Given u(x, y) = x^2 + 2y where x (r, t) = r sin(t) and y(r,t) = sin^2(t), determine the value of ∂u / ∂r and ∂u / ∂t using the chain rule.
Plz show step
surely you know about the chain rule.
∂u/∂r = ∂u/∂x ∂x/∂r + ∂u/∂y ∂y/∂r
= 2x sin(t) + 2*0
= 2x sin(t)
similarly for ∂u/∂t
still dunno
To find the partial derivatives ∂u/∂r and ∂u/∂t using the chain rule, we will first express u in terms of r and t. Let's substitute the expressions for x and y into u:
u(r, t) = (r sin(t))^2 + 2(sin^2(t))
Expanding and simplifying, we get:
u(r, t) = r^2 sin^2(t) + 2sin^2(t)
Now, let's calculate the partial derivative ∂u/∂t using the chain rule:
Step 1: Differentiate u with respect to t, treating r as a constant:
∂u/∂t = 2r^2 sin(t) cos(t) + 4sin(t)cos(t)
Step 2: Apply the chain rule for t in terms of x and y:
∂t/∂t = 1
Step 3: Multiply the partial derivatives from steps 1 and 2:
∂u/∂t = (2r^2 sin(t) cos(t) + 4sin(t)cos(t)) * 1
= 2r^2 sin(t) cos(t) + 4sin(t)cos(t)
Therefore, ∂u/∂t = 2r^2 sin(t) cos(t) + 4sin(t)cos(t).
Next, let's calculate the partial derivative ∂u/∂r using the chain rule:
Step 1: Differentiate u with respect to r, treating t as a constant:
∂u/∂r = 2r sin^2(t)
Step 2: Apply the chain rule for r in terms of x and y:
∂r/∂r = 1
Step 3: Multiply the partial derivatives from steps 1 and 2:
∂u/∂r = (2r sin^2(t)) * 1
= 2r sin^2(t)
Therefore, ∂u/∂r = 2r sin^2(t).
To summarize:
∂u/∂r = 2r sin^2(t)
∂u/∂t = 2r^2 sin(t) cos(t) + 4sin(t)cos(t)