integration

👨🏻‍🏫 學術台數學

Larry

2011-02-18 3:28 am

圖片參考：http://img706.imageshack.us/img706/5817/51216278.jpg

In the figure, the slope at any point (x,y) of the curve is given by dy/dx=2x+1. The equation of the normal to the curve at point P is x-9y+77=0.

更新1:

a)Find the coordinates of P. b)Find the equation of the curve.

更新2:

The slope of the normal why=1/9?

回答 (1)

myisland8132

2011-02-18 3:44 am

✔ 最佳答案

(a) The slope of the normal = 1/9

So, the slope of the tangent at P = 2x + 1 = -9 => x = -5

Then y = (x + 77)/9 = 8

P(-5,8)

(b) dy/dx = 2x + 1

y = x^2 + x + C

Sub. P(-5,8) => C = -12

The equation of the curve: y = x^2 + x - 12

2011-02-19 16:27:43 補充：
The slope of Ax + By + C = 0 is -A/B

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