Show that the curves whose equations are y = Squ(3x+1) and y = Squ(5x-x^2) have the same slope at their point of intersection. Find the equation of the common t

Question

• Show that the curves whose equations are y = Squ(3x+1) and y = Squ(5x-x^2) have the same slope at their point of intersection. Find the equation of the common tangent at this point. ???

Answer 1

The curves intersect when √(3x+1) = √(5x-x²)

 

Square both sides:  3x+1 = 5x-x²

 

                           x²-2x+1 = 0

 

                            (x-1)² = 0

 

                                x = 1

 

The curves intersect at the point (1, √(3(1) + 1)) = (1, 2).

 

Slope of tangent line is the derivative of y = √(3x+1) evaluated at

x = 1.

 

     y' = (1/2)(3x+1)^-1/2(3)

 

     So, when x = 1, y' = 3/4 = slope of tangent line

 

The tangent line has slope 3/4 and goes through the point (1,2).

 

    Using the point-slope form of a line, an equation of the line is:

 

       y-2 = (3/4)(x-1)

 

          y = (3/4)x + 5/4

 

 

 

Answer 2

Set the two curves equal to each other and solve for x.  I assume that "squ" means square, and not square-root.

 

(3x + 1)² = (5x - x²)²

 

 

Square-root both sides of the equation.

 

3x + 1 = 5x - x²

 

 

Move all terms to the left side.

 

x² - 2x + 1 = 0

 

 

Factor using FOIL.

 

(x - 1)(x - 1) = 0

 

x = 1

 

 

The intersection occurs at x=1.

 

 

For the first equation:

 

y = (3(1) + 1)²

y = 4²

y = 16

 

 

For the second equation:

 

y = (5(1) - 1²)²

y = 4²

y = 16

 

 

The point of intersection is (1, 16).  Now all we have to do is find the derivative of each equation.  Then plug in the values into the derivative.  Derivative is the slope of the tangent line.

 

y = (3x + 1)²

 

y = 9x² + 6x + 1

 

dy/dx = 18x + 6

 

dy/dx = 18(1) + 6

          = 24

 

For the second equation:

 

y = (5x - x²)²

y = (5x - x²)(5x - x²)

y = 25x² - 10x³ + x⁴

 

dy/dx = 50x - 30x² + 4x³

         = 50(1) - 30(1)² + 4(1)³

         = 50 - 30 + 4

         = 20 + 4

         = 24

 

They both have the same slope of 24 at the point of intersection.

 

To find the equation of the tangent line at this point use the point-slope form.

 

y - 16 = 24(x - 1)

 

Answer 3

Find the point of intersection of the two curves.

 

Find the derivative of each equation.

 

Check the point of intersection in each derivative equation showing the slope is the same for that point.

 

Then use the intersection point and the slope to get the equation of the line.