Question

If anyone completed exercise 2C-10 and 2E-5, I am not sure if the answer provided is correct. Would much appreciate if anyone can confirm my answer instead

Answer 1

for 2C-10, they should be using cosine (rather than sine)
It is true this is a version of Snell's Law, but in the case of Snell's Law, the angle is measured with respect to the normal (line perpendicular to the shore)... which leads to the sine. Whoever wrote up the answer did not check the details of how the angles are being measured, and assumed sine.

Answer 2

The answer to 2E-5 (pulley problem) is correct, walking speed = \( 2 \sqrt{5} \)

The idea is the length of the hypotenuse (labeled z, in the figure) is changing at a rate of 4 ft/sec

using x as the distance of the person from beneath the pulley, from geometry
x^2 + 10^2 = z^2

take the derivative with respect to time "t" (understanding that both x and z are changing with time):
\[ 2x\ \dot{x} = 2z\ \dot{z}\]
where e.g. \( \dot{x} \) represents dx/dt (rate of x changing with time)

When x is 20 ft, we use z^2 = x^2 + 10^2 to find
\[ z= \sqrt{20^2+100^2} = 10 \sqrt{5} \]
also remember \( \dot{z}= 4\ ft/sec \)
we get
\[ 2x\ \dot{x} = 2z\ \dot{z} \\
2\cdot 20 \ \dot{x} = 2 \cdot 10 \sqrt{5} \cdot 4 \\
\dot{x} = 2 \sqrt{5}
\]

Answer 3

We've had so many questions about 2C-10 that I've created a corrected version.

Answer 4

With regards to 2C-10, I don't understand the following:
Given: \[T=\frac{\sqrt{100^2+x^2}}{5}+\frac{\sqrt{100^2+(a-x)^2}}{2}\]
Then\[T' = \frac{1}{5}\frac{U'}{2\sqrt{U}}+\frac{1}{2}\frac{V'}{2\sqrt{V}}\]
Where:\[U=100^2+x^2,V=100^2+(a-x)^2\]
I agree with the solution's U, U' and V, but for V' I get:
\[V'=\frac{d}{dx}100^2+(a-x)^2=\frac{d}{dx}100^2+a^2-2ax+x^2=-2a+2x=-2(a-x)\]
In the solution it says:
\[V'=-1(a-x)\]
What am I doing wrong?

Answer 5

Actually. I get:
\[U'=2x\]
So, ultimately the T' in the solution is half the size og my T'. I know that won't change when T'=0, but still found it interesting.

Answer 6

I believe you're right. Looking at this again for the first time in a long time, it now appears that there's a mistake in the solution, albeit one that doesn't affect the ultimate result because it cancels out when we set T' equal to 0..

Answer 7

Here is the solution they posted, and it is ok except they should write cosine in place of sine.
Here are the details on taking the derivative of the second term
\[ \frac{d}{dx} \frac{\sqrt{100^2+(a-x)^2}}{2} = \frac{1}{2}\frac{d}{dx} \left(100^2+(a-x)^2\right)^\frac{1}{2}\\
= \frac{1}{2}\cdot \frac{1}{2}\left(100^2+(a-x)^2\right)^{-\frac{1}{2}}\frac{d}{dx}\left(100^2+(a-x)^2\right) \\
= \frac{1}{2}\cdot \frac{1}{2}\left(100^2+(a-x)^2\right)^{-\frac{1}{2}}\left(\cancel{\frac{d}{dx}100^2}+\frac{d}{dx}(a-x)^2\right) \\
= \frac{1}{2}\cdot \frac{1}{2}\left(100^2+(a-x)^2\right)^{-\frac{1}{2}} 2(a-x)\frac{d}{dx}(a-x)\\
= \frac{1}{2}\cdot \frac{1}{2}\left(100^2+(a-x)^2\right)^{-\frac{1}{2}} 2(a-x)\left( \cancel{\frac{d}{dx} a}- \frac{d}{dx} x\right) \\
= \frac{1}{2}\cdot \frac{1}{2}\left(100^2+(a-x)^2\right)^{-\frac{1}{2}} 2(a-x)(-1)
\]
which simplifies to
\[ \frac{-(a-x)}{2 \sqrt{100^2+(a-x)^2} }\]