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Hello,
r=5(1+cos t)
r'=5(-sin t)
r²+r'²= 25[(1+cos t)²+(-sin t)²]=50(1-cos t)=50 sin² (t/2)
Between 0 and π, sin x>0 ==>|sin x|=sin x
![l= 2*5* \int\limits^{\pi}_0{sin( \frac{t}{2} )} \, dt= 5[-cos (t/2)]_0^{\pi}\\\\ =5(0+1)=5](https://tex.z-dn.net/?f=l%3D%202%2A5%2A%20%5Cint%5Climits%5E%7B%5Cpi%7D_0%7Bsin%28%20%5Cfrac%7Bt%7D%7B2%7D%20%29%7D%20%5C%2C%20dt%3D%205%5B-cos%20%28t%2F2%29%5D_0%5E%7B%5Cpi%7D%5C%5C%5C%5C%0A%3D5%280%2B1%29%3D5)
Here is the method but i may have make some mistakes.
r=5(1+cos t)
r'=5(-sin t)
r²+r'²= 25[(1+cos t)²+(-sin t)²]=50(1-cos t)=50 sin² (t/2)
Between 0 and π, sin x>0 ==>|sin x|=sin x
Here is the method but i may have make some mistakes.
Answer:
(a) Decreasing on (0, 1) and increasing on (1, ∞)
(b) Local minimum at (1, 0)
(c) No inflection point; concave up on (0, ∞)
Step-by-step explanation:
ƒ(x) = x² - x – lnx
(a) Intervals in which ƒ(x) is increasing and decreasing.
Step 1. Find the zeros of the first derivative of the function
ƒ'(x) = 2x – 1 - 1/x = 0
2x² - x -1 = 0
( x - 1) (2x + 1) = 0
x = 1 or x = -½
We reject the negative root, because the argument of lnx cannot be negative.
There is one zero at (1, 0). This is your critical point.
Step 2. Apply the first derivative test.
Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.
(1) x = ½
ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1
ƒ'(x) < 0 so the function is decreasing on (0, 1).
(2) x = 2
ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½ = ⁵/₂
ƒ'(x) > 0 so the function is increasing on (1, ∞).
(b) Local extremum
ƒ(x) is decreasing when x < 1 and increasing when x >1.
Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.
(c) Inflection point
(1) Set the second derivative equal to zero
ƒ''(x) = 2 + 2/x² = 0
x² + 2 = 0
x² = -2
There is no inflection point.
(2). Concavity
Apply the second derivative test on either side of the extremum.
The function is concave up on (0, ∞).
