Find the coordinates of the point on the curve [tex]$y=2 x^2-x+1$[/tex] at which the tangent is parallel to the line [tex]$y=3 x+4$[/tex].
Answer (1)
Find the derivative of the curve y = 2 x 2 − x + 1 , which is d x d y = 4 x − 1 .
Set the derivative equal to the slope of the line y = 3 x + 4 , which is 3 , so 4 x − 1 = 3 .
Solve for x , which gives x = 1 .
Substitute x = 1 into the equation of the curve to find the y -coordinate: y = 2 ( 1 ) 2 − ( 1 ) + 1 = 2 . The coordinates of the point are ( 1 , 2 ) .
Explanation
Problem Analysis We are given the curve y = 2 x 2 − x + 1 and the line y = 3 x + 4 . We want to find the point on the curve where the tangent line is parallel to the given line.
Finding the Slope The slope of the line y = 3 x + 4 is 3 . Since the tangent line is parallel to this line, the slope of the tangent line must also be 3 .
Calculating the Derivative To find the slope of the tangent line to the curve y = 2 x 2 − x + 1 , we need to find the derivative of y with respect to x . Using the power rule, we have d x d y = 4 x − 1
Solving for x Now, we set the derivative equal to the slope of the line, which is 3 , and solve for x :
4 x − 1 = 3 4 x = 4 x = 1
Solving for y Now that we have the x -coordinate, we can find the y -coordinate by plugging x = 1 into the equation of the curve: y = 2 ( 1 ) 2 − ( 1 ) + 1 = 2 − 1 + 1 = 2
Final Answer Therefore, the point on the curve where the tangent line is parallel to the line y = 3 x + 4 is ( 1 , 2 ) .
Examples
Imagine you're designing a roller coaster. The track's curve is defined by a function, and you want a specific section to be parallel to a straight support beam. By finding the point on the curve where the tangent is parallel to the beam, you ensure the track aligns perfectly with the support, providing a smooth and safe ride. This problem demonstrates how calculus helps engineers align curves and lines in real-world applications.
