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I wanted to check a couple of these changing rate problems, since these are kind of tricky to me still:

1.

(a) If A is the area of a circle with radius r and the circle expands as time passes, find $\frac{dA}{dT}$ in terms of $\frac{dR}{dT}$.

(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 30m?

Answer Attempt

(a) $$A = \pi \cdot r^2$$ so: $$\frac{dA}{dT} = 2 \cdot \pi r \cdot \frac{dR}{dT}$$ (b) if r = 30m and $\frac{dR}{dT} = 1,$ then:

$$ \frac{dA}{dT} = 2 \cdot \pi \cdot (30m) \cdot 1 \frac{m}{s}$$ $$ = 60 \frac{m^2}{s}$$

2.

If $x^2 + y^2 + z^2 = 9$, $\frac{dx}{dt} = 5$, and $\frac{dy}{dt} = 4$, find $\frac{dz}{dt}$ when (x, y, z) = (2, 2, 1)

Answer Attempt:

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} + 2x\frac{dz}{dt} = 0$$

$$4 \cdot 5 + 4 \cdot 4 + 2 \cdot \frac{dz}{dt} = 0$$

$$2\frac{dz}{dt} = 36$$

$$\frac{dz}{dt} = 18$$

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You've forgotten to multiply $\pi$ to your final rate in #1, which should be $60\pi\frac{m^2}{s}$. Besides that, you've made an algebra mistake in #2, between these steps:

$$4 \cdot 5 + 4 \cdot 4 + 2 \cdot \frac{dz}{dt} = 0$$

$$2\frac{dz}{dt} = 36$$

If you simplify $4 \cdot 5 + 4 \cdot 4 + 2 \cdot \frac{dz}{dt} = 0 \,$ you get $36 + 2\frac{dz}{dt}=0 \,$ which should give you this:

$$2\frac{dz}{dt} = -36$$ $$\frac{dz}{dt} = -18$$

Which makes sense because if both $x \left(\frac{dx}{dt} = 5\right)$ and $y \left(\frac{dy}{dt} = 4\right)$ are increasing, $z$ must decrease rapidly $\left(\frac{dz}{dt} < 0\right)$ to keep the equation true.

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$1 b.$: you have forgotten your $\pi$ in your answer.

For your question $2$: the first line of your attempt, one of the $x$ is supposed to be $z$.

$\frac{dz}{dt}$ is suppose to be negative.

Good job besides some careless mistakes.

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