What is the tangent of y3-xy2+x3=5 ?

The tangent of y3-xy2+x3=5 is y= 1/((3-2a_{0))^2}x+(6a_{0}^2-20a_{0}+15)/((-2a_{0)+3)^2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y3-xy2+x3=5

tangent of

y 3 - x y 2 + x 3 = 5

y = \frac{1}{( 3 - 2 a _{0} ) ^{2}} x + \frac{6 a _{0}^{2} - 20 a _{0} + 15}{( - 2 a _{0} + 3 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{5 - 3 a _{0}}{3 - 2 a _{0}})

Find the slope of

y 3 - x y 2 + x 3 = 5 : \frac{d y}{d x} = \frac{1}{( 3 - 2 x ) ^{2}}

Compute the slope of

y 3 - x y 2 + x 3 = 5

x = a_{0} : m = \frac{1}{( 3 - 2 a _{0} ) ^{2}}

Find the line with slope m=

\frac{1}{( 3 - 2 a _{0} ) ^{2}}

and passing through

(a_{0}, \frac{5 - 3 a _{0}}{3 - 2 a _{0}}) : y = \frac{1}{( 3 - 2 a _{0} ) ^{2}} x + \frac{6 a _{0}^{2} - 20 a _{0} + 15}{( - 2 a _{0} + 3 ) ^{2}}

y = \frac{1}{( 3 - 2 a _{0} ) ^{2}} x + \frac{6 a _{0}^{2} - 20 a _{0} + 15}{( - 2 a _{0} + 3 ) ^{2}}

n = 1 \sum \infty (\frac{n}{n + 1})^{n}

\int 6 sin^{2} (x) cos^{3} (x) d x

\int \frac{1}{- t ^{2} + 4 t + 5} d t

\int \frac{1}{5 - x} d x

\int 256 - x^{2} d x

What is the tangent of y3-xy2+x3=5 ?
The tangent of y3-xy2+x3=5 is y= 1/((3-2a_{0))^2}x+(6a_{0}^2-20a_{0}+15)/((-2a_{0)+3)^2}