Question

Ying is a professional deep water free diver. His altitude (in meters relative to sea level), xxx seconds after diving, is modeled by: D(x)=\dfrac{1}{36}(x-60)^2-100D(x)= 36 1 ​ (x−60) 2 −100D, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, 36, end fraction, left parenthesis, x, minus, 60, right parenthesis, squared, minus, 100 How many seconds after diving will Ying reach his lowest altitude?

Answer 1

Answer:

60 seconds

Step-by-step explanation:

$D(x) = \dfrac{1}{36}(x-60)^2-100$

When Ying reaches his lowest altitude, his velocity is 0.

The velocity, v(x) is the tiome-derivative of D(x).

$v(x) = \dfrac{d}{dx}D(x)$

$v(x) = \dfrac{2}{36}(x-60)\times1 = \dfrac{x-60}{18}$                   (This was done by using chain rule)

Equating v(x) to 0,

$\dfrac{x-60}{18} = 0$

$x - 60 = 0$

$x = 60$

Answer 2

Answer:

60 seconds

Step-by-step explanation:

I did it on Khan :)