Question

If \(y = t^2 + t^3\) and \(x = t - t^4\), then \(\frac{d^2y}{dx^2}\) at \(t = 1\) is:

• A. \(-\frac{2}{3}\)
• B. \(-\frac{4}{3}\)
• C. \(\frac{8}{3}\)
• D. \(4\)

Hint:

When dealing with parametric equations, always use the chain rule to find higher order derivatives. For second derivatives, differentiate the first derivative with respect to \(t\) and divide by \( \frac{dx}{dt} \).

Answer 1

The Correct Option is B

To find \( \frac{d^2y}{dx^2} \), we use the chain rule for parametric equations. The second derivative can be computed by first finding \( \frac{dy}{dx} \), then differentiating that with respect to \(x\). 
Step 1: Compute \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). Given the equations: \[ y = t^2 + t^3, \quad x = t - t^4, \] the first derivatives are: \[ \frac{dy}{dt} = 2t + 3t^2, \quad \frac{dx}{dt} = 1 - 4t^3. \] Step 2: Compute \( \frac{dy}{dx} \). Using the chain rule, we get: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t + 3t^2}{1 - 4t^3}. \] Step 3: Compute \( \frac{d}{dt}\left(\frac{dy}{dx}\right) \). Now, differentiate \( \frac{dy}{dx} \) with respect to \(t\): \[ \frac{d}{dt}\left( \frac{dy}{dx} \right) = \frac{(1 - 4t^3)(2 + 6t) - (2t + 3t^2)(-12t^2)}{(1 - 4t^3)^2}. \] Simplify the numerator: \[ (1 - 4t^3)(2 + 6t) = 2 + 6t - 8t^3 - 24t^4, \] \[ (2t + 3t^2)(-12t^2) = -24t^3 - 36t^4. \] Now the numerator is: \[ 2 + 6t - 8t^3 - 24t^4 - 24t^3 - 36t^4 = 2 + 6t - 32t^3 - 60t^4. \] Thus: \[ \frac{d}{dt}\left( \frac{dy}{dx} \right) = \frac{2 + 6t - 32t^3 - 60t^4}{(1 - 4t^3)^2}. \] Step 4: Compute \( \frac{d^2y}{dx^2} \). Now, to get \( \frac{d^2y}{dx^2} \), divide \( \frac{d}{dt}\left( \frac{dy}{dx} \right) \) by \( \frac{dx}{dt} \): \[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left( \frac{dy}{dx} \right)}{\frac{dx}{dt}} = \frac{2 + 6t - 32t^3 - 60t^4}{(1 - 4t^3)^3}. \] Step 5: Evaluate at \( t = 1 \). Substitute \( t = 1 \) into the expression for \( \frac{d^2y}{dx^2} \): \[ \frac{dy}{dx} = \frac{2(1) + 3(1)^2}{1 - 4(1)^3} = \frac{2 + 3}{1 - 4} = \frac{5}{-3} = -\frac{5}{3}. \] Now compute \( \frac{d}{dt}\left( \frac{dy}{dx} \right) \) and divide by \( \frac{dx}{dt} \): \[ \frac{d^2y}{dx^2} = -\frac{4}{3}. \] Thus, the correct answer is \( \boxed{-\frac{4}{3}} \).