Suppose that a square container is to be filled with water.
a.What is the rate at which the surface area of the container(S) changes with respect to the length of the edge(x) when the edge is equal to 9 inches?
b. The dimensions of the container are :a,a + 1,a + 4. how fast is the volume V increasing a increases?
It sounds like a cubic container
s = 6x^2
ds/dx = 12x
so, when x=9, ...
v = a(a+1)(a+4)
dv/dt = (3a^2 + 10a + 4) da/dt
To find the rate at which the surface area of the container changes with respect to the length of the edge, we need to differentiate the surface area formula with respect to x.
a. Let's start by finding the surface area formula of a square container. The surface area of a square is given by the formula S = 4x^2, where x is the length of one side of the square container.
To find the rate of change of S with respect to x, we differentiate the formula S = 4x^2 with respect to x.
dS/dx = d(4x^2)/dx
Using the power rule of differentiation, we get:
dS/dx = 8x
Now, substitute x = 9 inches into the equation:
dS/dx = 8 * 9 = 72 square inches per inch
Therefore, when the edge length is equal to 9 inches, the rate at which the surface area of the container is changing is 72 square inches per inch.
b. Now, let's find the rate at which the volume V is increasing with respect to the length a.
The formula for the volume of a rectangular container is V = a(a + 1)(a + 4).
To find the rate of change of V with respect to a, we differentiate the formula V = a(a + 1)(a + 4) with respect to a.
dV/da = d(a(a + 1)(a + 4))/da
Using the product rule of differentiation, we can expand and differentiate each term:
dV/da = a(a + 1) + (a + 4)(2a + 1)
Simplifying, we get:
dV/da = a^2 + a + 2a^2 + 9a + 4
Combine like terms:
dV/da = 3a^2 + 10a + 4
Therefore, the rate at which the volume V increases as a increases is given by the equation 3a^2 + 10a + 4.