Answer to Question #183850 in Calculus for Chayzel Cayanga

Question

1. Starting from the same point, Reden starts walking eastward at 60 cm/s while Neil starts running towards the south at 80 cm/s. How fast is the distance between Reden and Neil increasing after 2 seconds?
2. A balloon, in the shape of a right circular cylinder, is being inflated in such a way that the radius and height are both increasing at the rate of 3 cm/s and 8 cm/s, respectively. What is the rate of change of its total surface area when its radius and height are 60 cm and 140 cm, respectively?

Answer 1

1.

Let r and n be the distance traveled by Reden and Neil respectively.

For Reden, $\frac{dr}{dt}=60 cm/s$ and for Neil, $\frac{dn}{dt}=80 cm/s$.

At time t=2s, the distance from the starting point are,

$r=2 \times60=120cm\newline n=2\times 80=160cm$

Since, their movement form a right angle triangle.

Therefore, the distance between Reden and Neil after 2s is p and rate of increasing is given by $\frac{dp}{dt}$ .

By pythagorean theorem,

$p^2=r^2+n^2\newline p=\sqrt{r^2+n^2}\newline \frac{dp}{dt}=\frac{1}{2}(r^2+n^2)^{-1/2}(2r\frac{dr}{dt}+2n\frac{dn}{dt})$

Putting the values,

$\frac{dp}{dt}=\frac{1}{2}(120^2+160^2)^{-1/2}(2.120.60+2.160.80)\newline \hspace{0.4cm}=\frac{1}{2}\sqrt{120^2+160^2}\newline \hspace{0.4cm}=100 cm/s$

Thus, the required rate is 100cm/s.

2.

Given, the rate of increasing of radius and height of the right circular cylinder is 3cm/s and 8cm/s.

Total surface area, a=$2 \pi rh+2\pi r^2$ .

Rate of total surface area,

$\frac{da}{dt}=2π(r\frac{dh}{dt}+h\frac{dr}{dt})+4πr\frac{dr}{dt}\newline \hspace{0.45cm}=2π(60.8+140.3)+4π60.3\newline \hspace{0.45cm}=7916cm^2/s$

Thus, the rate of change in total surface area is 7916cm²/s.