CALC: How do i determine the numbers btw 0 and 2pi where the line tangent to a curve is horizontal???

Question

I was sick for lecture this day and the book doesn't explain it.
Homework due tomorrow.
Please help!
To be more specific the function they give is y=cotx - 2cscx.

Thanks a ton

Also. I know the derivative is y'=-csc(x)^2 + 2cotxcscx

How do i solve that for zero? He expects us to know these things. It simplifies from -cscx^2 + 2cotxcscx to cscx/2cotx yah?

in the long run that's 2=1/cosx, correct? i just don't' know how to get the x in cosine to equal 2 X_X

Answer 1

Every time you see "tangent horizontal", it means that the derivative of this function will be 0 at this point.

You begin by finding the derivative, as you did right:

y'=-csc(x)^2 + 2cotxcscx

You now have to put it equal to 0 and solve the equation. A much easier way to do this is to rewrite the derivative as follow:

(2cos(x) -1) / sin(x)^2

to be equal to 0 the numerator must be 0. This means that cos(x) equal 1/2.

By looking at the trigonomertic circle you'll find that this is true at

pi/3 and 5pi/3

that is your answer,
Good luck !

Answer 2

If you want to do this with algebra, find the derivative of the function, and set it equal to 0. Then solve for x. Make sure to only write down the solutions for x that are between 0 and 2*Pi. Graphically, use a calculator to graph the derivative of the function. Then note all the points in your specified interval where the curve crosses the x-axis. The derivative of the function simplifies to (2cos(x) - 1)(csc^2(x)). Set this equal to 0. cos(x) = 1/2 and/or csc(x) = 0. So x = pi/3. cos(x) also = 1/2 at 5pi/3 (300 degrees), and this is still in the range. Now we have two solutions; x = pi/3 and 5pi/3. We can't get any solution from the term csc(x) = 0, since that means 1/sin(x) = 0, and there's no way to get that term to = 0. This shows that x = 0 is not in the domain of arccsc(x).

Answer 3

Set the derivative = 0 and solve. The solutions between 0 and 2 pi are your answer(s).

Answer 4

of direction, you want the derivative to be 0. The derivative is cos(x) - sin(x), so clean up cos(x) - sin(x) = 0. Rearrange to get sin(x)/cos(x) = a million (be conscious cos(x) isn't 0 on the answer). that's tan(x) = a million and is the midsection of looking those ideas. tan^-a million(a million) = x, for those x in -pi/2, pi/2). this provides x=pi/4. you may desire to renowned that tan(x) has a era of pi, so the subsequent answer is 5pi/4. The y values could be no problem. It in all probability could help at situations to renowned that Acos(x) + Bsin(x) = (A^2+B^2)^.5 cos(x-T), for a cost of T in [0, 2pi). So the amplitude is basic. to discover T, enhance cos(x-T) = cos(x)cos(-T)-sin(x)sin(-T)=cos(T)cos(x)... to get cos(T) = A/(A^2+B^2)^.5 and sin(T) = B/(A^2+B^2)^.5 (A^2+B^2)^.5. you may desire to get the two a form of to comply with get T splendid. that's the indicators of A and B that could impact which quadrant T is in. truthfully, cos^-a million(A/(A^2+B^2)^.5), that's in [0,pi], is the two T or T-pi, further, sin^-a million(B/(A^2+B^2)^.5), that's in [-pi/2, pi/2} is T, T-pi, or T-2pi. those be sure T uniquely. comparable formula Acos(x) + Bsin(x) = (A^2+B^2)^.5 sin(x - S) holds. For this problem, sinx+cosx = 2^.5 cos(x - pi/4) and this cosine hits its extremes while x-pi/4 is 0 or pi. this provides an analogous solutions.

Answer 5

A horizontal line has slope = 0, and derivative = slope, so solve the equation y'=0 (you need to find the derivative)

I see you did find it. Just make it = to zero, factor out a csc and go from there