What is the tangent of (10-x)y^2=x^3,(2,1) ?

The tangent of (10-x)y^2=x^3,(2,1) is y= 13/16 x-5/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of (10-x)y^2=x^3,(2,1)

tangent of

(10 - x) y^{2} = x^{3}, (2, 1)

y = \frac{13}{16} x - \frac{5}{8}

Solution steps

Find the slope of

(10 - x) y^{2} = x^{3} : \frac{d y}{d x} = \frac{3 x ^{2} + y ^{2}}{2 y ( 10 - x )}

Compute the slope of

(10 - x) y^{2} = x^{3}

(2, 1) : m = \frac{13}{16}

Find the line with slope m=

\frac{13}{16}

and passing through

(2, 1) : y = \frac{13}{16} x - \frac{5}{8}

y = \frac{13}{16} x - \frac{5}{8}

\int \frac{1}{cos ( θ ) - 1} d θ

\frac{d}{d x} (cos (0.5 x))

y^{^{'''}} - 9 y^{^{''}} + 15 y^{^{'}} + 25 y = 0

\frac{\partial}{\partial x} (\frac{x}{v})

x \to 8 + lim (\frac{∣ x - 8 ∣}{x - 8})

What is the tangent of (10-x)y^2=x^3,(2,1) ?
The tangent of (10-x)y^2=x^3,(2,1) is y= 13/16 x-5/8