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Question: If $x^k + y^k = a^k (a, k > 0)$ and $\frac{dy}{dx} + (\frac{y}{x})^{\frac{1}{2}} = 0$ then k has the...

If xk+yk=ak(a,k>0)x^k + y^k = a^k (a, k > 0) and dydx+(yx)12=0\frac{dy}{dx} + (\frac{y}{x})^{\frac{1}{2}} = 0 then k has the value:

A

13\frac{1}{3}

B

23\frac{2}{3}

C

14\frac{1}{4}

D

27\frac{2}{7}

Answer

The calculated value of k is 12\frac{1}{2}. However, this value is not among the given options. There might be an error in the question or the options provided.

Explanation

Solution

To find the value of kk, we use the given equations:

  1. xk+yk=akx^k + y^k = a^k (where a,k>0a, k > 0)
  2. dydx+(yx)12=0\frac{dy}{dx} + \left(\frac{y}{x}\right)^{\frac{1}{2}} = 0

Step 1: Differentiate the first equation with respect to xx.

Given xk+yk=akx^k + y^k = a^k. Differentiating both sides implicitly with respect to xx:

ddx(xk)+ddx(yk)=ddx(ak)\frac{d}{dx}(x^k) + \frac{d}{dx}(y^k) = \frac{d}{dx}(a^k)

Using the power rule ddx(un)=nun−1dudx\frac{d}{dx}(u^n) = n u^{n-1} \frac{du}{dx}:

kxk−1+kyk−1dydx=0k x^{k-1} + k y^{k-1} \frac{dy}{dx} = 0

Since k>0k > 0, we can divide the entire equation by kk:

xk−1+yk−1dydx=0x^{k-1} + y^{k-1} \frac{dy}{dx} = 0

Now, solve for dydx\frac{dy}{dx}:

yk−1dydx=−xk−1y^{k-1} \frac{dy}{dx} = -x^{k-1}

dydx=−xk−1yk−1\frac{dy}{dx} = -\frac{x^{k-1}}{y^{k-1}}

dydx=−(xy)k−1\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{k-1}

Step 2: Use the second given equation to express dydx\frac{dy}{dx}.

Given dydx+(yx)12=0\frac{dy}{dx} + \left(\frac{y}{x}\right)^{\frac{1}{2}} = 0.

Solving for dydx\frac{dy}{dx}:

dydx=−(yx)12\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac{1}{2}}

Step 3: Equate the two expressions for dydx\frac{dy}{dx}.

From Step 1: dydx=−(xy)k−1\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{k-1}

From Step 2: dydx=−(yx)12\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac{1}{2}}

Equating them:

−(xy)k−1=−(yx)12-\left(\frac{x}{y}\right)^{k-1} = -\left(\frac{y}{x}\right)^{\frac{1}{2}}

Multiply by -1:

(xy)k−1=(yx)12\left(\frac{x}{y}\right)^{k-1} = \left(\frac{y}{x}\right)^{\frac{1}{2}}

Step 4: Solve for kk by comparing exponents.

We can rewrite (yx)12\left(\frac{y}{x}\right)^{\frac{1}{2}} as ((xy)−1)12=(xy)−12\left(\left(\frac{x}{y}\right)^{-1}\right)^{\frac{1}{2}} = \left(\frac{x}{y}\right)^{-\frac{1}{2}}.

So, the equation becomes:

(xy)k−1=(xy)−12\left(\frac{x}{y}\right)^{k-1} = \left(\frac{x}{y}\right)^{-\frac{1}{2}}

For the bases to be equal, the exponents must be equal:

k−1=−12k-1 = -\frac{1}{2}

k=1−12k = 1 - \frac{1}{2}

k=22−12k = \frac{2}{2} - \frac{1}{2}

k=12k = \frac{1}{2}

The calculated value of kk is 12\frac{1}{2}. However, this value is not present in any of the given options. This suggests a potential error in the question's options or the question itself.