## November 24, 2006

### scribe post

hey guys, its charlene your scribe for today

well as usual we started class with a few problems on the board

1) find Dy/Dx @ (1,1) given

....X^3 + X^4 + Y^3 + Y^2 = 4

....X^3 + X^4 + Y^3 + Y^2 = 4
....
....3X^2 + 4X^3 + 3Y^2 Y^1 + 2YY' = 0

....3Y^2Y' + 2YY' = -3X^2 - 4X^3
....
....Y' (3Y^2 + 2Y ) = -3X^2 - 4X^3

....Y' = -3X^2 - 4X^3
.............3Y^2 + 2Y
...
....Y' = -X^2 (3+4X)
........... Y (3Y + 2)...........---derivative of the function

TO FIGURE OUT THE DERIVATIVE @ (1,1) JUST PLUG IN x=1 AND y=1 INTO THE EQUATION

....Y' (1,1) = -(1)^2 (3 + 4(1))
.....................(1) ( 3(1) +2)

....Y' (1,1) = - 7/5

2) Find the equation of the line tangent to X^2 + Y^3 + 2X^3 + 8Y^2 = 3 @ the point (-1 , 2)

....X^2 3 Y*2 Dy/Dx + 2 X Y^3 + 6X^2 + 16Y Dy/Dx = 0

....Dy/Dx (3X^2 Y^2 + 16Y) = -2XY^3 - 6X^2

....Dy/Dx = -2X ( Y^3 + 3X)
...................Y (3X^2 Y +16)-- <--DERIVATIVE OF FUNCTION

PLUG IN X=-1 AND Y=2 INTO EQUATION

....Dy/Dx (-1 , 2) = -2(-1)(2^3 + 3(-1))
................................2(3(-1)^2 (2) +16)

....Dy/Dx (-1,2) = 5/22

TO GET THE EQUATION OF THE TANGENT YOU USE THE point-slope formula

....POINT = (-1,2)

....SLOPE = 5/22

....Y - 2 = 5/22 (X+1)

3) find the equation of the line tangent X^3 Y^2 = Y +2 @ X=1

...X^3Y^2 = Y+2

....LET X=1

....(1)^3 Y^2 - Y -2 = 0

....(Y-2) (Y+1) = 0

....Y= 2...Y= -1

TO FIGURE OUT THE DERIVATIVE OF THE EQUATION

....X^3Y^2 = Y + 2

....3X^2 Y^2 + X^3 (2Y) DY/Dx = Dy/Dx

....2X^3 Dy/Dx - Dy/Dx = -3X^2 Y^2

....Dy/Dx ( 2X^3Y - 1 ) = -3X^2 Y^2

....Dy/Dx = -3X^2 Y^2
.......................2X^3Y -1

NOW YOU HAVE TO PLUG IN THE POINTS (1,2) AND (1,-1)

....Dy/Dx (1,2) = -3(1)^2 (2)^2
............................2(1)^3 (2)(-1)

........................= -4

....EQUATION Y - 2 = -4 (X-1)

....Dy/Dx (1 -1) = -3(1)^2 (-1)^2
.............................2(1)^3 (-1)(-1)

.........................= 1

....EQUATION Y + 1 = X - 1

AFTER PROBLEMS ON THE BOARD WE LEARNED SOMETHING NEW...

picture a smaller square in the center of a larger square

rate = derivative

sidelength with respect to time

Ds / Dt = 2 cm / sec

between the two sqaures these following things will change

- perimeter

- area -> length of sides depends on time

-diagonal

to calculate the perimeter with respect to time

....Dp / Dt = ?

....P = 4S ** S= LENGTH OF SIDES

....Dp / Dt = 4 Ds / Dt

....Dp / Dt = 4 (2)

....Dp / Dt = 8 CM/S

area with respect to time

....Da / Dt = ?

....AREA = S^2

....Da / Dt = 2 S Ds/Dt

....Da / Dt = 4s

SUPPOSE S = 5CM

....Da/Dt = 4(5)

...............= 20 CM^2 / SEC

Mr. K also showed us a presentation about what we just learned. Mr. K will put a link of the website on the blog.

homework section 4.6, questions 1 - 12

the next scribe is CRYSTAL