Find the maximum area of a triangle formed in the first quadrant by the x-axis, y-axis, and a tangent line to the graph of y = (x+1)^-2
yes
Major slip-up near the top
"slope of tangent = -2(a+1)^3
so equation of tangent :
ax + (a+1)^3 y = c " should say
slope of tangent = -2/(a+1)^3
so equation of tangent :
2x + (a+1)^3 y = c
Notice the two changes.
Let me know if you are even reading this.
If you are, I will continue the solution.
last area line should be
(1/2)(a^2+a+1)^2 / (a(a+1)^3) , change is in the
denominator
To find the maximum area of a triangle formed in the first quadrant by the x-axis, y-axis, and a tangent line to the graph of y = (x+1)^-2, we first need to determine the coordinates of the points where the tangent line touches the graph.
Let's start by finding the derivative of the function y = (x+1)^-2 with respect to x.
Using the chain rule, we have:
dy/dx = -2(x+1)^-3 * (d/dx)(x+1)
= -2(x+1)^-3 * 1
= -2/(x+1)^3
The slope of the tangent line at any point on the graph is given by the derivative dy/dx.
Now, we want to find the points where the tangent line touches the graph, which means the slopes of the tangent line should match the slope of the graph at those points.
Setting the derivative dy/dx equal to the slope of the graph:
-2/(x+1)^3 = -2
Simplifying the equation:
1/(x+1)^3 = 1
Taking the reciprocal of both sides:
(x+1)^3 = 1
Taking the cube root of both sides:
x+1 = 1
x = 0
Now that we have the x-coordinate of the point where the tangent line touches the graph, we can substitute it back into the equation y = (x+1)^-2 to find the y-coordinate.
When x = 0:
y = (0+1)^-2
y = 1
Therefore, the coordinates of the point where the tangent line touches the graph are (0, 1).
Next, let's find the coordinates of the other two vertices of the triangle formed by the x-axis, y-axis, and the tangent line.
- The y-intercept is always at x = 0, so the y-intercept point is (0, 0).
- The x-intercept is always at y = 0. By substituting y = 0 into the equation y = (x+1)^-2:
0 = (x+1)^-2
Since the square of any real number is non-negative, the only solution to this equation is x = -1.
Therefore, the x-intercept point is (-1, 0).
Now, we have the coordinates of all three vertices of the triangle:
(0, 0), (0, 1), and (-1, 0).
To find the area of the triangle, we can use the formula for the area of a triangle:
Area = (1/2) * base * height
The base of the triangle is the distance between the points (-1, 0) and (0, 0), which is 1.
The height of the triangle is the distance between the points (0, 0) and (0, 1), which is 1.
Plugging these values into the formula:
Area = (1/2) * 1 * 1
Area = 1/2
Therefore, the maximum area of the triangle formed in the first quadrant by the x-axis, y-axis, and the tangent line to the graph of y = (x+1)^-2 is 1/2.
let the point of contact of the tangent be (a,b)
dy/dx = -2(x+1)^-3 = -2/(x+1)^3
at (a,b)
slope of tangent = -2(a+1)^3
so equation of tangent :
ax + (a+1)^3 y = c
but (a,b) lies on this
a(a) + (a+1)^3 (b) = c
equation of tangent:
ax + (a+1)^3 y = a^2 + b(a+1)^3
the height of the triangle is the y-intercept of the tangent
the base of the triangle is the x-intercept of the equation
x-intercept, let y = 0
ax = a^2 + b(a+1)^3
x = (a^2 + b(a+1)^3)/a
y-intercept, let x = 0
y = (a^2 + b(a+1)^3)/(a+1)^3
but remember since (a,b) is on the curve
b = 1/(a+1)^2
so the x intercept reduces to (a^2 + a + 1)/a
and the y-intercept reduces to (a^2 + a + 1)/(a+1)^3
Area of triangle = A
= (1/2)(a^2+a+1)^2 / (2a(a+1)^3)
.... long way from being done.... What a question!
Ok, now take the derivative of A using the quotient rule, set that equal to zero and solve for a
sub that value of a back into the area equation.