How do you find the length of the curve #x^(2/3)+y^(2/3)=1# for the first quadrant?

Answer 1

#L = 3/2#

We can write the curve in parametric form by posing:

#x^(2/3)=t#

so that:

#y^(2/3) = 1-t#
As in the first quadrant #x# and #y# are positive:
#{(x=t^(3/2)),(y=(1-t)^(3/2)):}#
where #0 <= t <= 1#.

The length of the curve is therefore:

#L= int_0^1 sqrt( (dx/dt)^2+(dy/dt)^2)dt#
#L= int_0^1 sqrt( (3/2t^(1/2))^2+(-3/2(1-t)^(1/2))^2)dt#
#L= int_0^1 3/2sqrt( t+(1-t))dt#
#L= 3/2 int_0^1dt = 3/2#

graph{x^(2/3)+y^(2/3) = 1 [-2.5, 2.5, -1.25, 1.25]}

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Answer 2

To find the length of the curve (x^{2/3} + y^{2/3} = 1) for the first quadrant:

  1. Write the curve in terms of (y) as a function of (x).
  2. Use the formula for arc length integration from (x = 0) to a certain value of (x) in the first quadrant.
  3. Integrate the square root of (1 + (dy/dx)^2) with respect to (x) over the given interval.
  4. Evaluate the integral.

Let's proceed with these steps:

  1. Rewrite the equation in terms of (y) as a function of (x): (y(x) = (1 - x^{2/3})^{3/2})

  2. Use the formula for arc length integration: (L = \int_{0}^{x_1} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx)

  3. Find (dy/dx) and substitute into the formula: (\frac{dy}{dx} = -\frac{2}{3}x^{-1/3}(1 - x^{2/3})^{1/2}) (L = \int_{0}^{x_1} \sqrt{1 + \left(-\frac{2}{3}x^{-1/3}(1 - x^{2/3})^{1/2}\right)^2} , dx)

  4. Evaluate the integral.

This process will give the length of the curve in the first quadrant.

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Answer 3

To find the length of the curve (x^{2/3} + y^{2/3} = 1) for the first quadrant, we can use the arc length formula for parametric curves. First, we need to parameterize the curve. We can do this by letting (x = t) and (y = (1-t^{2/3})^{3/2}), where (0 \leq t \leq 1).

Then, we differentiate (x) and (y) with respect to (t) to find (dx/dt) and (dy/dt).

[ \frac{dx}{dt} = 1 ] [ \frac{dy}{dt} = \frac{-2}{3}t^{-1/3}(1-t^{2/3})^{1/2} ]

Next, we compute (ds/dt), where (ds) represents an infinitesimal arc length element along the curve. It is given by:

[ ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt ]

[ ds = \sqrt{1 + \left(\frac{-2}{3}t^{-1/3}(1-t^{2/3})^{1/2}\right)^2} dt ]

Now, we integrate (ds) from (t = 0) to (t = 1) to find the total arc length:

[ \text{Arc Length} = \int_{0}^{1} \sqrt{1 + \left(\frac{-2}{3}t^{-1/3}(1-t^{2/3})^{1/2}\right)^2} dt ]

This integral can be quite complex to solve analytically, but it can be evaluated numerically using computational tools or software. Alternatively, you can try simplifying the expression inside the square root or using special techniques for handling integrals involving square roots.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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