1)Suppose Q = x^3 y^2 . If [dQ/dt] = −1 and [dx/dt] = 2, find [dy/dt] when
x = 4 and y = 5 .
[dy/dt] =______
2)Let A be the area of a circle with radius r. If [ dr / dt ] = 2, find [ dA / dt ] when r = 3
I will do one, and be happy to check your thinking.
q=x^3 y^2
dq/dt= 2x^2 y dy/dt+ 3y^2x^2 dx/dt
put in dq/dt, dx/dt, and find dy/dt ..
i don't get it..
dq/dt isnt it 3x^2y
and i don't get the part where you put 3y^2x^2 how many exponents are there
To solve these questions, we can use the chain rule of differentiation. The chain rule states that if we have a composite function, such as Q(x,y) = f(g(x, y)), then the derivative of Q with respect to t (dt) can be found by multiplying the partial derivative of Q with respect to x (dQ/dx) by the derivative of x with respect to t (dx/dt), and adding it to the product of the partial derivative of Q with respect to y (dQ/dy) and the derivative of y with respect to t (dy/dt).
Now, let's apply the chain rule to each question:
1) Q = x^3 y^2
Given [dQ/dt] = -1, [dx/dt] = 2, x = 4, and y = 5.
We want to find [dy/dt].
Using the chain rule, we have:
[dQ/dt] = (dQ/dx)(dx/dt) + (dQ/dy)(dy/dt)
Substituting the given values, we have:
-1 = (3x^2 y^2)(2) + (2x^3 y)(dy/dt)
Plugging in x = 4 and y = 5, we get:
-1 = (3(4)^2 (5)^2)(2) + (2(4)^3 (5))(dy/dt)
-1 = 1440 + 1280(dy/dt)
-1 - 1440 = 1280(dy/dt)
-1441 = 1280(dy/dt)
(dy/dt) = -1441/1280
Therefore, [dy/dt] = -1441/1280.
2) A = πr^2
Given [dr/dt] = 2, and r = 3.
We want to find [dA/dt].
Using the chain rule, we have:
[dA/dt] = (dA/dr)(dr/dt)
Taking the derivative of A = πr^2 with respect to r, we get:
dA/dr = 2πr
Plugging in r = 3 and [dr/dt] = 2, we get:
[dA/dt] = (2π(3))(2)
[dA/dt] = 12π
Therefore, [dA/dt] = 12π.