AP Calculus question

<p>Hello, people. I desperately need your help here.
How do you guys do these problems?</p>

<li>y=sin t, x=t^2 (take the derivative)</li>
<li>y=sin^2 t, x=t^3 (find d^2/dx^2)</li>

<p>Please gimme idea how to solve these problems, and also don't forget to post answers.
Thank you.</p>

<li>you take the deriv of each part with respect to t</li>

<p>dy/dt = cost, dx/dt = 2t</p>

<p>then apply chain rule: dy/dx = dy/dt * dt/dx = cost * 1/(2t)</p>

<li>y = sin2t, x = t^3 same concept, apply chain rule first to get dy/dx, then take the derivative of dy/dx to get d^2y/dx^2</li>

<p>dy/dt = 2cos2t, dx/dt = 3t^2</p>

<p>dy/dx = dy/dt * dt/dx = 2cos2t * 1/(3t^2) = 2cos2t/(3t^2)</p>

<p>d^2y/dx^2 = derivative of dy/dx, apply the quotient rule</p>

<p>= [(3t^2)<em>(-4sin2t) - 2cos2t</em>(6t)] / (3t^2)^2</p>

<p>= [-12t^2<em>sin2t - 12t</em>cos2t]/ 9t^4</p>

<p>derivative of a parametric is defined by dy/dx = (dy/dt)/(dx/dt), in effect to cancel the t's</p>

<li><p>dy/dt= cos t dx/dt=2, therefore dy/dx= (cos t)/2, plug in appropriate t if necessary</p></li>
<li><p>same deal, except now the second derivative is regarded as (dy'/dt)/(dx/dt), note the prime of y' is derived once again</p></li>

<p>therefore, dy/dt= 2(sin t)(cos t) ; (using chain rule), which is equal to y', so u derive this yet again 2 [(sin t)(-sin t) + (cos t)(cos t)]</p>

<p>take that entire thing and divide it by dx/dt, which is 3t^2, cancel and plug in appropriately</p>

<p>hope this helps...</p>

<p>lee, its not sin2t, its sin^2t, as in (sin t)^2</p>

<p>someone verify our solutions please, i wouldn't want to give treasure the wrong solutions</p>

<li>y=sin t, x=t^2 (take the derivative)
dy/dt=cos t, dx/dt=2t, (dy/dt)/(dx/dt)=dy/dx</li>


<li>y=sin^2 t, x=t^3 (find d^2/dx^2)
same concept....
dy/dt=2sint*cost, dx/dt=3t^2


<p>heh, 3 different solutions, that ought to confuse you</p>

<p>guys, I truly appreciate for your responses. I respect your answers and different approaches.
I just wanna make sure that there is only one right answer. Help me!!!
Thank you for your helps again.</p>