How do you find the constants a and b?

#y# = #a/sqrt"1-bx"#

Given #x# = 3, #y# = 1 and #dy/dx# = #-1/8#

1 Answer
iamgadmarconi
Feb 4, 2018

#a# = #12#
#b# = #-143/3#

Explanation:

We are given the equation #y# = #a/sqrt(1-bx#
We know that x = 3, and that y = 1.

We can, therefore, plug the number's in to form the equation

#1# = #a/sqrt(1-3b#

If we move the variables around using algebra we get

#sqrt(1-3b# = #a#

If we solve for B and re-arrange the variables we get

#(1-a^2)/3# = #b#

The problem also gives us #a'# (derivative of #a#) = #-1/8#

We can differentiate #a# = #sqrt(1-3b# by using the chain rule ( F'(x)=f'(g(x))g'(x) )

The derivative of #sqrt(x)# is #1/(2*sqrt(x)#

We therefore get #a'# = # -3/(2*sqrt(1-3b#

If we rearrange the variables, we get

#-3/(a')# = #2*sqrt(1-3b#

= #-3/(2a')# = #sqrt(1-3b#

= #-3/(2a')^2# = #1-3b#

= #9/(2a')^2# = #1-3b#

= #1- 9/(2a')^2# = #3b#

= #1/3 - 9/(12(a')^2)# = #b#

If we recall our equation from earlier: #(1-a^2)/3# = #b# we can set the two equal to each other to produce

#(1-a^2)/3# = #1/3 - 9/(4(a')^2)#

If we simplify the expression #1/3 - 9/(12(a')^2)# we get #1/3 - 3/(4(a')^2)#

Our new equation is therefore #1/3 - 3/(4(a')^2)# = #(1-a^2)/3#

#(1-a^2)/3# is the same as #1/3 -a^2/3#

The #1/3#'s cancel out and we are left with

#- 3/(4(a')^2)# = #-a^2/3#

If we cancel the negatives we get

# 3/(4(a')^2)# = #a^2/3#

We can now get the #a#'s on one side

#(4(a')^2) * a^2/3# = 3

= #(4(a')^2) * a^2# = 9

We can now substitute the #-1/8# for #a'#

#4*(-1/8)^2 * a^2# = 9

#(4a^2) / 64# = 9

If we simplify the equation we get

#(a^2) / 16# = 9

#a^2# = 144

#a# = 12

Now that we found #a#, we can plug it in back in #b#

#(1-12^2)/3# = #b#

If we solve for #b# we get

#b# = #-143/3#.

We can now check our solutions!

#1# = #12/(sqrt(1-(-143/3)*3))#

#1# = #12/(sqrt(1-(-143))#

#1# = #12/(sqrt(144)#

#1# = #12/12#

Easy :)

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