help! calculus!?

Question

1. point A moves along a unit circle at the rate of 2 units/second counterclockwise. What is the rate of change of the perimeter of the triangle whose vertices are points A, B(1,0) and C(-1,0) when A is at the point with coordinate (1/2, (sqrt3)/2).

2. Find the dimensions of the largest rectangle that cen be inscribed in a right triangle with sides of length 3 in., 4 in. and 5in. if two sides of the rectangle lies on the ;egs of the triangle.

3. Find the area of the largest rectangle that can be inscribed in an isosceles right triangle with legs of length a, if one side of the rectangle lies on the hypotenuse.

4. A fence 6 meters high is 6 meters away from a building that is 78 meters high. What is the length of the shortest ladder that can reach the building if one end of the ladder rests on the ground outside the fence.

5. Show that sinΘ≈Θ whenever Θ≈0 using differentials.

Answer 1

i only do the first and the last parts since Steve did the rest

ABC is a right angle triangle whose hypotenuse is the diameter of the circle
now, assume at time t the coordinates of C is (x,y)
since it lies on the circle
x^2 + y^2 = 1
perimeter of the triangle (P)= AB + BC +CA
= 2 + sqrt((1-x)^2+y^2) + sqrt((1+x)^2+y^2)

now we find the angular speed, assume c makes w angle with the positive x axis
dw/dt = r* v = 1* 2 = 2 rad/ s



first substitute x= cosw, y= sinw
then p = 2 + sqrt((1-cosw)^2+sinw^2) + sqrt((1+cosw)^2+sinw^2)
p = 2 + sqrt (2-2cosw) + sqrt (2+ cosw)
p-2 = sqrt (2-2cosw) + sqrt (2+ cosw)
square both side
(p-2)^2 = 2(1-cosw) + 2(1+cosw) + 4sqrt(1-cos^2w)
= 2(1-cosw) + 2(1+cosw) + 4sinw
now differentiate both side with respect to w
2(p-2)dp/dw = 2sinw -2sinw +4cosw = 4cosw
dp/dw = 2cosw/(p-2) ---eqn(1) where p-2 = sqrt (2-2cosw) + sqrt (2+ 2cosw)

question asks dp/dt,
by chain rule dp/dt = (dp/dw)* (dw/dt)

the give point C makes pi/3 with the positive x axis substitute w = pi/ 3 in eqn (1) for the answer.


5) we know the function sinx/ x approaches 1 as x approaches 0
so that sinx approximately equal to x for the small values of x
rigorous proof of the limit (sinx/x), x-->0 can be found in any books on elementary calculus.

or else we also know that dsinx/dx = cosx
ie limit ( sin(x + dx) - sinx)/ dx, dx---->0 = cos x

for small values of dx

[sin( x+ dx) -sinx]/ dx ~ cos x
if x = 0
(sindx - 0)/dx ~1
sindx ~ dx for small values of dx

ie sinx ~ x when x~0

Answer 2

2.
Draw the right-angled triangle with the right angle at the origin. The short side is along the y-axis, the long side is on the x-axis. The equation of the hypoteneuse of this triangle is:

y=-3/4x+3

(you get the slope by finding the slope between (0, 3) and (4, 0))

The area of the inscribed rectangle will be:

A=length x width
A= x (y)
A= x (-3/4x+3)
A= -3/4x^2 + 3x

Note that this is an upside down parabola (-3/4 < 0), so it will have a maximum.

Then you find the derivative, set it equal to zero, and you have the x-coordinate of the maximum. Sub it into the equation for the line and you will have the y-coordinate.

3. This question is exactly like question 2, except you draw the hypoteneuse centered on the origin, on the x-axis. This makes the right angle on the y-axis.

Because of symmetry, you can use half of the triangle to solve the problem. The dimensions of the half-triangle are:

hypoteneuse: a
base: a x sqrt(2)/2
(use pythagorean theorem on the original triangle, then half it because you are only using half the hypoteneuse as your base)
height: sqrt( a^2 - 1/2 a^2 ) = a * sqrt(3) / 2

Therefore, you follow the procedure from question 2 with the equation:

slope of line = - a x sqrt(3)/2 / a x sqrt(2)/2
= - sqrt(3) / sqrt (2)

y = -x sqrt(3/2) + a * sqrt(3)/2

4. I'm guessing that it has to reach the top of the building? Well, if you put the top of the building at (0, 78) and the top of the fence at (6, 6), then the shortest ladder will go through both of those points. It can be solved without calculus.

y = mx + 78
6 = 6m +78
6m = -84
m = -14

Therefore, y= -14x +78 is the equation of the shortest ladder.

The ladder touches the building at (0, 78), and the ground at:

0 = -14x + 78
x = 39/7

Therefore the length of the ladder is:
sqrt ( 78^2 + (39/7)^2)
=39 sqrt(197)/7

Answer 3

i might say shape is a factor of engineering meaning that a physically powerful draw close on math is mandatory, pre-cal is is like algebra to the subsequent point with the Pi chart and understanding approximately radians and attitude measures (which looks significant) yet Calculus is the learn of derivatives, (the slope of a line at a undeniable factor) type of pointless no rely what field of workd your going to, in case you question me.