CHAZ2006

a boat leaves a dock at 2:00 pm and travels due south at 20 km/h

another boat travelling due east at 15 km/h arrives at the same dock at 3:00 pm. at what time were the two boats closest together?

I am picturing a right triangle where I am minimizing the hypotenuse

but I am stuck from that point on.

(x-15)/15 for eastward boat

(x-20)/20 for southward boat

x^2+y^2= r^2 ????

for example I know the limit at infinity of tanh (e^x-e^-x/e^x+e^-x) is 1, however do compute it algebraically?

To compute a limit at infinity you divide the numerator and denominator by the highest power in the denominator. and 1/x=0

Is there a rule that I am missing?

I have what I thought was the correct answer, the answer the book gives is the following steps

g'(x)= (1+4x)^5*8(3+x-x^2)^7(1-2x)+(3+x-x^2)*5(1+4x)^4*4

4(1+4x)^4(3+x-x^2)^7[2(1+4x)(1-2x)+5(3+x-x^2)]

4(1+4x)^4(3+x-x^2)^7[(2+4x-16x^2)+(15+5x-5x^2)]

g'(x)= 4(1+4x)^4(3+x-x^2)^7(17+9x-21x^2)

What happened to the "8" in the derivative of 2nd function and where did the 2 come from?

Any help would be greatly appreciated, been stuck on this for two days.

I know it's the product rule with chain rule

As well as vertical and horizontal asymptotes?

I know the area of a triangle is 1/2base*height ; I'll call the sides of triangle "x"

1. 1/2x(h) = area

2. I think the only way to get "h" in terms of x is to use the pythagorean theorem: 1/2x²+h² = x²

I get the 1/2x by bisecting the base of the triangle

Solving for H = sqrt x²- ½x²

Pluging back into area forumula and solving gives me

sqrt x³/4

However the correct answer is sqrt 3x³/4

But the only way for that the be correct is the is to add instead of substract on the pythagorean step. Please help, been stuck on this for a day

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