## Wednesday, November 12, 2008

### Functions Defined Implicitly

Minor adjustment: Justus' position as scribe has been zephed. I am zeph. zeph is scribe. I have implied that I am scribe. Makes sense? Rolling along...

Today we learned how to find the slope of the tangent line at a point intersecting a function.

Find the slope of the tangent to x^2+y^2=25 at (3,4).

From the diagram, we know that the slope of the tangent is -3/4 using rise over run.

Note the circle is not a function, but the circle can be seen as two functions graphed on the same graph.

Solution 1:

Using the equation of the circle, solve for y, then use the Chain Rule.

x^2+y^2=25....................................solve for y

y=(25-x^2)^(1/2).............................differentiate

y'={1/[2(25-x^2)^(1/2)]}(-2x)............plug in value

y'={1/[2(25-3^2)^(1/2)]}(-2*3)

y'=-3/4

Since x=3 at the intersection between the tangent and the radius at (3,4), plug in x=3 in the formula to get y'=-3/4, which is the slope of the tangent at (3,4).

Solution 2:

Using the equation of the circle, x^2+y^2=25, let h(x)=x^2+y^2, and let g(x)=25. Therefore, h(x)=g(x). Then differentiate both sides of the equation to get 2x+2y*y'=0 and solve for y'.

x^2+y^2=25.........let h(x)=left side, let g(x)=right side

h(x)=x^2+y^2.......h(x)=left side

g(x)=25................g(x)=right side

h(x)=g(x)..............differentiate

2x+2y*y'=0...........solve for y'

y'=-x/y.................plug in x-value

y'=-3/4

Since x=3 at the intersection between the tangent and the radius at (3,4), plug in x=3 in the formula to get y'=-3/4, which is the slope of the tangent at (3,4).

Find the slope of the tangent to y^2-2x=y at (1,-1).

y^2-2x=y ..................differentiate

2y*y'-2=y' ................solve for y'

y'*(2y-1)=2 y'=2/(2y-1)

Plug in y = -1 into y' = 2/(2y-1), and so the slope at the tangent at (1, -1) is y' = -2/3.

Find the slope of the tangent to x^2*y+siny=2*pi at (1, 2*pi).

x^2*y+sin y=2*pi .................................differentiate

x^2*y'+y*2x+cos y*y'=0 .....................x=1 and y= 2*pi, then solve for y'

(1)^2*y'+2*pi*2(1)+cos(2pi)y'=0

2y'=-4*pi

y'=-2*pi

The slope of the tangent to x^2*y+siny=2*pi at (1, 2*pi) is -2*pi

END NOTES:

Homework: 4.5 #3, 5, 7, 9, 11, 13, 15, 17, 21, 25

Next scribe is Yinan.

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