微积分.A curve is such that dy/dx = 16/x^3,and (1,4)is a point on th_其他解答

微积分.
A curve is such that dy/dx = 16/x^3,and (1,4)is a point on the curve.
(i) Find the equation of the curve.
(ii)A line with gradient -1/2 is a normal to the curve.Find the equation of this curve,giving your answer in the form ax+by=c
(iii)Find the area of the region enclosed by the curve,the x-axis and the lines x=1 and x=2

人气：264 ℃　时间：2020-07-10 06:26:20

解答

这题得用积分.
这个函数的微分是16/x^3=16x -3.the intergral of the derivative of the original function should be the original function which is the equation of the curve.Thus ∫16/x^3 dx =-8/x^2 +c then we take (1,4) into the equation and solve for c.the result is c=12.So the equation of the curve is -8/x^2 +12.
the line is normal to the curve means its sloap is reciprocal to the sloap of the curve.i am sure you can deal with the rest of it.
∫from 1 to 2 of 16/x^3 dx =-6
the area is thus 6