<p>HEY GUYS,
WOULD U MIND ME ASK SOME QUESTIONS ABOUT MATH?
HERE WE GO,</p>
<p>Q1:a point P is moving along the curve whose equation is y=√x. suppose that x is increasing at the rate of 4 units/s when x=3, how fast is the angle of inclination of the line segment from P to (2,0) changing at this instant?</p>
<p>Q2:a projectile is launched upwaed from ground level with an initial speed of 60m/s
(a)how long does it take for the projectile to reach its highest point?
(b)how high does the projectile go?
(c)how long does the projectile go?
(d)what is the speed of the projectile when it hits the ground?
Use the results of these questions to make a conjecture about the relationship between the initial and final ground level and returns to ground level. Prove your conjecture.</p>
<p>Q3:a closed, cylindrical can is to have a volume of V cubic units. Show that the can of min surface area is achieved when the height is eaual to the diameter of the base.</p>
<p>Q4:a box-shaped wire frame consists of two identical wire squares whose vertices are connected by four straight wires of equal length. if the frame is to be made from a wire of length L. what should the dimensions be to obtain a box of greatest volume?</p>
<p>We know x=3 and dx/dt=4 so plugging these in we get</p>
<p>dθ/dt=-5√3/6 rad/s</p>
<p>Q2: The equation is -4.9t^2 + 60t.
a) Highest point is when v=0, so take derivative and set equal to zero
-9.8t+60=0
t=6.12s
b)Plug a) back into x(t)
-4.9(6.12)^2+60(6.12)=183.7m
c)Just twice a), so 12.24s
d)v(12.24)=-9.8(12.24)+60=-59.95m/s
Velocity at t=0 and t=12.24 have same magnitude, just different directions. Indeed v(0)=59.95m/s and v(12.24)=-59.95m/s.</p>
<p>Hopefully I did those right lol. Sorry they weren't very thorough explanations.</p>
<p>Q3: V=π(r^2)h, so h=V/(πr^2).
Substitute this into surface area for h (A=2πrh+2πr^2) and we get A=2V/r+2πr^2.
This we differentiate and set equal to zero b/c we are looking for a critical point:
A'=-2V/r^2+4πr=0
We can multiply by r^2 because we know r cannot =0:
-2V+4πr^3=0 and solve for r
r=cbrt(V/2π).
This we substitute back into h=V/(πr^2) for r and simplify and get h=cbrt(4V/π).</p>
<p>If we multiply r by 2 (because we want d) we see that h=d.</p>
<p>Q4: Call sides of squares s and the width joining the two squares w.
So we have equation L=8s+4w, solving for w:
w=(L-8s)/4
We can see that V=ws^2, and substituting the previous equation for w into this and simplifying yields:
V=Ls^2/4-2s^3
Differentiate, set equal to zero:
V'=Ls/2-6s^2=0
And solve for s.
s=L/12
Substituting back into the first equation for s gives:
w=L/12</p>
<p>So dimensions are L/12 x L/12 x L/12.</p>
<p>Which makes sense because a cube has maximized volume.</p>