Question

If $x = 2 + t^3, y = 3t^2$ and $\left(\frac{d^2y}{dx^2}\right)/\left(\frac{dy}{dx}\right)^n$ is a constant then the value of $343n$ must be

Answer 1

1372

Answer 2

Given the parametric equations:

$x = 2 + t^3$ $y = 3t^2$

Step 1: Calculate the first derivatives with respect to $t$.

$\frac{dx}{dt} = \frac{d}{dt}(2 + t^3) = 3t^2$

$\frac{dy}{dt} = \frac{d}{dt}(3t^2) = 6t$

Step 2: Calculate the first derivative $\frac{dy}{dx}$ using the chain rule.

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6t}{3t^2} = \frac{2}{t}$ (for $t \neq 0$)

Step 3: Calculate the second derivative $\frac{d^2y}{dx^2}$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}$

First, calculate $\frac{d}{dt}\left(\frac{dy}{dx}\right)$:

$\frac{d}{dt}\left(\frac{2}{t}\right) = \frac{d}{dt}(2t^{-1}) = 2(-1)t^{-2} = -\frac{2}{t^2}$

Next, calculate $\frac{dt}{dx}$:

$\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{3t^2}$

Now, multiply these two results:

$\frac{d^2y}{dx^2} = \left(-\frac{2}{t^2}\right) \cdot \left(\frac{1}{3t^2}\right) = -\frac{2}{3t^4}$

Step 4: Substitute the calculated derivatives into the given expression $\frac{\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^n}$.

The expression is $\frac{-\frac{2}{3t^4}}{\left(\frac{2}{t}\right)^n}$.

Step 5: Simplify the expression.

$\frac{-\frac{2}{3t^4}}{\left(\frac{2}{t}\right)^n} = \frac{-\frac{2}{3t^4}}{\frac{2^n}{t^n}} = -\frac{2}{3t^4} \cdot \frac{t^n}{2^n} = -\frac{2 \cdot t^n}{3 \cdot 2^n \cdot t^4} = -\frac{2^{1-n}}{3} t^{n-4}$

Step 6: For the expression to be a constant, it must not depend on the parameter $t$. This means the exponent of $t$ in the simplified expression must be zero.

$n-4 = 0$ $n = 4$

Step 7: Calculate the value of $343n$ for $n=4$.

$343n = 343 \times 4 = 1372$

The value of $343n$ is 1372.