Let 𝑥(𝑡) = 𝑡^3 − 4𝑡 2 + 𝑡 and 𝑦(𝑡) = 𝑡^2 − 6𝑡 + 3. Find followings at 𝑡 = 1.

Question

a. x(1)

b. y(1)

c. dx/dt |t=1

d. dy/dt |t=1

e. dy/dx|t=1

My answers:

a. 1

b. -2

c. 3t^2-8t+1

d. 2t-6

e. 3t^2-18t+23/2(t-3)^2

Not sure if I did any of them correctly, to be frank. I forgot to consider t=1 for c-e. Thank you for the help in advance!

Answer 1

If x(t) = t³ - 4t² + t then x(1) = (1)³ - 4(1)² + (1) = 1 - 4 + 1 = -2

If y(t) = t² - 6t + 3 then y(1) = (1)² - 6(1) + 3 = 1 - 6 + 3 = -2

Since x(t) = t³ - 4t² + t, then x'(t) = dx/dt = 3t² - 8t + 1 so x'(1) = 3(1)² - 8(1) + 1 = 3 - 8 + 1 = -4 (note that you need to plug in t = 1 to your derivative function to get the answer they are seeking)

Since y(t) = t² - 6t + 3 then y'(t) = dy/dt = 2t - 6 so y'(1) = 2(1) - 6 = 2 - 6 = -4 (Again, (note that you need to plug in t = 1 to your derivative function to get the answer they are seeking)

To determine dy/dx, consider that the chain rule says dy/dt = (dy/dx)•(dx/dt) which means, dividing both sides by (dx/dt) that (dy/dt) / (dx/dt) = dy/dx

So, since dy/dx = (dy/dt)/(dx/dt) then for t = 1, dy/dx = (-4)/(-4) = 1

Answer 2

b is correct.

c) 3(1)^2 - 8(1) +1 or -4

d) 2(1) - 6 or -4

e) dy/dx is dy/ dt × dt/ dx or dy/dt ÷ dx/dt

dy/dx=-4÷-4 or 1

For a)

x(1)=1^3 - 4(1)^2 + 1

x(1)=-2

Answer 3

Let 𝑥(𝑡) = 𝑡³ − 4𝑡² + 𝑡 and 𝑦(𝑡) = 𝑡² − 6𝑡 + 3. Find followings at 𝑡 = 1.

a. x(1) b. y(1) c. dx/dt |t=1 d. dy/dt |t=1 e. dy/dx|t=1

Detailed Solution:

a) x(1) = (1)³ − 4(1)² + (1) = 1 − 4 + 1 = −2 ==> x(1) = −2

b) y(1) = (1)² − 6(1) + 3 = 1 − 6 + 3 = −2 ==> y(1) = −2

c) dx/dt = 3𝑡² − 8t + 1 ==> (dx/dt |t=1) = 3(1)² − 8(1) + 1 = 3 − 8 + 1 = −4 ==> (dx/dt |t=1) = −4

d) dy/dt = 2t − 6 ==> (dy/dt |t=1) = 2(1) − 6 = 2 − 6 = −4 ==> (dy/dt |t=1) = −4

e) dy/dx = (dy/dt) / (dx/dt)==> (dy/dx|t=1) = (dy/dt |t=1) / (dx/dt |t=1) = (−4) / (−4) = 1==> (dy/dx|t=1) = 1