Rates of Change (help) with picture

[asdfsystema]

hi my name is ken. i am a 11th grader taking calculus for my first year.

Here are 3 problems my teacher is giving us in preparation for the upcoming test... I'm pretty clueless ! hope you guys could clear up the confusion for me.

The pic:

1. So altitude is the height. What i get from this is that the height is increasing by 2 cm/min and the area is increasing by 4.5 cm/min. the question asks for what the rate of change is for the base when the height is 10.5cm and the area is 100cm. Do I start off by taking the differentiation?

2. Area of circle = pi*r^2
differentiate.

dA/dt = 2pi*r * dr/dt
dA/dt= 2pi(5) * 5
dA/dt= 50pi/sec

3. I am really confused about this one too. the equation is y= 3 sqrt(4x+4) and the poitn passes through (3,12). I tried graphing this and plotting the point and also graphing the line since its 3 units per second but i have no idea where to start this using calculus. my teacher does not allow us to use the calculator on tests.

Thank you.

 

[Mark44]

1) You're given that A = 1/2 * b * h
You can assume that A, b, and h are functions of time t.
Solve for b in the equation above.
Differentiate to get db/dt.
In your equation for db/dt, substitute the values of all other variables, including derivatives at the particular time in question.

2) Mahdiba11's solution is not quite right. dA/dt is not constant, as is implied in his or her solution.
First, find dA/dt, and then evaluation dA/dt at the time when r = 5.

3) How far is the point (x, 3sqrt(4x + 4)) from the origin? You want to find the rate of change of this distance with respect to time.
After you have found D (a function of x), calculate dD/dt. You'll need to use the chain rule to do this. Your equation should have dx/dt in it.
After you have found dD/dt, evaluate it at the moment when x = 3.

 

[asdfsystema]

thank you for verifying number 2 for me :) that is what i got as well ! can you help me with problem #1 and #3?

 

[Mark44]

thank you for verifying number 2 for me :) that is what i got as well ! can you help me with problem #1 and #3?

I don't think you saw my reply, since yours came only a minute after mine. See my reply for the steps you need to do for all three problems.

 

[asdfsystema]

thanks mark44 . i want to say that I'm taking this course in a college and that i miss one days out of 3 every week so sometimes i am not sure about the equations though i try to look at the textbook for the basics.

1. ok i solved for b = 2A/h and differentiated db/dt= 2*-h^-2 which is -2/h^2. I'm pretty sure that is wrong because i only need to plug in for h .. and that doesn't seem right. can i get more clues?

2. ok so i have dA/dt= 2pi*r. this time i just plugged in r=5 2pi(5)= 10pi ?


3. the distance from (3,12) to the origin (0,0) is sqrt(153). do I use differentiate the distanc formula for this problem ? i don't see where to use the chain rule here..


thanks for the help ! can you also check your inbox :)

 

[Mark44]

thanks mark44 . i want to say that I'm taking this course in a college and that i miss one days out of 3 every week so sometimes i am not sure about the equations though i try to look at the textbook for the basics.

1. ok i solved for b = 2A/h and differentiated db/dt= 2*-h^-2 which is -2/h^2. I'm pretty sure that is wrong because i only need to plug in for h .. and that doesn't seem right. can i get more clues?

2. ok so i have dA/dt= 2pi*r. this time i just plugged in r=5 2pi(5)= 10pi ?


3. the distance from (3,12) to the origin (0,0) is sqrt(153). do I use differentiate the distanc formula for this problem ? i don't see where to use the chain rule here..


thanks for the help ! can you also check your inbox :)

I checked my inbox, and replied to you.

1. A, b, and h are all changing as time changes, which means that all three are functions of t. When you differentiated 2A/h, you confused the idea of the derivative as a function of t with the value of the derivative at a particular time. For example, if f'(t) = 3t, you will get different values of f'(t) for different values of t. That is, f'(1) = 3, f'(2) = 6, and so on.

In your problem, you need to first find db/dt (which is a function of t), and then substitute in values of the derivatives at the specified time (they don't give you the value of t, but they give you the values of A, h, dA/dt, and dh/dt).

Since b = 2A/h, to find dh/dt, you need to use the quotient rule.
[tex]db/dt = 2(\frac{h dA/dt - A dh/dt}{h^2})[/tex]
Now, to find db/dt at the time in question, substitute the values you are given for A, dA/dt, h, and dh/dt.

2. 10pi is the answer, but the point I was trying to make is that dA/dt is a function of t, so will give different values at different times.

