What is the tangent of y= 1/(7t^3) ?

The tangent of y= 1/(7t^3) is y=-3/(7a_{0)^4}t+4/(7a_{0)^3}

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tangent of y= 1/(7t^3)

tangent of

y = \frac{1}{7 t ^{3}}

y = - \frac{3}{7 a _{0}^{4}} t + \frac{4}{7 a _{0}^{3}}

Solution steps

Compute the tangent line to the general point

t = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{7 a _{0}^{3}})

Find the slope of

y = \frac{1}{7 t ^{3}} : \frac{d y}{d t} = - \frac{3}{7 t ^{4}}

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{3}{7 a _{0}^{4}}

Find the line with slope m=

- \frac{3}{7 a _{0}^{4}}

and passing through

(a_{0}, \frac{1}{7 a _{0}^{3}}) : y = - \frac{3}{7 a _{0}^{4}} t + \frac{4}{7 a _{0}^{3}}

y = - \frac{3}{7 a _{0}^{4}} t + \frac{4}{7 a _{0}^{3}}

What is the tangent of y= 1/(7t^3) ?
The tangent of y= 1/(7t^3) is y=-3/(7a_{0)^4}t+4/(7a_{0)^3}