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# Calculus/calculus

Question
1.) The base of an isosceles triangle is 8ft long. If the altitude is 6 ft long and is increasing 3 inches per minute,at what rate are the base angles changing?

2.) The base diameter and altitude of a right circular cone are observed at a certain instant to be 10 and 20 inches respectively. If the lateral area is constant and the base diameter is increasing at a rate of 1 inch/minute, find the rate at which the altitude is decreasing.

1)  Let the angle be theta.  Let the altitude be h and the base be b.  The equation relating all the variables is
h/(0.5 b) = tan(t)    or 2h = b tan(theta) .
The question is to find    d(theta)/dt when h = 6, b = 8 and dh/dt=3/12. (assuming db/dt=0,  or that the base is not changing and is a constant).  Before that, note that at the precise moment when h=6, tan(theta) = 6/4 and the slanted side of the isosceles triangle is sqrt(52), so that
sec(theta)= sqrt(52)/4

Differentiating 2 h = b tan(theta)  = 8 tan(theta)   implicitly with respect to t, we get

2 dh/dt   = 8 sec^2(theta) *  d (theta)/dt

substituting, we get
2*3/12       = 8 * (sqrt(52)/4)^2  *  d (theta)/dt
=>     0.5        = 8 * 52 / 16 *  d (theta)/dt

Now all one needed to do is to solve for d (theta)/dt  which is the rate at which the base angle is changing.  since everything is measured in feet and time is minute, the answer will be in feet per minute.

2)  The most important formula in this question is to know the lateral surface area is
Area = pi  * radius * \sqrt (radius * radius  + height*height)

Let d be the base diameter and r be the base radius.  So 2r = d.  So dr/dt is half of the increasing rate of the diameter.
So the question is given dr/dt=0.5, find dh/dt, at the moment when r = 5 and h = 20.

Let the height (or altitude) be h.  The formula is
Area,  A = pi * r * sqrt(r^2 + h^2)

Since the lateral area is a constant, differentiating implicitly with respect to t (time), we get (using product rule)
0 =  pi*dr/dt* sqrt(r^2 + h^2)  + pi * r * 1/2 * (r^2 + h^2) * (2r * dr/dt+ 2h * dh/dt)

or  0 = dr/dt* sqrt(r^2 + h^2) + r(r^2 + h^2) (r dr/dt+ h  dh/dt)
Now substituting dr/dt=1/2, r = 5, h = 20, we get
0 = 1/2 *  sqrt(25 + 400)  + 5(25 + 400) (5*1/2 + 20 * dh/dt)

The challenge now would be to solve for dh/dt in the last equation above.  I leave that to you, as I trust that being now a calculus student solving a Related Rate problem, this algebraic difficulty is just a small one.  Good luck!
Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment THANK YOU SO MUCH!! THIS IS SUCH A BIG HELP.:))

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Calculus

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#### Amos

##### Expertise

I can answer all calculus question. I am a Math Lecturer and I teach Math in a College, usually Calculus 1, Calculus 2 and Calculus 3 and Linear Algebra.

##### Experience

I have been teaching in this college for more than 12 years. Prior to that, I taught for three years as Visiting Assistant Professor.

Education/Credentials
I got my doctorate from Univ. of Rochester in Algebraic Toppology. I got a MA from Univ of Rochester, an MA from York Univ. in Toronto and almost did my M.Sc in the National Univ. of Singapore.