Calculus Questions...?

Question

A few Q's...
1) What is the limit of [cos (pi/2 + h)] / h as h approaches 0?

2) If f(x) = 10^x and 10^1.04 = 10.96, which is the closest to f '(1)?

3) What is the area of the largest isosceles triangle that can be drawn with one vertex at the origin and with the others on a line parallel to and above the x-axis and on the curve y=27-x^2?

Any help is appreciated. Thank you in advance.

Sorry, I just realized that the multiple choice responses are necessary to answer #2.
A. 0.24 B. 0.92 C. 0.96 D. 10.5 E. 24

Answer 1

1) What is the limit of [cos (π/2 + h)] / h as h approaches 0?

cos (π/2 + h) = cos(π/2)cos(h) - sin(π/2) sin(h) = -sinh

So lim h→ 0 {[cos (π/2 + h)]/h}

= lim h→ 0 {[-sin(h)]/h}

= - lim h→ 0 {[sin(h)]/h}
= -1

2) If f(x) = 10^x and 10^1.04 = 10.96, which is the closest to f '(1)?

f'(1) ≈ {f(1.04) - f(1)}/{1.04 - 1}

≈ (10.96 - 10)/0.04

= 0.96/0.04

= 24 ← Option E.

NOTE:
f'(x) = ln(10) *10^x (as 10^x = e^(x ln10))

So f'(1) = ln10 * 10^1 ≈ 23.03

3) What is the area of the largest isosceles triangle that can be drawn with one vertex at the origin and with the others on a line parallel to and above the x-axis and on the curve y=27-x²?

y = 27 - x²

Let the base be the line joining the two points with the coordinates (x, 27 - x²) and (-x, 27 - x²) where |x| < √27

So its base length is 2x and its height is 27 - x²

So Area of triangle

A = ½ base * height

= ½ * 2x * (27 - x²)

= 27x - x³

dA/dx = 27 - 3x²

= 0 for stationary points

ie 3(9 - x²) = 0

So x = ±3

Ignoring x = -3 (as A < 0)

Thus maximum Area is when x = 3

ie maximum area = 27*3 - 3³

= 54 square units

This occurs when the base is on the line joining (-3, 18) to (3, 18) and the vertex is at the origin as defined.

Answer 2

1. Find the expression for the cosine of a sum of angles, and directly evaluate.

2. Note that 10^1.04 = f(1+0.04) .
f'(1) would be about [f(1.04)-f(1)]/0.04
I'm not too sure what's wanted here.

*** with added info: f'(1) is approx [10.96-10]/0.04 = .96/.04 = 24, which is one of the choices...

3. Needs better definition. Is that one vertex on the origin the 'unique' vertex, or one of the two?
Probably yes:
v1 on origin
v2 on a horizontal line
v3 on the parabola.

I drew a little sketch, and my intuition, suggests that ...
no, still not clear...

Answer 3

limit(h-->0) of [cos (pi/2 + h)] / h
= limit(h-->0) of [-sin (h)] / h
= -1
lim sin(h)/h when h--> 0 = 1