How do you find the equation of the tangent line to the curve at $(81, 9)$ $(81, 9)$ of $y = \sqrt{x}$ $y=sqrtx$ ?

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How do you find the equation of the tangent line to the curve at $(81, 9)$ $(81, 9)$ of $y = \sqrt{x}$ $y=sqrtx$ ?

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Answer 1 · 2018-01-17T06:39:08

See below

Explanation:

Given equation
$y^{2} = x$ $y^2 = x$

This is an equation of a parabola. graph{y^2 = x [-10, 10, -5, 5]}
For finding the tangent to the equation, we first differentiate it.

The goal here is to find the value of $\frac{d y}{d x}$ $dy/dx$ which will give the slope of the tangent to the parabola.

So,
$\frac{d}{d y} (y^{2}) = \frac{d x}{d y}$ $d/dy(y^2) = dx/dy$

$2 y = \frac{d x}{d y}$ $2y = dx/dy$

$\frac{1}{2 y} = \frac{d y}{d x}$ $1/(2y) = dy/dx$

Now, we want to evaluate the slope at the given point. So substituting y we get,

$\frac{1}{18} = \frac{d y}{d x}$ $1/(18) = dy/dx$

This is the slope of the tangent.

It passes through (81,9) [as stated in the question]

The general equation of a line is
$y = m x + c$ $y = mx +c$

m is the slope of the line.

For the line/ tangent that we got $m = \frac{d y}{d x}$ $m = dy/dx$ .

$y = \frac{x}{18} + c$ $y = x/(18) + c$

The point should satisfy the equation of the line/ tangent. So,

$9 = \frac{81}{18} + c$ $9 = 81/(18) + c$

So, $c = \frac{9}{2}$ $c = 9/2$

Finally substituting the value in our equation of the line/tangent,

$y = \frac{x}{18} + \frac{9}{2}$ $y = x/(18) + 9/2$

$18 y = x + 81$ $18y = x + 81$

Answer 2 · 2018-01-17T13:30:08

$y = \frac{1}{18} x + \frac{9}{2}$ $y=1/18x+9/2$

Explanation:

$∙ x m_{tangent} = \frac{d y}{d x} at x = 81$ $•color(white)(x)m_(color(red)"tangent")=dy/dx" at x "= 81$

$y = \sqrt{x} = x^{\frac{1}{2}}$ $y=sqrtx=x^(1/2)$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2} x^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$ $rArrdy/dx=1/2x^(-1/2)=1/(2sqrtx)$

$x = 81 \to \frac{d y}{d x} = \frac{1}{2 \sqrt{81}} = \frac{1}{18}$ $x=81tody/dx=1/(2sqrt81)=1/18$

$equation of tangent in point-slope form is$ $"equation of tangent in point-slope form is"$

$y - 9 = \frac{1}{18} (x - 81)$ $y-9=1/18(x-81)$

$\Rightarrow y = \frac{1}{18} x + \frac{9}{2}$ $rArry=1/18x+9/2$

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How do you find the equation of the tangent line to the curve at $(81, 9)$ $(81, 9)$ of $y = \sqrt{x}$ $y=sqrtx$ ?

2 Answers

Explanation:

Explanation:

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How do you find the equation of the tangent line to the curve at $(81, 9)$ $(81, 9)$ of $y = \sqrt{x}$ $y=sqrtx$ ?