# Calculus Problem By: Ashley Kim Period 6. Problem A curve is defined by x 2 y-3y 2 =48. A curve is defined by x 2 y-3y 2 =48. a) Verify that dy/dx = 2xy/6y-x.

## Presentation on theme: "Calculus Problem By: Ashley Kim Period 6. Problem A curve is defined by x 2 y-3y 2 =48. A curve is defined by x 2 y-3y 2 =48. a) Verify that dy/dx = 2xy/6y-x."— Presentation transcript:

Calculus Problem By: Ashley Kim Period 6

Problem A curve is defined by x 2 y-3y 2 =48. A curve is defined by x 2 y-3y 2 =48. a) Verify that dy/dx = 2xy/6y-x 2 a) Verify that dy/dx = 2xy/6y-x 2 b) Write an equation of the linearization of this curve at ( 5,3). b) Write an equation of the linearization of this curve at ( 5,3). c) Using your equation from part a, estimate the y-coordinate of the point on the curve where x=4.93. c) Using your equation from part a, estimate the y-coordinate of the point on the curve where x=4.93.

d) Find equations of all horizontal tangent lines. d) Find equations of all horizontal tangent lines. e) Find equations of all vertical tangent lines. e) Find equations of all vertical tangent lines.

First Problem a) x 2 y-3y 2 =48 a) x 2 y-3y 2 =48 Step #1: Differentiate Step #1: Differentiate x 2 (dy/dx)+ 2xy -6y(dy/dx)=0 x 2 (dy/dx)+ 2xy -6y(dy/dx)=0 Step #2 : Subtract each side by -2xy. Step #2 : Subtract each side by -2xy. (x 2 -6y) dy/dx =-2xy (x 2 -6y) dy/dx =-2xy Step #3 : Divide Step #3 : Divide dy/dx= 2xy/ 6y-x 2 dy/dx= 2xy/ 6y-x 2

Second Problem b) Write equation of the linearization of this curve at (5,3) b) Write equation of the linearization of this curve at (5,3) Step #1: Find slope using (5,3) Step #1: Find slope using (5,3) dy/dx = 2(5)(3)/ 6(3)-5 2 dy/dx = 2(5)(3)/ 6(3)-5 2 30/ 18-25 = -30/7  Slope!! 30/ 18-25 = -30/7  Slope!! Step #2: Use y-y 1 = m (x-x 1 ). Step #2: Use y-y 1 = m (x-x 1 ). y-3 = -30/7 ( x-5) y-3 = -30/7 ( x-5)

Third Problem c) Using your equation from part a, estimate the y-coordinate of the point on the curve where x= 4.93. c) Using your equation from part a, estimate the y-coordinate of the point on the curve where x= 4.93. Step #1: Just plug in 4.93 into x in y-3 = -30/7 ( x-5). Step #1: Just plug in 4.93 into x in y-3 = -30/7 ( x-5). y-3= -30/7( 4.93 -5)  y-3 =0.3 y-3= -30/7( 4.93 -5)  y-3 =0.3 y= 3.3 y= 3.3

Fourth Problem d) Find the equations of all horizontal tangent lines. d) Find the equations of all horizontal tangent lines. Step #1: Tangent Lines are horizontal when dy/dx = 2xy/6y-x 2 =0. This is only possible when x=0 or y=0 since 2xy=0. Step #1: Tangent Lines are horizontal when dy/dx = 2xy/6y-x 2 =0. This is only possible when x=0 or y=0 since 2xy=0. Step #2: For x=0 plug in 0 into the x. Step #2: For x=0 plug in 0 into the x. 0y-3y 2 =48, so it has no real solutions. 0y-3y 2 =48, so it has no real solutions.

Step #3: For y=0, plug in 0 into y. Step #3: For y=0, plug in 0 into y. x 2 (0) -3(0 2 ) =48, that doesn’t make sense and impossible. x 2 (0) -3(0 2 ) =48, that doesn’t make sense and impossible. So there are no horizontal tangents. So there are no horizontal tangents.

Fifth Problem Find equations of all vertical tangents. Find equations of all vertical tangents. Step #1: For vertical tangent lines the slope is undefined. So it’s only possible when 6y-x 2 =0. Step #1: For vertical tangent lines the slope is undefined. So it’s only possible when 6y-x 2 =0. Which is y= x 2 /6. Which is y= x 2 /6. Step #2: Substitute that into y into the original equation: x 2 y-3y 2 =48. Step #2: Substitute that into y into the original equation: x 2 y-3y 2 =48.

x 2 (x 2 /6) -3 (x 2 /6) =48. x 2 (x 2 /6) -3 (x 2 /6) =48. (x 4 /6)- ( x 4 /12)=48 (x 4 /6)- ( x 4 /12)=48 Step #3: Multiply each side by 12. Step #3: Multiply each side by 12. 2x 4 - x 4 = 48. 2x 4 - x 4 = 48. Step #4: x 4 =48 Step #4: x 4 =48 x= + or - (√ 24) x= + or - (√ 24) x= 4.898979 or – 4.898979 x= 4.898979 or – 4.898979

The End Hope you had fun learning this valuable math problem!! Hope you had fun learning this valuable math problem!!

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