Find the point on the curve y =2/3 〖√(18-x^2 )〗 (first quadrant) where a tangent line may be drawn so that the area of the triangle formed by the tangent line and the coordinate axes is a minimum.
we have
4x^2 + 9y^2 = 72
8x + 18y y' = 0
y' = -4x/9y
So, the line through (h,k) tangent to the curve is
y-k = (-4h/9k) (x-h)
That line has x-intercept at
(-4h/9k)x + 4h^2/9k + k = 0
x=(4h^2+9k^2)/(9k) * (9k/4h)
= (4h^2+9k^2)/(4h)
But, since 9k^2 = 72-4h^2, the x-intercept is at
(4h^2+72-4h^2)/(4h) = 1/(18h)
They-intercept is where x=0, or
y-k = (-4h/9k) (-h)
y = 4h^2/9k + k
= (4h^2+9k^2)/9k
= 1/(8k)
= 1/(8 * 2/3 √(18-h^2))
= 1/(12√(18-h^2))
The area of the triangle is 1/2 bh, or
a = (1/2)(1/18h)(1/(12√(18-h^2))
= 1/(432h√(18-h^2))
da/dh = (h^2-9)/(216h^2(18-h^2)^(3/2))
da/dh=0 when h=3.
So, the line through (3,2) defines the triangle with minimum area. That line has slope -2/3.
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D+2%2F3+%E2%88%9A%2818-x^2%29%2C+y+%3D+-2%2F3+%28x-3%29%2B2
Well, finding the point where the triangle formed by the tangent line and the coordinate axes has a minimum area is quite the mathematical challenge! But don't worry, I'm here to help... or clown around a bit!
To find this point, we need to determine the derivative of the function and set it equal to zero. So let's get started...
The given equation is y = (2/3)√(18 - x^2). Now, to find the derivative, we can use the chain rule. Remember kids, the chain rule is like the circus act of derivatives!
Taking the derivative of y with respect to x, we get:
dy/dx = (2/3) * (1/2)*(18 - x^2)^(-1/2) * (-2x)
= -4x/(3√(18 - x^2))
Now, let's set dy/dx equal to zero:
-4x/(3√(18 - x^2)) = 0
Hmm, I think we're stuck! It seems like there's no solution since x can't be zero and the denominator can't be zero either. It seems like the clown brigade is here to cause some mathematical chaos!
In this case, it appears that there is no point on the curve in the first quadrant where a tangent line can be drawn to form a triangle with the coordinate axes with a minimum area. It looks like we'll have to go back to the drawing board and come up with a new problem for our clown minds to tackle!
Did you hear about the circus lion who could solve calculus problems? They say he had a lot of integrated roar!
Well, don't worry if we couldn't find the minimum area point this time. There are plenty more math problems out there waiting for us to solve, or at least clown around with! If you have any more questions or need a good laugh, feel free to ask!
To find the point on the curve where a tangent line may be drawn so that the area of the triangle formed by the tangent line and the coordinate axes is a minimum, we need to find the minimum point of the curve y = (2/3)√(18-x^2).
Step 1: Find the derivative of the curve.
To find the derivative, we use the chain rule.
Let u = 18 - x^2.
Then, y = (2/3)√u.
Using the chain rule, the derivative of y with respect to x is:
dy/dx = (dy/du) * (du/dx)
To find dy/du, we can use the power rule:
dy/du = (2/3) * (1/2) * u^(-1/2) = (1/3) * (18 - x^2)^(-1/2)
To find du/dx, let's solve for u:
u = 18 - x^2
du/dx = -2x
So, combining both parts:
dy/dx = (1/3) * (18 - x^2)^(-1/2) * -2x
= -2x / (3√(18 - x^2))
Step 2: Find the x-coordinate of the critical point(s).
To find the critical points, we set the derivative equal to zero:
-2x / (3√(18 - x^2)) = 0
This equation is equal to zero when -2x = 0, which means x = 0.
Step 3: Find the y-coordinate of the critical point(s).
To find the y-coordinate, we substitute x = 0 into the original equation:
y = (2/3)√(18 - 0^2)
= (2/3)√18
= 2√2
So, the critical point is (0, 2√2).
Step 4: Check if it is a minimum.
To check if it's a minimum, we need to find the second derivative.
To find the second derivative, we differentiate the derivative:
d^2y/dx^2 = d/dx (dy/dx)
= d/dx (-2x / (3√(18 - x^2)))
= (6√(18 - x^2) + 2x^2 / √(18 - x^2)) / 9(18 - x^2)
To check if it's a minimum, we evaluate the second derivative at x = 0:
d^2y/dx^2 = (6√18 + 0 / √18) / 9(18)
= (6√18 / √18) / (9 * 18)
= 6 / (9 * 18)
= 1 / 27
Since the second derivative is positive, the critical point (0, 2√2) is a minimum.
Therefore, the point on the curve where a tangent line may be drawn so that the area of the triangle formed by the tangent line and the coordinate axes is a minimum is (0, 2√2).
To find the point on the curve where a tangent line can be drawn to form a triangle of minimum area, we should optimize the area of the triangle formed.
Let's start solving this problem step by step:
Step 1: Determine the equation of the curve.
The equation of the curve is given as y = (2/3)√(18-x^2). This is a semi-circle with center at the origin (0,0) and radius √18.
Step 2: Determine the area of the triangle.
The area of a triangle is given by the formula A = (1/2) * base * height.
In this case, the base of the triangle is the x-coordinate of the point of tangency, and the height is the y-coordinate of the point of tangency. Let's denote the point of tangency as (x, y).
So the area of the triangle can be expressed as A = (1/2) * x * y.
Step 3: Express y in terms of x.
Using the equation of the curve, we can express y in terms of x as:
y = (2/3)√(18-x^2).
Step 4: Substitute y into the formula for the area.
Substituting the value of y from step 3 into the area formula from step 2, we obtain:
A = (1/2) * x * (2/3)√(18-x^2).
Simplifying, we get A = (1/3) * x * √(18-x^2).
Step 5: Optimize the area A.
To optimize the area A, we need to find the point (x, y) on the curve where the derivative dA/dx is equal to 0.
Taking the derivative of A with respect to x, we get:
dA/dx = (1/3) * (√(18-x^2) - x * (1/2) * (1/2) * (18-x^2)^(-1/2) * (-2x)).
Simplifying the above expression, we get:
dA/dx = (1/3) * (√(18-x^2) + x^2/√(18-x^2)).
Setting dA/dx equal to 0 and solving for x, we have:
(1/3) * (√(18-x^2) + x^2/√(18-x^2)) = 0.
Multiplying both sides of the equation by √(18-x^2), we get:
√(18-x^2) + x^2 = 0.
Squaring both sides of the equation, we have:
18 - x^2 + x^4 = 0.
Simplifying further, we get:
x^4 - x^2 + 18 = 0.
Step 6: Solve for x.
This equation is a quadratic equation in terms of x^2. We can solve it by factoring or using the quadratic formula.
Unfortunately, there are no real solutions to this quadratic equation.
Therefore, there is no point on the given curve in the first quadrant where a tangent line can be drawn to form a triangle of minimum area.