Question

determine the values of a,b,c, and d so that the function is differentiable at all values of x on the interval [0,200]. Express all answer in exact form.

piecewise function:
2/5x 0 less than or equal to x which is less than or equal to 100
ax^2+bx-400 100 is less than x which is less than or equal to 120
c-dsin(pi*x/20) 120 is less than x which is less than or equal to 200

Answer 1

first check for continuity at each junction. f(x) is continuous at x=n iff\[\lim_{x\to n}f(x)\text{ exists and is equal to }f(n)\]

Answer 2

How can I check for continuity if there are 2 unknown values in the 2nd and 3rd pieces?

Answer 3

take the derivatives of both the second and 3rd pieces. In order for it to be differentiable, the derivatives must be equal at x=120. This will give you a system of equations. Then, since the two parts must be equal at 120 (actually, their limits must be equal), you can say that ax^2+bx-400=c-dsin(pi*x/20). Use the system and solve.

Answer 4

Thank you. I did. I got 14400a+120b-400=c, then I took the derivative of the three original functions. I set the der of the 1st equal to the der of the 2nd. But im stuck!

Answer 5

what do you have so far for the setting the derivatives equal?

Answer 6

2/5=2ax+b and 2ax+b=dcospi*x/20 (pi/20)

Answer 7

all right. Now, for the first one, evaluate for x=100, since that is the boundary point.
For the second one, evaluate for x=120., since that is the boundary point.

Answer 8

That should get it simplified in terms of a, b, and d. Then you can solve for d (it'll probably be easiest to do it that way), and plug it into the equations you did earlier.

Answer 9

I solved for the 1st and got 2/5=200a+b
I solved for the 2nd and got 240a+b=dcos6pi*pi/20

Answer 10

it is be 200(a+b) and 240(a+b), right?
This is gonna be a bit complicated, but you want to solve for (a+b). Once you have two equations with (a+b)=..., you can set them equal to each other.

Answer 11

Why is it 200(a+b) and 240(a+b)?
The unsubstituted equation is 2ax+b

Answer 12

oh... never mind. i thought it would've been easier that way. So. Then just solve for b in both equations, hen set them equal.

Answer 13

Sigh, for the 1st b=2/5-200a..for the 2nd b=dcos6pi*pi/20-240a

Answer 14

sorry this is such a tough problem.
now, just write down all the equations you know, and see if you can eliminate and reduce so you can solve for one...then another...then another...then the last one.

Answer 15

But I have 7 equations

Answer 16

ooh...any duplicates you can get rid of (alternate forms, etc.)?

Answer 17

440=10000a+100b
14400a+120b-400=c
2/5=200a+b
240a+b=dcos6pi*pi/20

Answer 18

OK...you can rearrange #1 and #3 so it is solving for a or b (b is probably easiest):
b=2/5+200a -- 100b=440-10000a
--> 100(2/5-200a)=440-10000a
-->40-20000a=440-10000a
-->-10000a=400
-->a=-.04
then, b=2/5+200(a) --> b=.4+8=8.4
so then, by that, c=14400(.04)+120(8)-400=1136
finally, I'll let you solve for d on you own.
These are some pretty weird values, so just double check all your (and my) work.

Answer 19

2/5-200a=b

Answer 20

oops...that mixes it up. Sry, got to go now. Good luck!

Answer 21

Thanks for all your help

Answer 22

y = 2/5x

y' = 2/5

-----------------------

y = ax^2+bx-400

y' = 2ax+b

In order for the two pieces to be differentiable (as one bigger function), the outputs of the derivative functions must be equal, so

2ax+b = 2/5

when x = 100, so we know that

2(100)a + b = 2/5

200a + b = 2/5

1000a + 5b = 2

===============================

Since we want the two pieces to join to form a differentiable function, this means that the two pieces must connect and form a continuous function (since a discontinuous function is not differentiable)

So if x = 100, then 2/5x = 2/5(100) = 40

which means that ax^2+bx-400 must equal 40 when x = 100


ax^2+bx-400

a(100)^2+b(100)-400 = 40

10000a + 100b - 400 = 40

10000a + 100b - 400 - 40 = 0

10000a + 100b - 440 = 0

10000a + 100b = 440

20(500a + 5b) = 440

500a + 5b = 440/20

500a + 5b = 22

============================================

We now have a system of 2 equations with 2 unknowns


1000a + 5b = 2
500a + 5b = 22

Solving for 'a' and b gives us



1000a + 5b = 2
500a + 5b = 22



1000a + 5b = 2
-500a - 5b = -22
--------------------
500a + 0b = -20


500a = -20

a = -20/500

a = -1/25


1000a + 5b = 2

1000(-1/25) + 5b = 2

-40 + 5b = 2

5b = 2+40

5b = 42

b = 42/5

Answer 23

So we now know that a = -1/25 and b = 42/5

Answer 24

To find c and d, you repeat what was done above (you force the two pieces to connect for both f and f') but you now you're forcing the 2nd and 3rd pieces

Answer 25

Ohh, so all ive done is completely wrong.

