Equation of Tangent Line and Normal Line

Calculus
Section 3.1

Can you do 42 as a guide for me to do 41?
Show steps along the way.

Thank you.

 

42.

y=x^(2/3) at point (1,1)

first find the slope of tangent line

(dy/dx)(x^(2/3))= 2/(3x^(1/3))

compute the slope at point (1,1)

m=2/(3*1^(1/3))=2/3

so far, tangent line is: y=(2/3)x +b

use point to calculate b:

1=(2/3)1+b

1=2/3+b

b=1-2/3
b=1/3


tangent line is: y=(2/3)x+1/3

 

Nice graphing work. Which line is the normal line?

 

opss, I didn't do normal line

if the slope of tangent line is (2/3) , the slope of a normal line is negative reciprocal -3/2

use point-slope formula

y-y1=m (x-x1) ..........given point is (1,1)

y-1=-(3/2)(x-1)
y-1=-(3/2)x+3/2
y=-(3/2)x+3/2 +1
y=-(3/2)x+5/2

 

Much better. In what way do the normal line and tangent line relate?

 

as I stated above, the slope of tangent line and the slope of a normal line are negative reciprocal to each other

 

Ok. Thanks. Interesting.

 

Can you provide the graph of both lines on the xy-plane?

 

I assume my answer is correct. Yes?

 

yes it is correct

 

Perfect. We move on to Section 3.2 aka the Product Rule and Quotient Rule. I haven't decided if to go back to the Riegelmann Boardwalk at Coney Island or try Prospect Park today. Something is pulling me toward the park.

Either way, I gotta go outside and take advantage of this heavenly day. By "heavenly day" I mean windy, sunny and low humidity. My weekend is almost over. Life sucks!! Too much at the job and very little time for the things I like to do.