Curve y = cos πx

Exercises 1.4

For part (a), I must evaluate the given trigonometric function at each x-value. Yes?
Once this is done, how do I find the slope of the secant line PQ?

I would like you to do part (d) using a different color per lines. Cool?

 

yes, evaluate the given trigonometric function at each x-value and find points

 

Sounds good. I will play with B question until all three parts.

 

Part (a)

When x = 0, y = 0.
When x = 0.4, y = -0.4
When x = 0.49, y = -0.49
When x = 0.499, y = -0.499
When x = 1, y = -1
When x = 0.6, y = -0.6
Whe x = 0.51, y = -0.51
When x = 0.501, y = -0.501

From the above, the following points are formed and their slopes:

P(0.5, 0)
Q(0, 0)

Slope of PQ = 0

P(0.5, 0)
Q(0.4, -0.4)

Slope of PQ = 4

P(0.5, 0)
Q(0.49, -0.49)

Slope of PQ = 49

P(0.5,0)
Q(0.499, -0.499)

Slope of PQ = 499

P(0.5, 0)
Q(1, -1)

Slope of PQ = -2

P(0.5, 0)
Q(0.51, -0.51)

Slope of PQ = -51

P(0.5, 0)
Q(0.501, -0.501)

Slope of PQ = -501

1. Is this right?

2. How do I use the the points and slopes, considering the information is right, to answer part (b)?

Forget parts (c) and (d) for now.

 

part a is correct

b. find average slope for tangent line at P

 

Can you set up part (b) for me? Just set it up for me to do.

Thanks.

 

use point P(0.5, 0)
find average slope m (choose two slopes from part a
use point slope formula y-y[1]=m(x-x[1]

 

1. Why do I need point P?

2. Do you mean (slope 1 + slope 2)/2?

3. Why use the point-slope formula here?

Stuck here....

 

1. Why do I need point P?
point slope formula y-y[1]=m(x-x[1]) -> so, you need a slope m (you have (slope 1 + slope 2)/2) and the coordinates of the point of tangency (x[1],y[1]) which is P(0.5, 0) to find equation of tangent

2. Do you mean (slope 1 + slope 2)/2?
yes
3. Why use the point-slope formula here?
explained in 1.

Point-slope form = y − y 1 = m ( x − x 1 ) , where ( x 1 , y 1 ) is the point given and m is the slope given.

 

Let's do this:

Complete this problem for me parts b through d. I will then use your reply to answer a similar question in the same exercise problem set. We can then move forward.

 

b.

P(0.5, 0)
Q(0, 0)

Slope of PQ = 0

P(0.5, 0)
Q(0.4, -0.4)

Slope of PQ = 4


m=(0 + 4)/2=2

c.

P(0.5, 0)

y-0=2(x-0.5)
y=2(x-1/2)
y=2x-1


d. Sketch the curve, two of the secant lines, and the tangent line.

y=cos(pi*x)
y=2x-1-> tangent line
the secant lines:
A secant line is simply a linear equation and with two given points you can find the equation.
The two points on the secant line are:
Q(0.4, -0.4)
Q(0.49, -0.49), and slope is PQ = -1
Next, solve for the y-intercept:

y=49x+b ....find b, substitute (0.4, -0.4)
-0.4=-1*0.4+b
b=0

y=-x

View attachment 649

 

What can I say? You are amazing. It is my hope that you will be my online calculus tutor just as you have been for precalculus. Let's step off the calculus 1 boat for now. One course at a time is the best choice for someone like me who took precalculus in 1993. Long ago, right?

 

right

 

Calculus is not going anywhere. After precalculus, I will step into calculus 1 and take it from there. Good to know what lies ahead.