Answer to Question #209555 in Calculus for Delmundo

Question

1. Determine the equation of the family of the curves whose slope is

3x – 5.

Also find the equation of the member whose curve passes through point

(1, 1).

2. A body is thrown vertically upward from the ground with an initial velocity

of 64 ft/sec. Find the maximum height attained by the body. Regardless of

frictional force, determine the total time the body returns to the ground.

(note: a = 32ft/sec2)

Answer 1

1. The slope of the curve is given by 

$slope=m=\dfrac{dy}{dx}=3x-5$

$dy=(3x-5)dx$

Integrate both sides with respect to $x$ 

$\int dy=\int(3x-5) dx$ $y=\dfrac{3}{2}x^2-5x+C$

The equation of the family of the curves is

$y=\dfrac{3}{2}x^2-5x+C$

As the curve passes through the point $(1,1)$ we have,

$1=\dfrac{3}{2}(1)^2-5(1)+C$ $C=\dfrac{9}{2}$

The equation of the curve pasiing through the point $(1,1)$ is

$y=\dfrac{3}{2}x^2-5x+\dfrac{9}{2}$

1.

$v(t)=v_0-at$

The equation of the motion of the body along the positive y-axis is

$y(t)=y_0+v_0t-\dfrac{at^2}{2}$

The body attains the maximum height when $v=0$

$v_0-at=0=>t=\dfrac{v_0}{a}$

Substitute

$y_{max}=y(\dfrac{v_0}{a})=y_0+\dfrac{v_0^2}{a}-\dfrac{v_0^2 }{2a}=y_0+\dfrac{v_0^2}{2a}$

Given $y_0=0, v_0=64\ ft/sec, a=32\ ft/sec^2$

$y_{max}=0 \ m+\dfrac{(64\ ft/sec)^2}{2(32\ ft/sec^2)}=64\ ft$

The body returns to the ground when $y(t)=0, t>0$

$v_0t-\dfrac{at^2}{2}=0, t>0$

$v_0-\dfrac{at}{2}=0$

$t=\dfrac{2v_0}{a}$

$t=\dfrac{2(64\ ft/sec)}{32\ ft/sec^2}=4\ sec$

The maximum height attained by the body is $64\ ft.$

The body returns to the ground $4\ sec$ after the start.