<p>FRQ #3<br>
dx/dt= .8
dy/dt= 3.6-9.8t</p>
<p>y=11.4 when t= 0
a) max y?
b)t when y=0
c) total distance travel
d) angle make with the x axis when y=0</p>
<p>my solutions
a) set dy/dt=0 find critical point t=.367
max y= 11.4+ intergral(3.6-9.8t,t,0,.367)
the answer is 12.0612</p>
<p>b) dy=(3.6+9.8t)dt
y=3.6t+4.9t^2+C
11.4= 0 + C ===> C=11.4
set 0=3.6t+4.9t^2 + 11.4
t=1.936</p>
<p>c) intergral( sqrt (.8^2+(3.6-9.8t)^2),t,0,1,0.936)
12.9423 meters i have a feeling that this is wrong…lol</p>
<p>d)this is the most confusing s*** of the questions</p>
<p>this is wat i did </p>
<p>slope dy/dx= (3.6-9.8t)/.8 right?</p>
<p>==take anti derivative of x and find x when t=1.936 (remember that initial condition of x is 0, so C=0) so x=1.549 when t=1.936 (omg i just check my cal again, i found out that i put 1.938 lol stupid me)</p>
<p>==y=0 when t=1.936 from part b)</p>
<p>slope of tangent line at t=1.936 is -19.216</p>
<p>equation of tangent line is y-0= -19.216(x-1.49) </p>
<p>find y intercept y=28.632</p>
<p>(draw a right triagle with hyp is the tangent line)</p>
<p>tanθ = 28.632/ 1.549
θ=tan-1 (28.632/1.549)=1.52 (pi/2 is 1.57) so i think this is a good guess lol…(f collegeboard if they dont give me some points, this is the best bs i can think of)</p>
<p>anyone need #6? thats the only one beside #3 i remember…</p>