How do you differentiate the following parametric equation: # x(t)=3(t+1)^2+2e^t, y(t)= (t+2)^2+t^2#?

Question

How do you differentiate the following parametric equation: #   x(t)=3(t+1)^2+2e^t,  y(t)= (t+2)^2+t^2#?

Charles Arocha · Accepted Answer

#(dy)/(dx) = (3(t+1))/(2t+1)#
Step 1: Find #(dx)/(dt)# Step 2: Find #(dy)/(dt)# 3) Step 3: #(dy)/(dt) = (dx)/(dt) * (dy)/(dx) hArr (dy)/(dx)= [(dy)/(dt)]/[(dx)/(dt)]# Step 1: Given #x(t)= 3(t+1)^2# #x'(t) = (dx)/(dt)= 3(2)(t+1)(1) # #color(red)((dx)/(dt)= 6(t+1))# Step 2: Given #y(t)= (t+2)^2 +t^2# #y'(t) = (dy)/(dt)= 2(t+2)(1)+2t# #(dy)/(dt)= 2t+4+2t = 4t+2# #color(blue)((dy)/(dt)= 4t+2 = 2(2t+1))# Step 3: #(dy)/(dt) = (dx)/(dt) * (dy)/(dx) hArr (dy)/(dx)= [(dy)/(dt)]/[(dx)/(dt)]# #(dy)/(dx)= color(red)(6(t+1))/color(blue)(2(2t+1))# #(dy)/(dx) = (3(t+1))/(2t+1)#

Emilia Aguilar · Answer

undefined To differentiate the given parametric equations $x(t)$ and $y(t)$, we will use the chain rule and the product rule where necessary.

$$
x(t) = 3(t+1)^2 + 2e^t
$$
$$
y(t) = (t+2)^2 + t^2
$$

$$
\frac{dx}{dt} = 3 \cdot 2(t+1) \cdot \frac{d(t+1)}{dt} + 2 \cdot \frac{d(e^t)}{dt}
$$
$$
\frac{dy}{dt} = 2(t+2) \cdot \frac{d(t+2)}{dt} + 2t
$$

$$
\frac{dx}{dt} = 6(t+1) + 2e^t
$$
$$
\frac{dy}{dt} = 2(t+2) + 2t
$$

So, the derivatives with respect to $t$ are:

$$
\frac{dx}{dt} = 6(t+1) + 2e^t
$$
$$
\frac{dy}{dt} = 2(t+2) + 2t
$$