<p>Hello. I have two questions from Arco. In the following problem, part D and E, I cannot solve for my answer to match the answer key.</p>
<p>The path of a particle from t = 0 to t = 10 seconds is described by the parametric equations x(t) = 4cos[(pi/2)t] and y(t) = 3sin[(pi/2)t].
A. Write a Cartesian equation for the curve defined by these parametric equations.
answer: [(x^2)/16] + [(y^2)/9] = 1.
B. Find dy/dx for the equation in part A.
answer: -9x/16y
C. not relevant to my question.
D. Demonstrate that your answers for part A and part B are equivalent.
Book: dy/dx = (dy/dt)/(dx/dt) = [(3pi/2)cos(t<em>pi/2)]/[(-2pi)sin(t</em>pi/2)]
= (-3/4)tan[(pi/2)t]
&
-9x/16y = [-36 cos((pi/2)t)]/[48sin((pi/2)t)] = (-3/4)tan[(pi/2)t]
Me: dy/dx = (dy/dt)/(dx/dt) = [(3pi/2)cos(t<em>pi/2)]/[(-2pi)sin(t</em>pi/2)]
=> (-3/4)cot[(pi/2)t]
&
-9x/16y = [-36 cos((pi/2)t)]/[48sin((pi/2)t)] = (-3/4)cot[(pi/2)t]
*<em>No matter how much I solved it, I keep getting cot instead of tan.
E. Write, but do not evaluate, an integral expression that would give the distance of the particle traveled from t = 2 to t = 6.
Book: L = integral from a to b of sqrt[(dx/dt)^2 + (dy/dt)^2]
= integral from a to b of sqrt[-2pi</em>sin((pi/2)t)^2 + ( (3pi/2)cos((pi/2)t) )^2]
Me: my answer is almost the same except that instead of squaring only the (pi/2)<em>t in the first part of the integral from a to b of sqrt[-2pi</em>sin((pi/2)t)^2 + ( (3pi/2)cos((pi/2)t) )^2], I squared -2pi*sin[(pi/2)t].
I am forgetting a rule? I am very confused. Please help me.</p>