Tutor profile: Mona M.
Expand x' and y' x' = r cos(a + b) y' = r sin(a + b) use x = r cos b and y = r sin b
x' = r cos(a + b) = r [cos a cos b - sin a sin b] after applying cosine rule for expansion, we need to expand r in each term = r cos a cos b - r sin a sin b y' = r sin(a + b) = r [sin a cos b + cos a sin b] same as x' using the sin expansion rule = r sin a cos b + r cos a sin b Using the x = r cos b and y = r sin b x' = (r cos b) cos a - (r sin b )sin a let's isolate the term that match x and y in parentheses = x cos a - y sin a replace terms by x and y appropriately y' = (r cos b) sin a + (r sin b) cos a = x sin a + y cos a note: I have some equipment that helps me draw a picture and write on a Microsoft whiteboard to each person that would like me to help remotely.
Derive the following: f(x)=6*x^(3)-9*x+4
For this case, we don't have a specific rule that applies. each term has an addition sign. so each term follows the general rule of the derivative: ( a*x^(2n) )' = a*2n*x^(2n-1) the ()' means the term inside the parentheses should be derived. apply this general formula on ( 6*x^(3) )' = 6*3*x^(3-1) = 18*x^(2) for (-9x )' = -9*1*x^(1-1) -> note: x=x^(1) every base to the power of 1 is equal the base = -9*x^0 -> note x^0 =1 every base to the power of zero is 1 =-9*1 =-9 for 4 is a constant the derivative of a constant is 0 why ? it is the same as 4*1 and 1 is same as x^0 because x^0 is 1 so (4)' =(4*x^0)' = 4*0*x^(0-1) 4*0=0 that means 4*0*x^(-1) =0 and therefore the derivative of a constant is always equal to zero final result : 18*x^2 -9 +0= 18*x^2 -9
simplify and factorize: 7a^2 +3b +6a -2a^2
answer: 7a^2 +3b +6a -2a^2= 5a^2 +6a +3b = a *( 5a+6)+3b First we group the parts that have the same variable with same exponent. In this example, we have 3 variables: a^2, b and a. Then, if it is an addition we add, id it is a subtraction we subtract, division we divide, multiplication we multiply. the variable a^2 : we have 7a^2-2a^2= 5a^2 3b 6a answer : 5a^2 +6a +3b Next, we have to factorize. We have to look up for the term that have a common variable. we doesn't care about the exponent for now. In this example we have a and b. the a terms get factorize, and the b term get factorize. a*a= a^2 so a(a) =a^2 too. in this principle we have to look at the equation calculated: 5a^2 +6a +3b 5a^2+6a= 5a(a) +6(a) : equation 6 = (a) ( 5a+6) : I factorized the (a) because it is in each term of the equation 6 for b term i have only one which is 3b. We can't factorize it should start with at least 2 terms. the finale answer is: a *( 5a+6)+3b
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