# Answer in Calculus for Delmundo #209555

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)

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.