Can someone help me on question 9 & 10. One Is just fine. Thank you
1 answer:
Answer:
9. y = x - 1/2 + π/4
10. y = (-2/3)(√6)x + (2/3)√3 + π/3
Step-by-step explanation:
9. From a table of derivatives,
... arctan'(x) = 1/(1+x²)
Then, using the chain rule, ...
... f'(x) = arctan'(2x) = 2/(1 +(2x)²) = 2/(1 +4x²)
For x=1/2, the slope of the tangent is ...
... f'(1/2) = 2/(1 +4·(1/2)²) = 1
Then in point-slope form, the equation of the tangent line is
... y = 1(x -1/2) +π/4
... y = x -1/2 +π/4
10. From a table of derivatives,
... arccos'(x) = -1/√(1 -x²)
Then, using the chain rule, ...
... arccos'(x²) = -2x/√(1 -x⁴)
For x=1/√2, the slope of the tangent is ...
... f'(1/√2) = -(2/√2)/√(1 -(1/√2)⁴) = (-√2)/√(3/4) = -2√(2/3)
... f'(1/√2) = (-2/3)√6
Then in point-slope form, the equation of the tangent line is
... y = (-2/3)(√6)(x -1/√2) +π/3
... y = (-2/3)(√6)x +(2/3)√3 +π/3
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Answer:
Option B.
Step-by-step explanation:
The given curve is
We need to find the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve .
Let the vertex in quadrant I be (x,y), then the vertex in quadrant II is (-x,y) .
Length of the rectangle = 2x
Width of the rectangle = y
Area of a rectangle is
Substitute the value of y from the given equation.
.... (1)
Differentiate with respect to x.
Equate , to find the critical points.
Divide both sides by 6.
The value of x can not be negative because side length can not be negative.
Substitute x=3 in equation (1).
The area of the largest rectangle is 108 square units.
Therefore, the correct option is B.