Questions: Consider the curve defined by (x^2=e^x-y) for (x>0). At what value of (x) does the curve have a horizontal tangent? (A) 1 (B) (sqrt2) C 2 D There is no such value of (x).

Consider the curve defined by (x^2=e^x-y) for (x>0). At what value of (x) does the curve have a horizontal tangent?
(A) 1
(B) (sqrt2)

C 2

D There is no such value of (x).
Transcript text: Consider the curve defined by $x^{2}=e^{x-y}$ for $x>0$. At what value of $x$ does the curve have a horizontal tangent? (A) 1 (B) $\sqrt{2}$ C 2 D There is no such value of $x$.
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Solution

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To find the value of \( x \) where the curve \( x^2 = e^{x-y} \) has a horizontal tangent, we need to determine where the derivative of \( y \) with respect to \( x \) is zero. This involves implicit differentiation of the given equation and solving for \( \frac{dy}{dx} = 0 \).

Step 1: Implicit Differentiation

Given the curve \( x^2 = e^{x-y} \), we differentiate both sides with respect to \( x \) implicitly: \[ \frac{d}{dx}(x^2) = \frac{d}{dx}(e^{x-y}) \] \[ 2x = e^{x-y} \left(1 - \frac{dy}{dx}\right) \]

Step 2: Solve for \(\frac{dy}{dx}\)

Rearrange the equation to solve for \(\frac{dy}{dx}\): \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} \left(1 - \frac{dy}{dx}\right) \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - 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