If y=x²-4x and x=√2t²+1, find dy/dt when t=√2

Answer 1

3 possible ways to interpret the problem as written

depending on whether x=sqr(2t^2+1) or x=sqr(2t^2)+1 or x=(sqr2)t^2 +1

the 3rd alternative is the way the problem is literally written,

but the scope of the square root sign is ambiguous and 1st or 2nd

alternatives are possible

1st alternative:

y=x^2 -4x

x = sqr(2t^2+1)

y= (sqr(2t^2+1)^2 -4sqr(2t^2+1)

y =2t^2 +1 -4(2t^2+1)^(1/2)

dy/dt = 4t -8t(2t^2+1)^(-1/2)

= 4t -8t/sqr(2t^2+1)

when t = sqr2, replace t by sqr2 to get:

dy/dt = 4(sqr2) - 8sqr2/sqr(2(sqr2)^2+1)

= 4sqr2 -8sqr2/sqr(2(2)+1)

= 4sqr2 -8sqr2/sqr(5)

= 4sqr2-8sqr.4

= about 5.656 - 5.060

= about 0.596

= about 0.6

2nd alternative:

y=x^2 -4x

x =sqr(2t^2) +1= (sqr2)t +1

then

y = [(sqr2)t +1)]^2 -4x

= 2t ^2+ (2sqr2)t +1 -4(sqr2)t+1)

=2t +(2sqr2)t + 1 -(4sqr2)t -4

= 2t -(2sqr2)t -3

=(2-2sqr2)t -3

dy/dt = 2 -2sqr2

= about 2(1-1.414)

=about 2(-.414)

=about -0.828

for t=sqr2

dy/dt doesn't change, it's always

= 2-2sqr2 = about -0.828

3rd alternative, which is literally how it's written

y=x^2 -4x

x = (sqr2)t^2 +1

y = [(sqr2)(t^2+1)]^2 -4[sqr2)t^2+1)]

= 2t^4 + (2sqr2)t^2 + 1 -(4sqr2)t^2 -4

dy/dt = 8t^3 + (4sqr2)t -(8sqr2)t

= 8t^3 -(4sqr2)t

for t=sqr2

dy/dt = 8(sqr2)^3 - (4sqr2)sqr2

= 16sqr2 -8

= about 16(1.414)-8

= about 22.63-8

= about 14.63

Answer 2

Use chain rule: dy/dt = (dy/dx)(dx/dt) = (2x-4)(2sqrt(2)t)

Now sub in t = sqrt(2) Let's work out what x(sqrt(2)) is: sqrt(2)(sqrt(2))²+1 = 2sqrt(2)+1

y = 2(2sqrt+1)-4)(2sqrt(2))(sqrt(2)) = (4sqrt(2)-2)(4) = 16sqrt(2)-8

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