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# Calc- parametric curves- finding tangents?

Hi I do not understand how to find the tangent from complex numbers so I was wondering if you could help?

1. find an equationof the tangent to the curve at the point corresponding to the given value of the parameter.

x= e^sq root of t , y=t-lnt^2 , t=1

2. find an equation of the tangent to the curve at the given point by (a) without eliminating the parameter (b) by first eliminating the parameter

x=tanx, y=secx; (1, sq root of 2)

3. find the length:

x=(e^t) - t , y=4(e^t/2) , t is between -8 and 3

Thanks for any help! Even if you can only answer 1 or 2 questions!

### 2 Answers

- Ed ILv 71 decade agoFavorite Answer
1. x = e^√t, y = t - ln t^2

dx/dt = e^√t/(2√t)

dy/dt = 1 - 2 ln t

dy/dx = (dy/dt)/(dx/dt) = (1 - 2 ln t)/(e^√t/(2√t))

At t = 1, dy/dx = (1 - 2 ln 1)/(e^√1/2√1)) = (1 - 0)/(e/2) = e/2

At t = 1, x = e^√1 = e, and y = 1 - ln 1^2 = 1 - ln 1 = 1

(e, 1)

y - 1 = (e/2)(x - e)

2. (a) x = tan t

y = sec t

dx/dt = sec^2 t

dy/dt = sec t tan t

dy/dx = (sec t tan t)/(sec^2 t) = tan t/sec t = (sin t/cos t)/(1/cos t) = sin t

x = tan t

1 = tan t

t = π/4

At t = π/4, dy/dx = sin π/4 = √2/2

y - √2 = (√2/2) (x - 1)

(b) tan^2 t + 1 = sec^2 t

x^2 + 1 = y^2

± √(x^2 + 1) = y

Since (1,√2) is on the curve, y = √(x^2 + 1) = (x^2 + 1)^(1/2)

dy/dx = (1/2)(x^2 + 1)^(-1/2) • 2x = x/√(x^2 + 1)

At x = 1, dy/dx = 1/√(1^2 + 1) = 1/√(1 + 1) = 1/√2 = √2/2

Same slope, same point ==> same equation for the tangent

3. x = e^t

y = 4e^(t/2)

dx/dt = e^t

dy/dt = 2e^(t/2)

(dx/dt)^2 + (dy/dt)^2 = (e^t)^2 + (2e^(t/2))^2 = e^(2t) + 4e^t

ds = √((dx/dt)^2 + dy/dt)^2) dt = √(e^(2t) + 4e^t)) dt

s = ∫ √(e^(2t) + 4e^t) dt {from t = -8 to t = 3} = 28.10726288... ≈ 28.107

Source(s): Honestly, I couldn't see an easy way to evaluate the integral in #3, so I used my TI-84 calculator to find the length.