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
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