The direction ratios of perpendicular to
x + 2y - z = 3
are:
1:2:-1 and so direction cosines are 1/(√6), 2/(√6) and -1/(√6) --------------------- (1); and
Similarly, the direction cosines of perpendicular to
2x - z = 0 are:
2/(√5), 0 and -1/(√5) ----------------------------------------…
Cosine of the acute angle, θ between their pependiculars or teh planes from (1) and (2) are given by
cos θ = |[{1/(√6)}*{ 2/(√5)}] + [{2/(√6)*0}]+[{-1/(√6)}*{ -1/(√5)}]| = 3/(√30) = 0.5477 or
Required θ = 56.79 or 56.8°
I am not able to solve the next part.
3. show that integral -1 to 0 of 2x + 1 / ( x - 2)( x^ 2 + 1) dx = -3/ 2 ln 2
expression not clear use parentheses properly.
4. integral of sec^2 (x) / tan x it is = log(tan x)
5. the angle theta satisfies the equation tan2 theta = sin theta or
(2tan θ)/(1 - tan² θ) = sin θ or
[(2sin θ)/{cos θ - ((sin² θ)/cos θ)}] = sin θ
So either sin θ = 0 ----------------------- (1) or dividing by sin θ both sides we get,
[(2cos θ)/{(cos² θ) - (sin² θ)}] = 1 or
cos θ= (2cos² θ - 1) [= cos (2θ)] ---------(2)
Hence proved what was required see (1) and (2).
6. the parametric equations of a curve are
x = t ^3 - e ^ -t, y = t ^ 2 - e ^ -2t^2
dx/dt = 3t² + e^(-t) and dy/dt = 2t+{(4t)*{e ^(-2t^2)}
dx/dt at t= 0, = 1 and dy/dt at t= 0, = 0
so dy/dx at t = 0, = [(dy/dt)/(dx/dt)] = 0
at t = 0, x = -1 and y = -1
So the equation of the tangent to the curve at the point where t = 0
y = -1 as its slope is 0.
Let A be (0, k)
x = 0 implies
t³ = e^(-t) ------------------ (1) or 3 ln t = -t substituting in
k= t² - e ^ -2t^2 = t² - [{e ^(-t)}^t]^2; substituting from (1) we get
k= = t² - [t³}^t]^2 = t² - t^(6t)
I am unable to proceed further.
the curve cuts the y axis at the point A
show that the value of t at A lies between 0 and 1