Math  /  Calculus

Question10. A triangle with base bb and height hh is expanding such that its area is increasing at the rate 4 m2/s4 \mathrm{~m}^{2} / \mathrm{s}. (A) dAdt=4\frac{d A}{d t}=4 B) dbdt=4\frac{d b}{d t}=4 C) dhdt=4\frac{d h}{d t}=4 D) dtdA=4\frac{d t}{d A}=4 dAdt=4 m2/s\frac{d A}{d t}=4 \mathrm{~m}^{2} / \mathrm{s}
11. A spherical rock erodes over time due to winds and rain. The radius of the rock is changing at a rate of 3in/3 \mathrm{in} / year. Find the rate that the volume of the rock is changing when the surface area (A) of the rock is 400in2400 \mathrm{in}^{2}. A) dAdt=3\frac{d A}{d t}=3 B) drdt=3\frac{d r}{d t}=3 (C) dVdt=3\frac{d V}{d t}=-3 D) drdt=3\frac{d r}{d t}=-3

Directions: Solve the following problems. Show all work clearly and neatly. (3r+1)1/2(3 r+1)^{1 / 2}
12. Let ss and rr be variables such that 2s=3r+12 s=\sqrt{3 r+1}. Find drdt\frac{d r}{d t} when dsdt=5\frac{d s}{d t}=5 and s=2s=2. 2dsdt=12(3r+1)123drdt(22)2=3r+1225=123r+13drdt16=3r+1203r+1=3drdt15=3r20r\begin{array}{ll} 2 \frac{d s}{d t}=\frac{1}{2}(3 r+1)^{-\frac{1}{2}} \cdot 3 \frac{d r}{d t} & (2 \cdot 2)^{2}=\sqrt{3 r+1}^{2} \\ 2 \cdot 5=\frac{1}{2 \sqrt{3 r+1}} \cdot 3 \frac{d r}{d t} & 16=3 r+1 \\ 20 \sqrt{3 r+1}=3 \frac{d r}{d t} & 15=3 r \\ 20 r \end{array} 3r+1=3drdt203r+13=drdt=20163=2043=803dx\begin{array}{l} \sqrt{3 r+1}=3 \frac{d r}{d t} \\ \frac{20 \sqrt{3 r+1}}{3}=\frac{d r}{d t}=\frac{20 \sqrt{16}}{3}=\frac{20 \cdot 4}{3}=\frac{80}{3} d x \end{array}
13. A particle is moving along the graph of xy2=36x y^{2}=36 where dydt=2\frac{d y}{d t}=-2. Find dxdt\frac{d x}{d t} when y=3y=-3. x(3)2=36dxdt2ydydt=0x(9)=36dxdt232x=4dxdt12=0\begin{array}{ll} x(-3)^{2}=36 & \frac{d x}{d t} \cdot 2 y \frac{d y}{d t}=0 \\ x(9)=36 & \frac{d x}{d t} \cdot 2 \cdot-3 \cdot-2 \\ x=4 & \frac{d x}{d t} \cdot 12=0 \end{array}
14. The radius of a circle is decreasing at a rate of 3 inches per second. At what rate is the area of the circle changing when the circumference is 16π16 \pi inches?

Studdy Solution

STEP 1

What is this asking? We need to find how fast the variable rr is changing given how fast ss is changing and a relationship between rr and ss, when ss is a particular value. Watch out! Don't forget to use the chain rule when taking derivatives with respect to time, and make sure to plug in the given values only *after* taking the derivative.

STEP 2

1. Rewrite the equation
2. Differentiate both sides
3. Find rr
4. Substitute and solve

STEP 3

To get rid of the square root, let's **square both sides** of the equation 2s=3r+12s = \sqrt{3r+1}.
This gives us (2s)2=(3r+1)2(2s)^2 = (\sqrt{3r+1})^2, which simplifies to 4s2=3r+14s^2 = 3r+1.
Now, isn't that much nicer to work with?

STEP 4

Now, we **differentiate both sides** of 4s2=3r+14s^2 = 3r+1 with respect to tt.
Remember the chain rule!
We get 42sdsdt=3drdt4 \cdot 2s \cdot \frac{ds}{dt} = 3 \cdot \frac{dr}{dt}, which simplifies to 8sdsdt=3drdt8s\frac{ds}{dt} = 3\frac{dr}{dt}.

STEP 5

We're given that s=2s=2, so let's plug that back into the *original* equation, 2s=3r+12s = \sqrt{3r+1}, to **find** rr.
We get 22=3r+12 \cdot 2 = \sqrt{3r+1}, which simplifies to 4=3r+14 = \sqrt{3r+1}.

STEP 6

Squaring both sides gives us 16=3r+116 = 3r+1.
Subtracting 1 from both sides gives 15=3r15 = 3r, and dividing by 3 gives us r=5r = 5.
Great!

STEP 7

We know that dsdt=5\frac{ds}{dt} = 5 and s=2s=2.
Let's **plug these values** into our differentiated equation, 8sdsdt=3drdt8s\frac{ds}{dt} = 3\frac{dr}{dt}.
This gives us 825=3drdt8 \cdot 2 \cdot 5 = 3\frac{dr}{dt}.

STEP 8

Simplifying, we get 80=3drdt80 = 3\frac{dr}{dt}.
Dividing both sides by 3 gives us drdt=803\frac{dr}{dt} = \frac{80}{3}.
Awesome!

STEP 9

The rate of change of rr with respect to tt is drdt=803\frac{dr}{dt} = \frac{80}{3}.

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