A line tangent to y=(x^2)+1 at x=a, a>0, intersects the x-axis at point P.
a) Write an expression for the area of the triangle formed by the tangent line, the x-axis, and the line x=a.
b) For what value of 'a' is the area of the triangle a minimum? Justify your answer.
Show all work. Thanks!
quite a question for this forum.
I worked it out, but will only give you some of the steps, leaving it up to you to fill in most of the steps.
1. The graph is a parabola with vertex at (0,1)
2. Let A(a,a^2 + 1) be the point of contact
3. the slope of the tangent is the derivative of the function which is y' = 2x
At the point A the slope is then 2a
3. find the equation of the tangent line with slope 2a and point (a,a^2+1)
let y = 0 and solve for x to get the x-intercept
(you should get x = (a^2 - 1)/(2a)
call that point P
let the other point of the triangle which is on the x-axis be Q
Of course Q is (a,0)
So the base of the triangle is PQ and the height is AQ
we need PQ which is a - (x-intercept)
= a - (a^2 - 1)/(2a)
you should get (a^2+1)/(2a), this is your base of the triangle
the height AQ of course is a^2+1 from the equation.
Now Area = (1/2)base x height
Differentiate that, set it equal to zero and solve for a
(I got a = 1/√3)
To find the expression for the area of the triangle formed by the tangent line, the x-axis, and the line x=a, we need to determine the coordinates of the vertices of the triangle.
Let's start by finding the point of tangency on the curve y=(x^2)+1 at x=a. The derivative of the curve will give us the slope of the tangent line at x=a.
1) Differentiate y=(x^2)+1 with respect to x:
dy/dx = 2x
2) Substitute x=a into the derivative to find the slope of the tangent line at x=a:
m = dy/dx (a) = 2a
Now we have the slope of the tangent line, and we know it passes through point (a, (a^2)+1). By using the point-slope form of a linear equation, we can determine the equation of the tangent line.
3) Using the point-slope form, we have:
y - ((a^2)+1) = (2a)(x - a)
Expanding this equation, we get:
y - (a^2+1) = 2ax - 2a^2
Now, let's find the x-coordinate at the intersection of the tangent line and the x-axis (point P). At this point, y-coordinate will be 0.
4) Setting y=0 in the equation, we get:
0 - (a^2+1) = 2ax - 2a^2
Simplifying the equation:
2ax - a^2 - 1 = 0
Now we can solve this equation for x.
5) Rearranging the terms, we get:
2ax = a^2 + 1
Dividing both sides by 2a:
x = (a^2 + 1)/(2a)
Now, we have the x-coordinate of point P, which is the intersection of the tangent line and the x-axis, and the coordinates of the point of tangency (a, (a^2)+1).
To find the area of the triangle, we need to compute the base and the height of the triangle.
The base is the distance between x=a and the x-coordinate of point P:
Base = P - a = [(a^2 + 1)/(2a)] - a
The height is the y-coordinate of the point of tangency:
Height = (a^2) + 1
Now we can calculate the area of the triangle using the formula: Area = 0.5 * base * height.
a) Expression for the area of the triangle:
Area = 0.5 * [(a^2 + 1)/(2a) - a] * [(a^2) + 1]
b) To find the value of 'a' that minimizes the area of the triangle, we can take the derivative of the area expression with respect to 'a' and set it equal to zero. This will help us find the critical points.
Let's differentiate the area expression with respect to 'a':
d/dA [0.5 * [(a^2 + 1)/(2a) - a] * [(a^2) + 1]] = 0
After simplifying, we can solve for 'a' to find the critical points. Once we have the critical points, we can determine which one minimizes the area of the triangle by comparing the areas at those respective points.