The parametric equation of a curve are #x=t+e^-t#, #y=1-e^-t#, where t takes all real values. How do you find dy/dx in terms of t, and hence find the value of t, for which the gradient of the curve is 1, giving your answer in logarithmic form?

Question

Scarlett Allison · Accepted Answer

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

Cameron Alexander · Answer

undefined To find dy/dx in terms of t, we use the chain rule. First, we differentiate y with respect to t and then differentiate x with respect to t. Then, we divide dy/dt by dx/dt. This gives us dy/dx in terms of t.

Given x = t + e^(-t) and y = 1 - e^(-t), we differentiate both x and y with respect to t:

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

Now, we find dy/dx:

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

To find the value of t for which the gradient of the curve is 1, we set dy/dx equal to 1 and solve for t:

1 = (e^(-t)) / (1 - e^(-t))

Multiplying both sides by (1 - e^(-t)):

1 - e^(-t) = e^(-t)

Adding e^(-t) to both sides:

1 = 2 * e^(-t)

Dividing both sides by 2:

1/2 = e^(-t)

Taking the natural logarithm of both sides:

ln(1/2) = ln(e^(-t))

Using the property of logarithms ln(a^b) = b * ln(a):

ln(1/2) = -t * ln(e)

Since ln(e) = 1:

ln(1/2) = -t

Therefore, the value of t for which the gradient of the curve is 1 is t = -ln(1/2) or t = ln(2).