Answer 26

I'm not entirely sure what you've done so far, but you may have some correct parts

Answer 27

Ive done what is shown with BTaylor

Answer 28

i gotcha, he's on the right track since he got a = -0.04 (which is -1/25) and b = 8.4 (which is 42/5)

Answer 29

I just noticed that

Answer 30

I understand your work, but I am messing up with c and d

Answer 31

Since we know that a = -1/25 and b = 42/5, we can say

ax^2+bx-400

becomes

(-1/25)x^2+(42/5)x-400


So if y = (-1/25)x^2+(42/5)x-400

Then y' = -2/25x + 42/5

Answer 32

So then, how would I find c to then derive c-dsin(pi*x/20)?
I previously derived it first and c-dcos(pi*x/20) x pi/20

Answer 33

For the function to be differentiable on the second and third pieces, the output at the junction must be equal

so when x = 120, -2/25x + 42/5 is

-2/25x + 42/5

-2/25(120) + 42/5

-48/5 + 42/5

-6/5

This means that the derivative of the 3rd piece, which is -d*(pi/20)*cos(pi*x/20), must also equal -6/5 when x= 120


-d*(pi/20)*cos(pi*x/20) = -6/5

-d*(pi/20)*cos(pi*(120)/20) = -6/5

-d*(pi/20)*cos(6pi) = -6/5

-d*(pi/20)*1 = -6/5

-d*(pi/20) = -6/5

d = (-6/5)*(-20/pi)

d = 24/pi

Answer 34

Since d = 24/pi, we go from

c-dsin(pi*x/20)

to

c-(24/pi)*sin(pi*x/20)

Answer 35

To make the entire function differentiable over the interval [0, 200], it needs to be continuous on [0,200]

So at the junction, x = 120, the outputs of the 2nd and 3rd pieces must be the same for this to happen

When x = 120, the second piece is

(-1/25)x^2+(42/5)x-400

(-1/25)(120)^2+(42/5)(120)-400

(-1/25)(14400)+(42/5)(120)-400

-1/25+42/5-400

-9791/25

The output of the third piece must also be -9791/25 when x = 120

c-(24/pi)*sin(pi*x/20)

c-(24/pi)*sin(pi*120/20) = -9791/25

c-(24/pi)*sin(6pi) = -9791/25

c-(24/pi)*0 = -9791/25

c = -9791/25

Answer 36

So we can add that c = -9791/25 and d = 24/pi

Answer 37

Sigh, thank you soooo much. If youre not too busy, id appreciate if you give me a min to look over this and ask any questions I may have.

Answer 38

sure it's a lot to go over and take in

Answer 39

So, in a nutshell..

Answer 40

the der of the 1st and 2nd were set equal to eachother at x =100

Answer 41

then 100 was inputted into the original 1st func and output was 40, then the 2nd original func was solved equal to 40 at x equals 100

Answer 42

then the equations from setting the 1st and 2nd der and 2nd and 3rd der equal was solved as a system, this yielded the a value

Answer 43

then the a value was inputted into the equation found from setting the 1st and 2nd der equal to eachother, this yielded the b value

Answer 44

then the original 2nd function was substituted with the known a and b values, the der of that was taken and then solved at x=120

Answer 45

then the der of the 3rd function was taken, x was 120 and the equation was solved equal to -6/5 since -6/5 was the result of solving the der of the 2nd function at x=120

Answer 46

by setting the der of the 3rd function at x=120 equal to -6/5, d was solved for..yielding 24/pi

Answer 47

24/pi was then inputted into the original 3rd function

Answer 48

the original 2nd function was solved for x=120 with the known a and b values inputted, this yielded -9791/25

Answer 49

then the original 3rd function was solved for at x=120 with the known d value inputted set equal to -9791/25

Answer 50

this yielded that c is equal to -9791/25

Answer 51

You are correct

The general outline is

plug in x = 100 into the first two pieces, they should spit out the same output value

plug in x = 100 into the derivatives of the first two pieces, they should spit out the same output value

This gives you a system of 2 equations of 2 unknowns, which allows you to find 'a' and 'b'

---------------------------------------

plug in x = 120 into the last two pieces, they should spit out the same output value

plug in x = 120 into the derivatives of the last two pieces, they should spit out the same output value

Again you'll get 2 equations with 2 unknowns and you can use them to find c and d.

Answer 52

Yes!! So a=-1/25, b= 42/5, c=-9791/25, and d=24/pi

Answer 53

correct

Answer 54

keep in mind that -1/25 = -0.04 and 42/5 = 8.4

so BTaylor is correct, but the answer of -1/25 is preferred over -0.04 because they want the answers in exact form

Answer 55

Correct.

Answer 56

I am so thankful. You have no clue. Thank you!!!

Answer 57

you're welcome