**hey guys, its charlene your scribe for today**

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

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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