Find k in AP Calculus BC problem

[nomadreid]

Homework Statement

Given path (x,y) described by dx/dt = 1/(t+1), dy/dt = k*exp(kt), constant k > 0. At t = 2 the tangent is parallel to y = 4x + 3. Find k. The given solution: approx. 0.495 .

Homework Equations

(dy/dt)/(dx/dt) = dy/dx = slope of tangent of path

The Attempt at a Solution

At t=2, (dy/dt)/(dx/dt) = (2e^2k)/(1/3) = 6e^2k = 4 (from y=4x+3), giving k=0.5*ln(2/3), or approx -0.2.

 

[SteamKing]

Homework Statement

Given path (x,y) described by dx/dt = 1/(t+1), dy/dt = k*exp(kt), constant k > 0. At t = 2 the tangent is parallel to y = 4x + 3. Find k. The given solution: approx. 0.495 .

Homework Equations

(dy/dt)/(dx/dt) = dy/dx = slope of tangent of path

The Attempt at a Solution

At t=2, (dy/dt)/(dx/dt) = (2e^2k)/(1/3) = 6e^2k = 4 (from y=4x+3), giving k=0.5*ln(2/3), or approx -0.2.

At t = 2, dy/dt ≠ 2e^2k. Remember, dy/dt = k*exp(kt)

 

[nomadreid]

Oops. Thanks very much, SteamKing.