How do you find the equation of the tangent and normal line to the curve #y=x^(1/2)# at x=1?

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Answer 1 · 2018-06-10T20:52:00

Read below.

We find the tangent line to a curve by finding the derivative first.

#d/dx(x^n)=nx^(n-1)# given that #n# is a constant.

#f(x)=x^1/2#

#=>f'(x)=1/2*x^(1-1/2)#

#=>f'(x)=1/2*x^(-1/2)#

#=>f'(x)=1/2*1/sqrtx#

#=>f'(x)=1/(2sqrtx)#

#=>f'(1)=1/(2sqrt1)#

#=>f'(1)=1/2#

Let's now find the y value when #x=1#

#=>f(1)=sqrt1#

#=>f(1)=1#

We can now use these information to find the equation of the tangent line using the formula #m(x-x_1)=y-y_1#

#=>1/2(x-1)=y-1#

#=>1/2x-1/2=y-1#

#=>1/2x+1/2=y#

That is the equation of the tangent line at the point #(1,1)#

A normal line is perpendicular to the tangent line.

We can find the slope of the normal line by finding the negative reciprocal of the slope of the tangent line.

#1/2=>-2#

We use the formula #m(x-x_1)=y-y_1# once again.

#=>-2(x-1)=y-1#

#=>-2x+2=y-1#

#=>-2x+3=y#

That is the equation of the normal line at the point #(1,1)#