + y2 + x2yy′ = 3
2xy + x2y′ + y2 + 2xyy′ = 3.
Solve for y′ by placing all y′ terms on one side.
Rearranging so that all terms containing y′ are on one side, we have
x2y′ + 2xyy′ = 3 – 2xy – y2
y′ (x2 + 2xy) = 3 – 2xy – y2
Solve for y′ .
Solve for y′ .
You can factor out x from the bottom if you like, but it doesn't really simplify the answer.