Question

I just had some questions on the attachments below. On question 5, I'm completely clueless but for question 10, I saw area as l(2r) + πr^2, but I can't get the equation for the perimeter. Would anyone be able to explain?

Answer 1

Answer:

5. 1.55 ms^-2

10. πr(r - 2) + 1800

Step-by-step explanation:

Question 5

We can solve this question by finding the maximum extremum of the function within the given interval. This can be found by finding the derivative of the function and finding its zeros (as the slope of all extreme points is 0).

To find the derivative of the function, I'll be using the product rule. If given some function f(x) = g(x) * h(x), then the derivative of this function is:

$\frac{df(x)}{dx} = \frac{dg(x)}{dx}\times h(x) + g(x) \times \frac{dh(x)}{dx}$

Therefore,

$a(t) = e^t\cos{t}\\\\\frac{da(t)}{dt} = \frac{de^t}{dt}\times\cos{t} + e^t\times\frac{d\cos t}{dt}\\= e^t\cos t + e^t(-\sin t)\\= e^t(\cos{t} - \sin{t})\\\\\text{ Find the zeros of the derivative: }\\\\e^t(\cos t - \sin t) = 0\\\to e^t = 0 \to \text{No solutions}\\\\\to \cos t - \sin t = 0\text{ //}\cos t - \sin t = 1 - \tan t\\1 - \tan t = 0\\1 = \tan t\\t = \frac{\pi}4 + \pi k$

The only solution within the given interval is t = pi/4.

This extreme point is maximum as:

$f'(0) = e^0(\cos(0)-\sin(0)) = 1(1-0)=1 > 0\\f'(2)=e^2(\cos(2)-\sin(2)) \approx-9.8 < 0$

To find the maximum acceleration, we'll substitute this value of t back into a(t) to find the acceleration value at this time.

$a(\frac{\pi}4) = e^{\frac{\pi}4}\cos(\frac{\pi}4) \approx 1.55 \text{ ms}^{-2}$

The maximum acceleration of is 1.55 ms^-2.

Question 10

You were correct to say that the area of the track can be expressed using the formula l(2r) + πr^2. However, we were only asked to express this area using r and not using l. Thus, we'll find a way to define l using r.

The perimeter of the track can be found by summing the area of the two semicircles (or one full circle) with twice the length of the track (l).

The perimeter of a circle with radius r is 2πr. Therefore, the perimeter of the shape is 2πr + 2l = 2(πr + l).

We also know this perimeter is equal to 1800 meters. Thus, we can equate that expression to 1800 and solve for l.

$2(\pi r + l) = 1800\text{ //}\div2\\\pi r + l = 900\\l = 900 - \pi r$

We can now substitute l = 900 - πr into the expression for the track's area and simplify to get the expression in terms of r only.

$2lr + \pi r^2\\\\\to 2(900 - \pi r) + \pi r^2\\=1800 - 2\pi r + \pi r^2\\=\pi r(r - 2) + 1800$

The expression for the area of the track is πr(r - 2) + 1800.