Respuesta :
Answer:
The rate change of volume of the cylinder is [tex]\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})[/tex] cubic inch per second.
Step-by-step explanation:
Given that the radius of right circular cylinder is [tex]\sqrt{(t+6)}[/tex] and its height is [tex]\frac16 \sqrt t[/tex] where t is time in second and the dimension are inches.
[tex]\therefore r = \sqrt{(t+6)}[/tex]
The base area of the cylinder is A= [tex]\pi r^2[/tex]
[tex]=\pi (\sqrt{t+6})^2[/tex]
[tex]= \pi (t+6)[/tex]
[tex]\therefore A= \pi(t+6)[/tex]
Differentiating with respect to t
[tex]\frac{dA}{dt}=\pi[/tex]
[tex]\therefore h=\frac16\sqrt t[/tex]
Differentiating with respect to t
[tex]\frac{dh}{dt}=\frac16 \times \frac12(t)^{\frac12-1}[/tex]
[tex]\Rightarrow \frac{dh}{dt}=\frac1{12} (t)^{-\frac12}[/tex]
The volume of cylinder is V= Ah
∴V= Ah
Differentiating with respect to t
[tex]\frac{dV}{dt}=A\frac{dh}{dt}+h\frac{dA}{dt}[/tex]
[tex]=\pi (t+6). \frac1{12}t^{-\frac12} +\frac16\sqrt t . \pi[/tex]
[tex]=\pi. \frac1{12}.t^{\frac12}+\pi . 6.\frac1{12} t^{-\frac12} +\pi\frac16 \sqrt t[/tex]
[tex]=\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})[/tex]
The rate change of volume of the cylinder is [tex]\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})[/tex] cubic inch per second.
The rate of change of volume of the cylinder with respect to time is;
dV/dt = π/₄(√t) + (π/(2√t))
What is the rate of change of Volume?
We are given;
Radius of right circular cylinder; r = √(t + 6)
Height of Cylinder; h = ¹/₆√t
where;
t = time
The base of the cylinder is circular and the area formula is;
A = πr²
A = π * (√(t + 6))²
A = π(t + 6)
Differentiating with respect to t gives;
dA/dt = π
Similarly differentiating the height with respect to t gives;
dh/dt = ¹/₁₂ * ¹/√t
Volume of a cylinder can be expressed in terms of Area and height as;
V = Ah
Differentiating w.r.t t gives;
dV/dt = A(dh/dt) + h(dA/dt)
Plugging in the relevant values gives;
dV/dt = π(t + 6)(¹/₁₂ * ¹/√t) + (¹/₆√t) * π
dV/dt = (π/₁₂)(√t) + (π/₂)(¹/√t) + (¹/₆√t) * π
dV/dt = π/₄(√t) + (π/(2√t))
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