Finding how fast the area is changing of an equilateral triangle

[meeklobraca]

Homework Statement

The height h of an equilateral triangle is increasing at a rate of 3cm/min. How fast is the area changing when h is 5cm. Give to 2 decimal places.

Homework Equations

The Attempt at a Solution

Im kind of stumped with this one because the question doesn't give any other variable other than h. I am not sure if I need the base of the triangle but I don't know how to do this question without it.

Thank you for your help

 

[yyat]

An equilateral triangle (all sides have the same length) is completely determined by its height. Use the Pythagorean theorem to compute the length of the sides.

 

[lanedance]

or use the fact a triangle area is half a rectangle... set up in terms of h & deifferentiate

 

[meeklobraca]

How do I use pythagorean therom if the height is the only value I know?

Or in reference to lanedance's post does b=1/2h? I believe I can differentiate that, but I have to eliminate that b value in the formula right?

 

[yyat]

How do I use pythagorean therom if the height is the only value I know?

One of the sides has length exactly half of the hypotenuse, so the lengths are some number x and x/2. Plug that into the Pythagorean theorem (along with the height) and you can solve for x.

 

[Avodyne]

Step 1, what is the area A of an equilateral triangle whose height is h? Call this function A(h).

Step 2, differentiate A(h) with respect to time using the chain rule,

{dA\over dt}={dA\over dh}\,{dh\over dt}

You can compute dA/dh because you know A(h) from Step 1, and you are given the value of dh/dt.

 

[meeklobraca]

One of the sides has length exactly half of the hypotenuse, so the lengths are some number x and x/2. Plug that into the Pythagorean theorem (along with the height) and you can solve for x.

I did this and figured the side as being 2.9cm. Would that be correct?

Avodyne: My main problem with this question is what to do with the base part int he equation. The equation to solve this if I were solving for area is A=1/2bh right?

Doing some trial with figuring out what the base could be, I got b=2.9, h=5, dh/dt=3 therefore dA/dt=21.8 cm/min.

What do you guys think?

 

[yyat]

I did this and figured the side as being 2.9cm. Would that be correct?

Avodyne: My main problem with this question is what to do with the base part int he equation. The equation to solve this if I were solving for area is A=1/2bh right?

Doing some trial with figuring out what the base could be, I got b=2.9, h=5, dh/dt=3 therefore dA/dt=21.8 cm/min.

What do you guys think?

The length of the base should be about 5.77 cm, when h=5 cm, but keep in mind that the length of the base also changes when the height changes (because the triangle is supposed to be equilateral at all times, if I understand correctly)

 

[meeklobraca]

Upon further calculations I also got that the base is 5.77. We shouldn't be concerned about the triangle changing since were just tryiung to figure out when height = 5cm right? So for our variables we have

b=5.77
h=5
dh/dt=3

A=1/2bh

dA/dt=1/2(b)(h)(dh/dt)
=43.28

?

 

[HallsofIvy]

Write the formula for area in terms of h only before you differentiate. If an equilateraL triangle has height h, what is its base? (If the triangle has side length "s" then dropping a perpendicular from one vertex gives two right triangles with hypotenuse s and legs of length s/2 and h.)

 

[meeklobraca]

I believe that has been done no?

 

[Avodyne]

No. Your formula for A as a function of h is wrong, because is still has b in it. You must eliminate b (by writing b in terms of h) before you differentiate.

 

[meeklobraca]

So yyat's idea of using the h value to find out what b = is incorrect? Also. How do I find b in terms of h if I don't have any other values?

 

[lanedance]

no, you're just not doing what is suggested

example: say you are asked to find the rate of change of Area of a rectangle of height h & side b(h) = 2h. So we have h as our variable, note that b is a function of h.

so area as a function of h

A(h) = h.b(h) = h.2h = 2h^2
differentiating with either the product rule or straight differntiation

A'(h) = 1.b(h) + h.b'(h) = 4h

try again with this example in mind, keeping track of what is a function of h

 

[meeklobraca]

what?

what the hell does a rectangle have to do with anything?

where do you come up with this? height h & side b(h) = 2h

 

[lanedance]

its an example of a very similar problem, if you can see the similarities you should have no problem with your question

the key is, regardless of how you calculate the area, everything that can be, must be expressed in term of one dependent variable (in the rectangles case, height & width). The dependent variable represent the increase in scale of the triangle, noting that both dimsnions height & width increase proportinally at the same rate.

