Question

If A is the area of a circle with radius r and the circle expands as time passes, find dA/dt in terms of dr/dt. dA dt = dr dt (b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 m/s, how fast is the area of the spill increasing when the radius is 25 m?

Answer 1

Answer:

A) dA/dt = 2πr(dr/dt)

B) dA/dt = 50π m²/s

Explanation:

A) The formula for Area is;

A = πr²

Since, we want to find dA/dt. Thus,

(A)(1/dt) = (πr²)(1/dt)

Thus, differentiating;

(dA/dr)(1/dt) = 2πr(1/dt)

Multiply both sides by dr to obtain;

dA/dt = 2πr(dr/dt)

(b) we want to find rate of area (dA/dt) when r = 25m and dr/dt = 1 m/s

since we know that, dA/dt = 2πr(dr/dt), we can solve it as;

dA/dt = 2π(25)(1)

dA/dt = 50π m²/s

Answer 2

Answer:$157.1 m^2/s$

Explanation:

Given

A is the area of circle with radius r

$A=\pi r^2$

Differentiate w.r.t time

$\frac{\mathrm{d} A}{\mathrm{d} t}=\pi \times 2\times \frac{\mathrm{d} r}{\mathrm{d} t}$

Also $\frac{\mathrm{d} r}{\mathrm{d} t}=1 m/s at r=25 m$

$\frac{\mathrm{d} A}{\mathrm{d} t}=\pi \times 2\times 25\times 1=157.1 m^2/s$