a curve c is defined by the parametric equations x =1/(√(t+1)) and y = t/(t+1) for t≥0. rest in comments

Question

(a) find dy/dx in terms of t.

(b) find an equation of the tangent line to C at t=3.

(c) set up but do not evaluate a definite integral representing the length of the curve C on the interval 0≤t≤1.

(d) find an equation for the curve C in terms of x and y.

Answer 1

a)

Given x = 1/√(t+1) and y = t/(t+1) we can compute dy/dx using "chain-rule"...

dy/dx = dy/dt * dt/dx (where we treat the derivatives as ratios of small numbers)

dt/dx = 1 / (dx/dt)

So...

dy/dx = dy/dt / dx/dt

dy/dt = d/dt (t/(t+1)) = d/dt ( t * (t+1)^-1 ) = (t+1)^-1 - t*(t+1)^-2 --> 1/(t+1)²

dx/dt = -1/2 * (t+1)^-3/2

dy/dx = 1/(t+1)² / -1/2 * (t+1)^-3/2 ---> (t+1)^3/2 / (t+1)² * -2 ---> -2/√(t+1)

b)

dy/dx = y' = slope --> y'(3) = -2/2 = -1

c)

L = arc length = int(sqrt( 1+ (f'(x))²) * dx --> int_{lb=0}^{ub=1}(sqrt( (dx/dt)² + (dy/dt)²) * dt

Plugging in what we know from dy/dt and dx/dt yields...

L = int_{lb=0}^{ub=1}(sqrt( 1/(4(t+1)³) + 1/(t+1)⁴ )*dt

d)

We can use the definitions of x(t) and y(t) to invert and find y(x)...

x² = 1/(t+1) --> 1/x² = t+1 --> 1/x² - 1 = t --> (1 - x²)/x² = t

1/y = (t+1)/t = 1 + 1/t --> 1/y - 1 = 1/t --> (1-y)/y = 1/t --> y/(y-1) = t

Now equate the two expressions since they both represent 't'...

(1 - x²)/x² = y/(y-1)

And simplify...

If you want, solve for t in terms of x and then plug into the representation for y to find y(x) quicker...

y = ((1 - x²)/x²) / ((1 - x²)/x² + 1 )

And simplify again if so desired...