# Curve y = cos πx

Discussion in 'Calculus' started by nycmathguy, Oct 10, 2021.

1. ### nycmathguy

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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? nycmathguy, Oct 10, 2021

2. ### MathLover1

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

MathLover1, Oct 10, 2021
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3. ### nycmathguy

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

nycmathguy, Oct 10, 2021
4. ### nycmathguy

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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.

nycmathguy, Oct 10, 2021
5. ### MathLover1

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part a is correct

b. find average slope for tangent line at P

MathLover1, Oct 10, 2021
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6. ### nycmathguy

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

Thanks.

nycmathguy, Oct 11, 2021
7. ### MathLover1

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

MathLover1, Oct 11, 2021
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8. ### nycmathguy

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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....

nycmathguy, Oct 11, 2021
9. ### MathLover1

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1. Why do I need point P?
point slope formula y-y=m(x-x) -> so, you need a slope m (you have (slope 1 + slope 2)/2) and the coordinates of the point of tangency (x,y) 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.

MathLover1, Oct 11, 2021
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10. ### nycmathguy

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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.

nycmathguy, Oct 11, 2021
11. ### MathLover1

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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

Last edited: Oct 11, 2021
MathLover1, Oct 11, 2021
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12. ### nycmathguy

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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?

nycmathguy, Oct 11, 2021
13. ### MathLover1

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right

MathLover1, Oct 12, 2021
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