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tangent sqrt(9-t)

解答

切线

9 - t

解答

y = - \frac{1}{2 9 - a _{0}} t + 9 - a_{0} + \frac{a _{0}}{2 9 - a _{0}}

求解步骤

计算在一般点

t = a_{0}

的切线

找到切点:

(a_{0}, 9 - a_{0})

计算

y = 9 - t

的斜率:

\frac{d y}{d t} = - \frac{1}{2 9 - t}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{1}{2 9 - a _{0}}

找到斜率 m=

- \frac{1}{2 9 - a _{0}}

且穿过

(a_{0}, 9 - a_{0})

的直线:

y = - \frac{1}{2 9 - a _{0}} t + 9 - a_{0} + \frac{a _{0}}{2 9 - a _{0}}

y = - \frac{1}{2 9 - a _{0}} t + 9 - a_{0} + \frac{a _{0}}{2 9 - a _{0}}

作图

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