Calculus Question

Question

If 2y²-xy-x² = 0, find dy/dx.

 

And, determine the slope of the tangent at x =1. 

Answer 1

This has to be differentiated implicitly as follows:

 

d/dx(2y²) - d/dx(xy) - d/dx(x²) = d/dx(0)

 

4y dy/dx - (y + x dy/dx) - 2x = 0

 

4y dy/dx - y - x dy/dx - 2x = 0

Combine like terms and factor:

 

dy/dx (4y - x) -y-2x=0

 

dy/dx (4y - x) = y+2x

Thus: 

dy/dx = (y+2x)/(4y-x)

 

To obtain the corresponding y-coordinate (s), substitute x=1 in the given equation.

 

Therefore, 2y² - y - 1 =0

Factor this equation to obtain (2y + 1)(y - 1) = 0

 

Thus 2y + 1 = 0 or y - 1 = 0; This gives us the values of y as -1/2 or 1

 

Substituting for x and y in the dy/dx equation yields:

For x = 1, y = -1/2:

 

dy/dx = (-1/2 + 2)/(-1/2*4 - 1)

dy/dx = (3/2)/(-3) = -1/2 = -0.5

 

For x = 1, y = 1:

dy/dx = (1 +2)/(4 - 1) = 3/3 = 1

 

Therefore, there are two slopes at x = 1: -1/2 and 1.

 

Comment

This shows that the graph represented by this relation is not a function because it fails the vertical line test; it is definitely not a one-to-one and not even a function. It has two values in the range (y) for one value in the domain (x).

 

Method 2:

You can also use an alternative method by factoring the original equation to give you the product of two binomial factors. If you rewrite this equation as x² + xy - 2y² = 0 and factor, you will obtain (x + 2y)(x - y) = 0

 

Then, differentiate this implicitly by using the product rule.

 

Thus: d/dx[(x + 2y)(x - y)] = d/dx(0)

 

(1 + 2 dy/dx)(x-y) + (1 - dy/dx)(x +2y) = 0

 

x - y +2x dy/dx - 2y dy/dx + x +2y - x dy/dx - 2y dy/dx = 0

 

Combine like terms and simplify:

 

x dy/dx -4y dy/dx + 2x + y = 0

 

2x + y = (4y - x) dy/dx

 

dy/dx = (2x + y)/(4y - x)

 

Complete the rest and compare your answer to the one above.

 

 

 

 

 

Answer 2

at x = 1 solve for y:
2y² - (1)y - 12 = 0
2y² - y - 1 = 0
(2y +1 )(y -1) = 0
2y + 1= 0 or y - 1 = 0
y = -1/2 or y = 1
So, at x=1, there are two y's (1, -1/2) and (1,1)

Now find implicit derivative:
4y(dy/dx) -[ 1y + x(dy/dx) ] - 2x = 0
4y(dy/dx) - y - x(dy/dx) - 2x = 0
4y(dy/dx) - x(dy/dx) = y + 2x
dy/dx (4y - x) = y + 2x
dy/dx = (y + 2x)/(4y - x)

 

 

Since there are two points of tangency: (1, -1/2) and (1,1)
there will be two slopes.

Case 1: (1, -1/2)
dy/dx = (-1/2 + 2(1) )/(4(-1/2) - 1) = (3/2) / (-3) = -1/2

Case 2: (1,1)
dy/dx = (1 + 2(1) )/(4(1) - 1) = (3) /(3) = 1

Answer 3

The slope of the tangent at x = 1 is the same as dy/dx evaluated at x = 1.  

 

When x = 1, 2y² -(1)y - (1)²  = 0

 

       2y² - y - 1 = 0

 

       (2y + 1)(y - 1) = 0

 

       y = -1/2 or y = 1

 

So, when x = 1, there are 2 points on the graph:  (1,-1/2) and (1,1).

 

Differentiating implicitly, we have 4y(dy/dx) - y - x(dy/dx) -2x = 0

 

(dy/dx) (4y - x) = 2x + y     So, dy/dx = (2x + y)/(4y - x)

 

At the point (1, -1/2), dy/dx = (3/2)/(-3) = -1/2.

 

At the point (1,1), dy/dx = 3/3 = 1.