How do you find the equation of the tangent and normal line to the curve
#y=sqrtx# at x=9?

Question

How do you find the equation of the tangent and normal line to the curve 
#y=sqrtx# at x=9?

Isaiah Allen · Accepted Answer

eqn. tgt: #" "x-6y+27=0#

eqn. normal #" "6x+y-57=0# To find both the equation of the tangent and normal we will use #y-y_1=m(x-x_1)# where #m=#gradient; #(x_1,y_1)=# the coordinate in question just considering the # +sqrt# #y=sqrtx, y_1=sqrt9=3# #(x_1,y_1)=(9,3)# the problem now is to find the gradients tangent #m_t=(dy)/(dx)=d/(dx)(x^(1/2))=1/2x^(-1/2)# #m_t(9)=((dy)/(dx))_(x=9)=1/2xx1/sqrt9=1/6# eq tgt #y-3=1/6(x-9)# #6y-18=x-9# #6y-x-27=0# #x-6y+27=0# normal the normal is perpendicular to the tangent, by definition. So we can use the fact that the product of the gradients of perpendicular lines is #(-1)# #m_txxm_n=-1# #1/6xxm_n=-1# #m_n=-6# eqn normal #y-3=-6(x-9)# #y-3=-6x+54# #6x+y-57=0#

Isabella Ackerman · Answer

undefined To find the equation of the tangent and normal lines to the curve $ y = \sqrt{x} $ at $ x = 9 $, we first need to find the slope of the tangent line, which is given by the derivative of the function $ y = \sqrt{x} $ evaluated at $ x = 9 $.

The derivative of $ y = \sqrt{x} $ with respect to $ x $ is $ \frac{1}{2\sqrt{x}} $. Evaluating this derivative at $ x = 9 $, we get:

$$ \frac{dy}{dx} = \frac{1}{2\sqrt{9}} = \frac{1}{6} $$

So, the slope of the tangent line at $ x = 9 $ is $ \frac{1}{6} $.

The slope of the normal line (perpendicular to the tangent line) will be the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is $ -6 $.

Now, we have the slope of the tangent line ($ \frac{1}{6} $) and the slope of the normal line ($ -6 $). We also have the point $ (9, \sqrt{9}) = (9, 3) $ on the curve.

Using the point-slope form of a line equation $ y - y_1 = m(x - x_1) $, where $ m $ is the slope and $ (x_1, y_1) $ is a point on the line, we can find the equations of both the tangent and normal lines.

For the tangent line:
$$ y - 3 = \frac{1}{6}(x - 9) $$

For the normal line:
$$ y - 3 = -6(x - 9) $$

These equations can be simplified as needed for specific formats, such as slope-intercept form ($ y = mx + b $).

How do you find the equation of the tangent and normal line to the curve #y=sqrtx# at x=9?