differential

[Eng. Ahmed Said]

please answer me

find the point(s) on the curve Y = tan half X, X belongs to]0,360[ at which the tangent is perpendicular to the straight line Y=8-X

if the tangent to the curve of the circle X square+Y square=25 at the point (3,4) intersects the X-axis at the point C and intersectsthe Y -axis at the point D . prove that area of triangle COD equals 625/24 unit of area where O is the origin

If Y =2tanx/tan square x-1 Find
Y at X=15,then determine the measure of the posivite angle which the tangent to the curve at X=15 makes with the posivite direction of the Y-axis

if Y=(X square-2/X square+2)n, then prove that dy/dx =8nxy/X square-4
find dy/dx of each of the following composite functions

if V(t)=t square - 2t+3 is the relation between the velocity of a body V in cm/sec . and the time t in seconds.
calculate the average rate of change of the velocity of the body during the fifth second and evaluate the rate of change of the velocity at t=4

[MAN UTD]

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يعنى

قطة

[Eng. Ahmed Said]

أرجوكم اى حد يرد بسرعة

[نعمةمحمد محمود]

لا اله االا الله لحنا مش لغات

[geokenda]

دى كتاب النماذج
the answer
y=tanhalf x
y'=half sec square half x
m=1
cos square half x= half cos half x =+or-1/root2
x=90 or 270

[MathPrince]

(1) y' = (1/2) sec^(2)[x/2]
y=8-x (so) m = 1 (so) 1 = (1/2) sec^(2)[x/2] (so) 2 = sec^(2)[x/2] (so) sec [x/2] = sq. r. (2) (so) x/2 = 45 (so) x = 90 / 270

(2) y = sq. r. [25 - x^(2)] (so) y' = [-2x] / 2 sq. r. [25 - x^(2)] (so) y' (at x=3) = -6 / 8 (so) m = -2/3
eq. of tang. [y - 4] / [x - 3] = -2/3 (so) 3y + 2x - 6 = 0
at x=0 "intersection with y-axis" y = 2 point is (0,2)
at y=0 "intersection with x-axis" x = 3 point is (3,0)
the area of the triangle = (1/2) (2) (3) = 3 unit square

(3) y = - [tan x] / [1- tan^(2) x] at x= 15 (so) y = - tan 2x = - tan 30 = - 1 / sq. r. (3)

(5) v(h) = [(5+h)^2 - 2(5+h) +3] - [25 - 10 + 3] = h^(2)
A(h) = v(h) / h = h
you can get the averae as an exercise

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