Finding the equation of the normal to the curve?
回答 (5)
2021-04-18 7:13 am
y=4cos^-1(x/2)
=>
cos(y/4)=x/2
=>
-sin(y/4)(y'/4)=1/2
=>
y'=-2csc(y/4)
At x=sqr(3),
y(sqr(3))=4cos^-1(sqr(3)/2)=4(pi/6)=2pi/3
y'(sqr(3))=-2csc(pi/6)=-4
=>
the slope of the normal line=1/4. Thus
the equation of the normal line is
y-2pi/3=(x-sqr(3))/4
=>
3x-12y+(8pi-3sqr(3))=0
=>
cos(y/4)=x/2
=>
-sin(y/4)(y'/4)=1/2
=>
y'=-2csc(y/4)
At x=sqr(3),
y(sqr(3))=4cos^-1(sqr(3)/2)=4(pi/6)=2pi/3
y'(sqr(3))=-2csc(pi/6)=-4
=>
the slope of the normal line=1/4. Thus
the equation of the normal line is
y-2pi/3=(x-sqr(3))/4
=>
3x-12y+(8pi-3sqr(3))=0
2021-04-17 6:55 pm
y = 4arcos(x/2)
x/2 = cos(y/4)
dx/dy = -(1/2)sin(y/4)
sin(y/4) = √[1 – cos^2(y/4)]
dx/dy = -(1/2)√[1 – x^2/4] = -(1/4)√[4 – x^2]
dy/dx = -4/√(4 – x^2)
When x = √(3), y = 4arcos[√(3)/2] = 4*π/6
Also, dy/dx = -4 = m the slope of the tangent there
Slope of the normal, -1/m = 1/4
If y = mx + c is the normal
2π/3 = (1/4)√(3) + c
y = x/4 + 2π/3 - √(3)/4
Bryce found this result first
x/2 = cos(y/4)
dx/dy = -(1/2)sin(y/4)
sin(y/4) = √[1 – cos^2(y/4)]
dx/dy = -(1/2)√[1 – x^2/4] = -(1/4)√[4 – x^2]
dy/dx = -4/√(4 – x^2)
When x = √(3), y = 4arcos[√(3)/2] = 4*π/6
Also, dy/dx = -4 = m the slope of the tangent there
Slope of the normal, -1/m = 1/4
If y = mx + c is the normal
2π/3 = (1/4)√(3) + c
y = x/4 + 2π/3 - √(3)/4
Bryce found this result first
2021-04-17 4:08 pm
y'= -4/√(4 - x²)
At x= √3, y'= -4
-1/m= 1/4
y=(1/4)(x - √3) + 2π/3
At x= √3, y'= -4
-1/m= 1/4
y=(1/4)(x - √3) + 2π/3
2021-04-17 11:20 am
First, find the slope of the curve at that point. Can you do that part?
2021-04-17 1:27 pm
No, thank you - not today, sweetie. I do not help people appear smarter than they actually are.
收錄日期: 2021-04-24 08:52:24
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