Calc AB Questions

Question

1. A particle moves in a straight line and has acceleration given by a(t) = 2cost + sec²t on the interval -π/2 < 2 < π/2. The velocity at π/3 seconds is 4 m/s. Find the velocity function v(t) for the particle.
2. Find the coordinates for the absolute maximum and absolute minimum of f(x) = ln(4x²+2x+1) over the interval [-1, 1].

Evaluate the following limits. Show appropriate calculus work.

1. lim x--> ∞ x+lnx/xlnx
2. lim x--> 0⁺ (2x+1)^cotx

Answer 1

1. A key concept to review is the relationship between distance, s(t); velocity, v(t); and acceleration, a(t). Velocity is the derivative of s(t): v(t) = s'(t), and acceleration is the derivative of velocity: a(t) = v'(t). In particular, v is the antiderivative of a. So to find v(t), we must compute the indefinite integral of a(t):

v(t) = ∫(2cos(t) + sec²(t))dt = 2∫cos(t)dt + ∫sec²(t)dt = 2sin(t) + tan(t) + C

Now, we are not done yet as there is a unknown constant, C. To find C, we must utilize the other piece of given information, namely, v(π/3)=4m/s. Plugging this condition into our function, we have

4 = v(π/3) = 2sin(π/3)+tan(π/3)+C ⇒ C = 4-2sin(π/3)-tan(π/3) = 4-2(√(3)/2) - √3 = 4-2√3.

1. Right off the bat, we are considering a function which is continuous on the closed interval [-1,1], and so must attain an absolute max and min, potentially at the endpoints of the interval. First, let's find the critical points of the function. By using the chain rule, we have

f'(x) = (8x+2)/4x²+2x+1.

Notice the denominator is not zero for any real number. So f'(x)=0 if and only if 8x+2=0 if and only if x=-1/4. So x=-1/4 is our only critical point. Picking a test point to the left and right of this value, and evaluating the test point at f', we discover that f' is negative (-∞,-1/4) and positive on (-1/4,∞) implying f is decreasing on

(-∞,-1/4) and increasing on (-1/4,∞). So x=-1/4 is a relative min, and, in fact, an absolute minimum! Now, we still need to find the absolute max. Given the behavior of f, we know the absolute max must occur at x=-1 or x=1. Simply plug these values into the function and whichever yields the largest value is your absolute max.

The limit problems require the use of various forms of L'Hopital's Rule:

lim_(x->∞) (x+ln(x))/(xln(x)) = ∞/∞

Apply L'Hopital's Rule, by taking derivative of the numerator and denominator, and taking the limit again:

lim(x->∞) (1 + (1/x))/(ln(x) + 1) = 0. So the original limit is zero.

The second limit is a little more complicated. Evaluating the limit as presented, we result in the indeterminate form 1^∞. The trick here is to apply the exponential form of L'Hopital,

lim(x->0⁺) (2x+1)^cot(x) = e^{lim(x->0+) (cot(x)ln(2x+1))}

However, when we evaluate the limit in the exponential, we result in ∞ · 0, which is another indeterminate form. In this case, we need to rewrite the expression to get an indeterminant form of type 0/0 or ∞/∞, so we can use L'Hopital again. This requires us to move either cot(x) or ln(2x+1) into the denominator via its reciprocal (ie. ln(2x+1) is equivalent to 1/(1/ln(2x+1)) or cot(x) is equivalent to 1/(1/cot(x))) . As 1/(1/cot(x)) = 1 / tan(x) and the derivative of tan(x) is simple, we'll go this route.

lim_x->0+ cot(x)ln(2x+1) = lim_x->0+ln(2x+1)/tan(x) = 0/0

Using L'Hopital's Rule, we have

lim_x->0+(2/2x+1)/(sec²x) = 2.

Hence, going back to the exponential, we have determined that the limit in the exponential is 2. Thus the original limit is

lim_x->0+ (2x+1)^cot(x) = e²