Question

Part 1: Find the tangent line approximation to cos x at x=Ï€/4.

Part 2: To one decimal place, estimate the maximum value of the error on the interval 0≤x≤π/2.

please explain! Thanks :)

Answer 1

Do you know how to find the tangent of \(\cos x\) at a given point?

Answer 2

umm not sure :/

do you mean the tangent equation at cos x when x=pi/4? if so, would it be what it says in this link? http://www.wolframalpha.com/input/?i=tangent+equation+of+cos+x+at+x%3Dpi%2F4

Answer 3

Yes, that's what I mean. As I recall, you're taking a calculus class, right?

Answer 4

yes i am :)

and for this, it's linear approximation, so i'd be using the formula
f(a+∆x) roughly equals f(a)+f'(a)(x-a)?

Answer 5

Yes.

Answer 6

so how would I find part 1 and part 2? via linear approximation?

Answer 7

You start by finding \(f'(x)\)

Answer 8

of the equation cos x? or of the result from that link?

Answer 9

\[f(x) = \cos x\]\[f'(x) =\]

Answer 10

-sin x?

Answer 11

You are trying to approximate \(f(x) = \cos x\). To do that, you need to find the derivative, which gives you the slope of the tangent line at any point.

Answer 12

Yes. Now, what does \(-\sin(x) =\) when \(x = \pi/4\)?

Answer 13

ahh okay...

so -sin(pi/4) = -1/(√2) ?

Answer 14

Yes, it does. So that's the slope of the tangent line to \(\cos(x)\) at \(x=\pi/4\).

Now, we can do things from first principles, and construct the equation of the line through that point with that slope, and use that as our approximation, or we can use the formula which has already done so.

Answer 15

oh yay!

so do we use y-y1=m(x-x1) now? :/ or are we using f(a+∆x) = f(a)+f'(a)(x-a)?

Answer 16

Well, those are the two choices. However, if you think about it, they are equivalent.

Answer 17

More a matter of how you understand the situation — you get the same answer both ways.

Answer 18

ohh okay.. in this case, which would you recommend?

Answer 19

I tend to remember the basic constructs and re-derive more complicated ones as necessary, rather than trust my memory to be 100% accurate. So in this case, I would probably go the former route, and then say "oh yeah" after doing so and realizing that I could have remembered the other.

But really, they amount to the same thing.

Answer 20

okay so we have y-0=(-1/√2)(x-(pi/4)) ?

Answer 21

So, we know that \(x_1 = \pi/4\) and \(y_1 = \cos(x_1) = \cos(\pi/4) = \frac{1}{\sqrt{2}}\)

\[y - \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}}(x-\frac{\pi}{4})\]\[y = -\frac{1}{\sqrt{2}}x + \frac{\pi}{4\sqrt{2}} + \frac{1}{\sqrt{2}}\]
which is exactly what WolframAlpha gave us.

Answer 22

ohhh okay so that is the tangent line approximation? part 1 is: what you wrote above?

Answer 23

so our approximation could be written as \[y = g(x) = -\frac{1}{\sqrt{2}}x + \frac{\pi}{4\sqrt{2}} + \frac{1}{\sqrt{2}}\]

Answer 24

ahhh okay haha answered my question :P

woo! so part 2 now?

Answer 25

Here's a graph, I trust you can figure out which line is which!

Answer 26

Now we want to estimate the maximum error over that region. Any idea how we can compute the error at any given point?

Answer 27

the red one right?

ermm not sure... plug a number in?

Answer 28

what is the error? What does that term mean to you?

Answer 29

would that mean what is the exact equation/number/solution?

because what we just got was the approximation, so would we have to find the exact and see what the difference is or something like that?

Answer 30

Yes, the function - the approximation gives you the error.

Answer 31

okay :)

how can i find that ? do i use that equation we got above and plug in pi/4 wherever there's an x?

Answer 32

You're not getting this, are you?

\[Error(x) = f(x) - g(x)\]where \(g(x)\) is the function that we determined for our approximation. At \(x=\pi/4\), the error is 0. Everywhere else on that graph, the error is some number other than 0.

Answer 33

Here's a graph that includes the error.

Answer 34

yeah calc kinda confuses me :(

but i see now.. so g(x) was what we found earlier and f(x) is the exact equation?

Answer 35

oh the error is the curve at the bottom?

Answer 36

Yeah.

Answer 37

okay:)

so how would we find the max value rounded to one decimal place?

because it has to be an exact number right?

Answer 38

First, you need to figure out where the error is going to be a maximum, yes?

Answer 39

When you know where, then you can calculate the actual error.

Answer 40

is it from 0.5-1?

Answer 41

Even if you ignore the graph where I showed the error, it shouldn't be too terribly difficult to figure out the value of \(x\) where our approximation is furthest from the actual value of \(\cos x\)

Answer 42

pi/4√2 ?

Answer 43

what's that? your shoe size?

Answer 44

Look at the graph!

Answer 45

hahaha :P

at -0.2 ?

Answer 46

Where are the purple and blue lines close together? Where are they far apart?

Answer 47

1.0 and 1.2 ?

Answer 48

^farthest apart

Answer 49

My graph starts at \(x = 0\). It goes to \(x = \pi/2 \approx 1.5708\). At what value of \(x\) are the blue and purple lines separated by the largest horizontal distance?

Answer 50

That separation is the value of the error at that point.

Answer 51

sorry, vertical distance, not horizontal distance.

