1. Find the derivitive of 3x^2tan(x)/(sec(x))
I put ((6xtan(x))+sec^2(x)3x^2(sec(x))-(sec(x)tan(x))(3x^2tan(x)))/(sec^2(x))
Not sure where I messed up...
2. (1 pt) If a projectile is fired from ground level with initial velocity of 3775 ft/sec and an inclination angle , and if air resistance can be ignored, then its range---the horizontal distance it travels---is about R=(1/16)02sincos
feet. What value of (in radians) maximizes the range?
I do not know how to do this problem..
Hello, lmao!
. . . . . . . . . . . . . . . . .3x²·tan(x)
(1) Differentiate: .y .= .------------
. . . . . . . . . . . . . . . . . . sec(x)
. . . . . .sec(x)·[6x·tan(x) + 3x²·sec²(x)] - 3x²·tan(x)·sec(x)·tan(x)
y' . = . ---------------------------------------------------------------------
. . . . . . . . . . . . . . . . . . . . sec²(x)
. . . . . .6x·sec(x)·tan(x) + 3x²·sec³(x) - 3x²·sec(x)·tan²(x)
y' . = . ------------------------------------------------------------
. . . . . . . . . . . . . . . . . . . sec²(x)
. . . . . . . . . . . 3x·sec(x)·[2·tan(x) + x·sec²(x) - x·tan²(x)]
Factor: .y' . = . ---------------------------------------------------
. . . . . . . . . . . . . . . . . . . . .sec²(x)
. . . . . . . . . . . . 3x·[2·tan(x) + x[sec²(x) - tan²(x)]]
Reduce: .y' . = . -------------------------------------------
. . . . . . . . . . . . . . . . . . . sec(x)
. . . . . . . . . . . . . .3x·[2·tan(x) + x]
Therefore: .y' . = . ---------------------
. . . . . . . . . . . . . . . . . sec(x)
I'm presuming the formula is R=(1/16)*2sincos ?
Let's call alpha = q.
Note that 2sin(q)cos(q) = sin(2q) so
R = (1/16)*sin(2q)
The non-Calculus way:
I'm afraid you are reduced to graphing R as a function of q. However, note that R is merely a sine function, so we can easily tell where the maximum is. Note that the sine function reaches its maximum value at an angle of pi/2 rad (90 degrees). Thus the maximum R value will be for
2q = pi/2 rad
q = pi/4 rad (45 degrees)
The Calculus way:
What value of q maximizes R? Take the first derivative of R with respect to q and set it equal to 0:
dR/dq = (1/16)*2*cos(2q) = 0
Thus
cos(2q) = 0
2q = pi/2 rad
q = pi/4 rad.
-Dan
Hello, lmao!
Your post has some glitches; I have to guess what's missing.
And what can be seen is confusing . . .
2. If a projectile is fired from ground level with initial velocity of 3775 ft/sec
and an inclination angle θ, then its range is about: R = [???]
What value of θ (in radians) maximizes the range?
The horizontal displacement is: .x .= .(3775·cosθ)t
The vertical displacement is: .y .= .(3775·sinθ)t - 16t²
The projectile is on the ground when y = 0.
. . (3775·sinθ)t - 16t² .= .0 . → . t[3775·sinθ - 16t] .= .0
. . Hence: .t = 0 .and .t = 3775·sinθ/16
The range is: .x .= .3775·cosθ·(3775·sinθ/16) .= .(3775²/16)·sinθ·cosθ
. . x .= .(3775²/32)·(2·sinθ·cosθ) .= .(3775²/32)·sin(2θ)
The range is a maximum when: .sin(2θ) = 1 . → . 2θ = π/2 . → . θ = π/4