This is the variable you differntiate with respect to, to find the change in rate

try reading & thinking through what is suggested in the whole post before replying... then attempt the problem and show your working aagin

 

[meeklobraca]

What is the area of an equillateral triangle? a=1/2bh or a=sqrt3/4 * s^2 or does it matter for this question?

 

[HallsofIvy]

Do you know what an equilateral triangle is? I would hope so, but then I don't understand why, by post #18 you still have not done what was suggested in the first response to your question.

 

[meeklobraca]

For god sakes, I have. Read post 9. Do you guys read the previous responses when you respond? I have two different people trying to tell me to do it two different ways without giving any insight as to my questions to other posts.

 

[lanedance]

ok I have re-read the posts, I think your question was answered, but you missed the key point... can be confusing when a few people are trying to help

say you are going to use

A(h) = b(h).h/2

the key thing you have to imaginge is an expanding equilateral triangle, as h expands so does b, so to capture this expansion effect on the Area, you have to solve for b in terms of h, ie b(h).

In key - you are putting in number too early, you need to solve for b(h) first

can you do this? draw an equilateral triangle it always has an angle of 60dgress or pi/3 radians - how can you get the height b(h) in terms of a side length (h) from this?

then before you put in any numbers for h, differentiate A(h) with respect to h. Then, finally you can put in the numbers

 

[Redbelly98]

So yyat's idea of using the h value to find out what b = is incorrect? Also. How do I find b in terms of h if I don't have any other values?

But you do have other values. You have the angles in the triangle.

 

[meeklobraca]

Again, again, again. Which is it I do? find b in terms of h, or use the pythag. therom to find b?

 

[lanedance]

Again, again, again. Which is it I do? find b in terms of h, or use the pythag. therom to find b?

the options you gave aren't exclusive

you want to find b in terms of h by either:
- use the pythgorean theorem
- use an angle relation (cos, sin, tan etc.)

they will both work - why not try them both?

I suggest using the tangent angle relation for the easiest calc

 

[meeklobraca]

Well he means, is that I can find a numerical value for b by using tan 60 = 5/adj. I thought I've done that. Just another case of different people not reading previous posts before posting. READ PREVIOUS POSTS BEFORE POSTING. YOU CONFUSE THE **** OUT OF PEOPLE.

Thank you.

 

[meeklobraca]

I get b=(2/3)(sqrt3)h
A=(1/3)sqrt3h^2
dA/dt=(2/3)(sqrt3)(h)dh/dt
=(2/3)(sqrt3)(5)(3)
=8.66cm^2/min

 

[lanedance]

I get a slightly different value for b, though your derivative is looking good now

Imagine one half of the equilateral triangle, by Pythagorean
b^2 = (b/2)^2 + h^2
3b^2/4 = h^2
then
b^2 = 4.h^2/3
so
b(h) = \frac{2h}{\sqrt{3}}

so the area is
A(h) = \frac{h.b(h)}{2} = \frac{2h^2}{2 \sqrt{3}}= \frac{h^2}{ \sqrt{3}}

differentiating
\frac{d}{dt} A(h) = \frac{d}{dt} \frac{h^2}{ \sqrt{3}} = \frac{2h}{ \sqrt{3}}\frac{dh}{dt}

 

[meeklobraca]

Your b value may look different from mine, but as far as I can tell by checking h=5, both our b values are exactly the same.

 

[lanedance]

yep - you're correct, i wasn't 100% how to read ur post, got it now

but as
\frac{2.\sqrt{3}}{3} = \frac{2}{ \sqrt{3}}

 

[HallsofIvy]

I, on the other hand would have used the Pythagorean theorem, mainly because that is how I remember the cosine of 60 degrees! The perpendicular from one vertex to the opposite side divides an equilateral triangle into two right triangles haveing hypotenuse of length b and two legs of lengths h and b/2. b^2= h^2+ b^2/4 or h^2= b^2- b^2/4= (3/4)b^2 so h= (\sqrt{3}/2) b or b= (2\sqrt{3}/3)h.