Find (d y)/(d x) and (d^2 y)/(d x^2) given that [ x=a t^2 y=2 a t ]

To find \( \frac{d y}{d x} \) and \( \frac{d^{2} y}{d x^{2}} \), we first need to find \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \).<br /><br />Given:<br />\[ x = a t^2 \]<br />\[ y = 2 a t \]<br /><br />First, let's find \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \):<br /><br />\[ \frac{d x}{d t} = \frac{d}{d t} (a t^2) = 2 a t \]<br />\[ \frac{d y}{d t} = \frac{d}{d t} (2 a t) = 2 a \]<br /><br />Now, we can find \( \frac{d y}{d x} \) by using the chain rule:<br /><br />\[ \frac{d y}{d x} = \frac{d y}{d t} \cdot \frac{d t}{d x} \]<br /><br />To find \( \frac{d t}{d x} \), we take the reciprocal of \( \frac{d x}{d t} \):<br /><br />\[ \frac{d t}{d x} = \frac{1}{\frac{d x}{d t}} = \frac{1}{2 a t} \]<br /><br />Now, substitute \( \frac{d y}{d t} \) and \( \frac{d t}{d x} \) into the equation for \( \frac{d y}{d x} \):<br /><br />\[ \frac{d y}{d x} = (2 a) \cdot \left( \frac{1}{2 a t} \right) = \frac{1}{t} \]<br /><br />Next, let's find \( \frac{d^{2} y}{d x^{2}} \) by differentiating \( \frac{d y}{d x} \) with respect to \( x \):<br /><br />\[ \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} \left( \frac{d y}{d x} \right) = \frac{d}{d x} \left( \frac{1}{t} \right) \]<br /><br />Using the chain rule again, we have:<br /><br />\[ \frac{d^{2} y}{d x^{2}} = \frac{d}{d t} \left( \frac{1}{t} \right) \cdot \frac{d t}{d x} \]<br /><br />\[ \frac{d}{d t} \left( \frac{1}{t} \right) = -\frac{1}{t^2} \]<br /><br />Now, substitute this into the equation for \( \frac{d^{2} y}{d x^{2}} \):<br /><br />\[ \frac{d^{2} y}{d x^{2}} = \left( -\frac{1}{t^2} \right) \cdot \left( \frac{1}{2 a t} \right) = -\frac{1}{2 a t^3} \]<br /><br />So, the correct answers are:<br />\[ \frac{d y}{d x} = \frac{1}{t} \]<br />\[ \frac{d^{2} y}{d x^{2}} = -\frac{1}{2 a t^3} \]