
posted by bobpursley
Respond to this Question
Similar Questions

math
Eliminate the parameter (What does that mean?) and write a rectangular equation for (could it be [t^2 + 3][2t]?) x= t^2 + 3 y = 2t Without a calculator (how can I do that?), determine the exact value of each expression. cos(Sin^1 
calculus
Find complete length of curve r=a sin^3(theta/3). I have gone thus (theta written as t) r^2= a^2 sin^6 t/3 and (dr/dt)^2=a^2 sin^4(t/3)cos^2(t/3) s=Int Sqrt[a^2 sin^6 t/3+a^2 sin^4(t/3)cos^2(t/3)]dt =a Int 
Calculus
Consider the function f(x)=sin(1/x) Find a sequence of xvalues that approach 0 such that (1) sin (1/x)=0 {Hint: Use the fact that sin(pi) = sin(2pi)=sin(3pi)=...=sin(npi)=0} (2) sin (1/x)=1 {Hint: Use the fact that sin(npi)/2)=1 
Calculus
Consider the function f(x)=sin(1/x) Find a sequence of xvalues that approach 0 such that (1) sin (1/x)=0 {Hint: Use the fact that sin(pi) = sin(2pi)=sin(3pi)=...=sin(npi)=0} (2) sin (1/x)=1 {Hint: Use the fact that sin(npi)/2)=1 
TRIG!
Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1  (3/4)sin^2 2x work on one side only! Responses Trig please help!  Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + 
Calculus
Please look at my work below: Solve the initialvalue problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/ Sqrt(4^24(1)(6)))/2(1) r=(16 +/ Sqrt(8)) r=8 +/ Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, 
Calculus
Consider the function f(x)=sin(1/x) Find a sequence of xvalues that approach 0 such that sin (1/x)=0 sin (1/x)=1 sin (1/x)=1 Is sin sin (1/x)=0 and sin (1/x)=1 does not exist. What is sin (1/x)=1 then. 
Calculus
Hello, Could somebody kindly check my answer for the following question? Find the derivative of the following function: h(x)=3e^(sin(x+2)) h'(x)=3'(e^(sin(x+2))+3(e^(sin(x+2))' h'(x)=0(e^(sin(x+2))+3(e^(sin(x+2))(cos(1)) 
tigonometry
expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by b and using that cos(b)= cos(b) sin(b)= sin(b) gives: sin(ab) = sin(a)cos(b)  cos(a)sin(b) 
calculus
Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x)