# The position of a particle at time t is given by s the velocity fraction {ds}{dt}. 1) s^3 - 4 st...

## Question:

The position of a particle at time {eq}t {/eq} is given by {eq}s {/eq} the velocity {eq}\frac {ds}{dt}. {/eq}

{eq}1) \ s^3 - 4 st + 2t^3 - 5t = 0 \\ 2) \ 2s^2 + \sqrt {st} - 4 = 3t {/eq}

## Derivative:

To find the derivative we will use the power rule where we will decrease the power of x by 1 and divide it by the same decreased power and then we will rearrange the expression to get the derivative.

To find the derivative we will use the power rule:

{eq}s^{3}-4st+2t^{3}-5t=0 {/eq}

Differentiating it we get:

{eq}3s^{2}\frac{\mathrm{d} s}{\mathrm{d} t}-4t\frac{\mathrm{d} s}{\mathrm{d} t}-4s+6t^{2}-5=0 {/eq}

Now the derivative becomes:

{eq}\frac{\mathrm{d} s}{\mathrm{d} t}=\frac{4s-6t^{2}+5}{3s^{2}-4t}\\ b) 2s^{2}+\sqrt{st}-4=3t {/eq}

Differentiatng again:

{eq}4s\frac{\mathrm{d} s}{\mathrm{d} t}+\frac{\sqrt{s}}{2\sqrt{t}}+\frac{\sqrt{t}}{2\sqrt{s}}\frac{\mathrm{d} s}{\mathrm{d} t}=3 {/eq}

After rearranging we get:

{eq}\frac{\mathrm{d} s}{\mathrm{d} t}=\frac{3-\frac{1}{2}\frac{1\sqrt{s}}{2\sqrt{t}}}{4s+\frac{1}{2}\frac{\sqrt{t}}{\sqrt{s}}} {/eq}