Find the equation of the tan line to the curve at (81;9) of y=x

y=x means that y2=x So, ddy(y2)=dxdy 2y=dxdy 12y=dydx Substitute 118=dydx Fo the line that we have m=dydx y=x18+c 9=8118+c y=x18+92 18y=x+81

mtangent=dydx at x=81 y=x=x12 ⇒dydx=12x−12=12x x=81⇒dydx=1281=118 Equation of tan in point-slope form is: y−9=118(x−81) y=118x+92

y=x means that y2=x So, ddy(y2)=dxdy 2y=dxdy 12y=dydx Substitute 118=dydx Fo the line that we have m=dydx y=x18+c 9=8118+c y=x18+92 18y=x+81

mtangent=dydx at x=81 y=x=x12 ⇒dydx=12x−12=12x x=81⇒dydx=1281=118 Equation of tan in point-slope form is: y−9=118(x−81) y=118x+92

"Find the equation of the tan line to..." solved and explained

High School
Calculus and Analysis
Calculus 2
Parametric equations

David Lewis

Answered question

2021-12-11

Find the equation of the tan line to the curve at

(81; 9)

y = \sqrt{x}

Answer & Explanation

levurdondishav4

Beginner2021-12-12Added 38 answers

y = \sqrt{x}

means that

y^{2} = x

So,

\frac{d}{d y} (y^{2}) = \frac{d x}{d y}

2 y = \frac{d x}{d y}

\frac{1}{2 y} = \frac{d y}{d x}

Substitute

\frac{1}{18} = \frac{d y}{d x}

Fo the line that we have

m = \frac{d y}{d x}

y = \frac{x}{18} + c

9 = \frac{81}{18} + c

y = \frac{x}{18} + \frac{9}{2}

18 y = x + 81

braodagxj

Beginner2021-12-13Added 38 answers

$m_{t a n g e n t} = \frac{d y}{d x}$ at $x = 81$
$y = \sqrt{x} = x^{\frac{1}{2}}$
$\Rightarrow \frac{d y}{d x} = \frac{1}{2} x^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$
$x = 81 \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{81}} = \frac{1}{18}$
Equation of tan in point-slope form is:
$y - 9 = \frac{1}{18} (x - 81)$
$y = \frac{1}{18} x + \frac{9}{2}$

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