Answer 52

atx=0 ?

not sure if i'm saying that right... but this is where i see the largest vertical distance... is that right?

Answer 53

yes. at x = 0. Now, estimate, to 1 decimal place, the error. The convention is that you take the actual value of the function and subtract the approximated value to get the error.

Answer 54

0.2 ?

Answer 55

What is the value of \(\cos 0 =\)

Answer 56

cos0=1 ?

Answer 57

You tell me.

Answer 58

I'll wait until you're sure.

Answer 59

okay

yeah cos 0 = 1 :)

Answer 60

I may wander off and go to bed, but I'll wait :-)

Answer 61

haha :P

Answer 62

Okay. What is the value of the approximation at x = 0?

Answer 63

thanks!! yeah sorry this is taking so long!! i'm not understanding this too well :/ more practice and studying is needed!!

i got 1.817827516....

is that right?

Answer 64

so the error is 0.8 ?

Answer 65

No. Show me how you got that number. Does it look like the value from the graph?

Answer 66

\[y = g(x) = \frac{1}{\sqrt{2}} + \frac{\pi}{4\sqrt{2}} - \frac{x}{\sqrt{2}}\]\[g(0) = \frac{1}{\sqrt{2}} + \frac{\pi}{4\sqrt{2}} - \frac{0}{\sqrt{2}} \approx \frac{1}{1.414} + \frac{3.14}{4*1.414} =\]

Answer 67

no :/

umm \[-\frac{ 1 }{ \sqrt{2} }(0)+\frac{ \pi }{ 4\sqrt{2} }+\frac{ 1 }{ \sqrt{2} }\]

Answer 68

ohh i accidentally plugged the wrong way.. oops

so from there you should actually get 0.5551626591 ?

so the error is (when rounded to one decimal place) 0.6?

Answer 69

So that's about 0.707 + 3.14/5.656

Answer 70

oh so you get 1.262162659

and 1.262162659-1 = 0.2621626591

so the error is about 0.3 ?

Answer 71

This is an example of how the proliferation of calculators has destroyed the good number sense of today's students...and then if you can't operate the calculator accurately, you not only don't get the right answer, but you don't even have a sense of whether you should be questioning whether you got the right answer! :-(

Answer 72

Once again. Take the actual value of the function. Subtract the approximated value. What do you get?

Answer 73

yeah that and poor understanding of the subjects lol

i got 0.26216

:/ is that right?

Answer 74

What is the actual value of the function?

Answer 75

What is the approximated value of the function?

Answer 76

What is the actual value of the function - the approximated value of the function?

Answer 77

actual value is 1.26247
approx value is 1.26216

?

Answer 78

Actual value of \(\cos 0 = \)

Answer 79

oh 1-1.26 ? so the difference is 0.3?

Answer 80

I give up.

\[\cos 0 = 1\]\[g(0) = 1.26\]\[\cos 0 - g(0) = 1-1.26 = -0.26\]

Answer 81

oh oops i meant difference is 0.26 so about 0.3 ?

Answer 82

yes what you said haha :P

Answer 83

I really don't know how I could have been any more clear. What does the expression "take x and subtract y" mean to you?

Answer 84

it means \(x-y\)

No, the difference is NOT 0.26!!!

Answer 85

oh i forgot the - sign.... whoops!!! sorry!!

so -0.26, so about -0.3 ?

yikes... so sorry!!

Answer 86

Yes, thank you.

Answer 87

ah sorry again! but thanks so much for your help!!

Answer 88

One of the most valuable lessons to be gained from taking math classes is not the Pythagorean theorem, or the fundamental theorem of calculus, or anything like that. It's the importance of working and reading carefully and methodically.

Answer 89

yes i agree! i will definitely be reading and working a lot more carefully from now on!! sorry about that! i probably made you really exasperated at my thinking just now haha.. i apologize for that!! but thank you so much for continuing to explain it to me!!

Answer 90

You say that you forgot the minus sign, but I'll bet you help on another problem that you were subtracting 1.26 - 1 because the approximation is higher on the graph than the function...

Answer 91

I don't give up easily when I think the student is capable of understanding what I have to say :-)

Answer 92

But while many would have just told you the answer, I think you really learn it better if you figure it out yourself, even if it raises my blood pressure in the meantime :-)

Answer 93

It's too easy to look at what someone else did and convince yourself that you understand it, but then when they aren't there and you have to do it yourself, suddenly it doesn't seem nearly so clear...

Answer 94

hahaa i like the bet :P

but at first, i did do 1.26-1, but then i did it the x-y thingy... but i thought it couldn't be negative for some odd reason (<--my brain malfunctioned a bit there just now lol)

but maybe let's call that bet off? :P hahaa since technically i lost it as i did it incorrectly lol :P

thank you:) i really appreciate it!!!sorry about the blood pressure thing haha but i'm really glad you kept explaining it to me :) i think that helps a lot more in the long run :)

Answer 95

If you go look at the 2nd graph I provided again, you'll notice that the error is always negative, except at the point where it just kisses the x-axis at x = 0.707 (pi/4) and there it is 0.

Answer 96

it seems a bit counter-intuitive (and indeed, I had to go verify my recollection) that the error is negative when the approximation came out to be larger than the function, but that's the way it is. I don't doubt that this caused some of your desire to give me positive numbers for the maximum error :-)

Answer 97

and if the problem had said "the magnitude of the error" instead of "the maximum error" your answer would have been fine.