<p>Hello, I was solving some AP Calculus BC problems from Arco when I got confused on the following problem:
The rate at which air is leaking out of a tire is porportional to the amount of air in the tire. The tire was originally filled to capacity with 1,500 cubic inches of air. After one hour, there were 1,400 cubic inches of air left in it.
A. Express the amount of air in the tire in cubic inches as a function of time t in hours.</p>
<p>The book solved it in the following way:
A(t) = (N)(e^kt)
1400 = (1500)(e^k)
Ln(14/15) = k
=> A(t) = (1500)(e^t*Ln(14/15))
However, I though k should be negative.
Can anyone help me understand this problem, please?</p>
<p>k is ln(14/15) which is negative.</p>
<p>I'm guessing that answers your question.</p>
<p>No problem. It's an easy mistake to make. :)</p>
<p>Hello again. I have two more 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>
<p>I'll get back to you later, when I have a chance to write it down. I'm not very good at looking at computer notation for math.</p>