g(x)=sinx and its tangent line help?

Question

Tangent line of sinx?
1. Graph g(x)=sinx
a. Sketch a tangent line at the point (0,0). Write a formula for the slope of the secent line joining point (s, sinx) and (0,0)
b. Evaluate the formula for x=.1 and x=.01
c. Find exact slope of the tangent line to g at point (0,0)
d. Use dirivitive to confirm answer
e. Find equation of the tangent line to g at point (0,0)

Any help would be great, I had a family emergency and missed a ton of class, and this would help out a lot.


What formula to use???

Maybe walk me through this

Answer 1

I see what you're trying to do here. You're finding the derivative of g(x) = sin x at x=0 by a limit process. You have a typo in (a); it should read "point (x, sin x)" instead of "point (s, sin x)".

They gave you all the steps. I'll walk you through it, providing some details.

1. Graph one cycle of the sine function from (-pi) to pi, but spread it out, and on lined loose-leaf paper, let the max (+1) and min (-1) be about three lines above & below the x-axis. You need space for this. Also, mark off +/-pi/2 (max/min) on the x-axis.

(a) The tangent line will just be a 45-degree line through the origin. Formula is y=x. Very simple.

Your two points on the secant line are (0,0) and (x, sin x). The slope of the line connecting those two points is

m = (y-yo)/(x-xo) = (sin x - 0)/(x - 0) = (sin x)/x

(b) m(0.1) = (sin 0.1) / 0.1 = 10 sin 0.1 = 0.99833 (use radians!!)
m(0.01) = (sin 0.01) / 0.01 = 100 sin 0.01 = 0.99998

Notice that as x approaches zero, m (the slope) approaches one.

(c) The exact slope is the limit, as x approaches zero, of (sin x)/x. Here you have to be careful. If you plug in zero, you get (sin 0)/0 = 0/0, an indeterminate form. You can use L'Hopital's rule, differentiating top & bottom to get, for x --> 0,

lim(sin x)/x = lim(cos x)/1 = 1/1 = 1

but in the context of this problem, that may be cheating. There's a geometric proof of that limit in my calculus book, and if you need it, email me. Meanwhile, we'll use L'Hopital's shortcut, above, to say the exact slope of the tangent line at the origin is 1.

(d) If g(x) = sin x, then dg/dx = cos x, and dg/dx = 1 for x=0.

(e) The equation of the tangent line, using slope-intercept, is

y = mx + b = x

That's everything. The only question is whether you're allowed to use derivatives in part (c).

Answer 2

y = 2x³ ‒ 3x² dy/dx = 6x² ‒ 6x This represents the slope of a tangent line once the value of x is chosen. To be parallel two lines must have the same slope. So it must be the case that 6a² ‒ 6a = 6b² ‒ 6b → 6a² ‒ 6b² ‒ 6a + 6b = 0 This factors as 6(a – b)(a + b – 1) = 0 Therefore either a – b =0 → a = b OR a + b – 1 = 0 → a + b = 1 as suggested =============== It bothers me that you have a question that requires Calculus, but you do not know the characteristics of parallel lines.