Sam Longoria

2021-12-21

Find an equation of the tangent line to the curve at the given point.
$y=\sqrt{x}\left(81,9\right)$

Edward Patten

Expert

Step 1
Consider the given function:
$y=\sqrt{x}$
Step 2
Now differentiate the function to find the slope.
$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$
Step 3
Now find the slope at given point (81,9).
$m=\frac{dy}{{dx}_{\left(81,9\right)}}=\frac{1}{2\sqrt{81}}$
$=\frac{1}{18}$
Step 4
Write the equation for a line.
$y-{y}_{1}=m\left(x-{x}_{1}\right)$
Step 5
Put the values of known quantities.
$y-9=\frac{1}{18}\left(x-81\right)$
$18y-162=x-81$
$x-18y+81=0$

Laura Worden

Expert

Step 1
Let $f\left(x\right)=\sqrt{x}$
Then, by Eq. 3, the slope of the tangent at $\left(1,1\right)$ is
$m=\underset{h⇒0}{lim}\frac{f\left(81+h\right)-f\left(9\right)}{h}=\underset{h⇒0}{lim}\frac{\sqrt{81+h}-9}{h}$
$=\underset{h⇒0}{lim}\frac{\left(\sqrt{81+h}-9\right)\left(\sqrt{81+h}+9\right)}{h\left(\sqrt{81+h}+9\right)}=\underset{h⇒0}{lim}\frac{81+h-9}{h\left(\sqrt{81+h}+9\right)}$
$=\underset{h⇒0}{lim}\frac{h}{h\left(\sqrt{81+h}+9\right)}=\underset{h⇒0}{lim}\frac{1}{\sqrt{81+h}+9}=\frac{1}{18}$
Using the point-slope form of the equation of a line, we find that an equation of the tangent at the point $\left(1,1\right)$ is
$y-9=\frac{1}{18}\left(x-81\right)$
Which simplifies to $x-18y+81=0$

nick1337

Expert