3. You need to get D (the distance from the origin to an arbitrary point on the graph of the function that you were given), not the distance from the origin to (3, 12). I provided a summary of what you needed to do in a previous post. Here it is again:

How far is the point (x, 3sqrt(4x + 4)) from the origin? You want to find the rate of change of this distance with respect to time.
After you have found D (a function of x), calculate dD/dt. You'll need to use the chain rule to do this. Your equation should have dx/dt in it.
After you have found dD/dt, evaluate it at the moment when x = 3.

For this problem, you don't have an explicit value of t to work with, but you know at that moment that x = 3, y (= 3sqrt(4x + 4) ) = 12, and dx/dt = 3 units/sec.

Mark

 

[asdfsystema]

Thanks Mark :)

1. I understand how you got that equation now.

[tex]
db/dt = 2(\frac{h dA/dt - A dh/dt}{h^2})
[/tex]

h= 10.5
dA/dt= 4.5 sq cm / min
A= 100 sq cm
dh/dt= 2cm/min

plugging all those into the equation , I get a negative number -29.0952 ?


2. Ah i see what you were saying about this.



3. distance is sqrt(x^2+y^2)

a. i subbed in for y, D= sqrt (x^2 + (3*(4x+4)^1/2)^2)
b. D= sqrt (x^2+ 36x+36)

Then I differentiated. dD/dt= (x^2+36x+36)^1/2

and used the chain rule for this part. u= x^2+36x+36 du= 2x+36 *dx/dt

finally , 1/2*u^-1/2 --> dD/dt= 1/2(x^2+36x+36)^-1/2 * 2x+36 *dx/dt and I plug everything in the equation to find the answer ?

 

[Mark44]

Thanks Mark :)

1. I understand how you got that equation now.

[tex]
db/dt = 2(\frac{h dA/dt - A dh/dt}{h^2})
[/tex]

h= 10.5
dA/dt= 4.5 sq cm / min
A= 100 sq cm
dh/dt= 2cm/min

plugging all those into the equation , I get a negative number -29.0952 ?

I didn't check your numbers, but a negative value is not unreasonable. It's possible that the altitude of the triangle is increasing more rapidly than can be accounted for by the area increase. In that case, the base would have to be decreasing. In terms of this problem, and if you did your calculations correctly, this means that the base is decreasing at a rate of about 29.1 cm/min.

2. Ah i see what you were saying about this.



3. distance is sqrt(x^2+y^2)

a. i subbed in for y, D= sqrt (x^2 + (3*(4x+4)^1/2)^2)
b. D= sqrt (x^2+ 36x+36)

Then I differentiated. dD/dt= (x^2+36x+36)^1/2

Correction to the last line above: dD/dt= d/dt[(x^2+36x+36)^1/2]
Do you see what I did?

and used the chain rule for this part. u= x^2+36x+36 du= 2x+36 *dx/dt

finally , 1/2*u^-1/2 --> dD/dt= 1/2(x^2+36x+36)^-1/2 * 2x+36 *dx/dt and I plug everything in the equation to find the answer ?

Perfectamundo!
Let me make some minor adjustments - a couple of pairs of parens- in what you wrote:
dD/dt= 1/2(x^2+36x+36)^(-1/2) * (2x+36) *dx/dt

dD/dt is a function of t. You can evaluate dD/dt at t_0 to get a number.
We don't know t_0, but we know what x and dx/dt are at that time, so substitute these values in the expression above, and Bob's your uncle, as they say in Britain.

Mark

 

[asdfsystema]

Ah thank you so much. you seriously make things so much clearer to me !

 

[asdfsystema]

Ah sorry mark, to be bothering again... I emailed the professor and he told me my number 1 and number 3 is wrong. but he said i did a great job on number 3 and 4, so it's all thanks to you :) !

question number one answer isn't -29.0952... i keep getting this same answer but i might be doing a wrong calculation. Can you help me check what i am doing wrong ?

 

[Mark44]

1. A = .5 * b * h ==> b = 2A/h
db/dt = d/dt(2A/h) = 2(h * dA/dt - A * dh/dt )/(h^2), using the quotient rule.
At the moment in question, dh/dt = 2 cm/min, dA/dt = 4.5 cm^2/min, h = 10.5 cm, and A = 100 cm^2,
so dh/dt at that time is 2(10.5 * 4.5 - 100 * 2)/10.5^2 = -2.770975 cm/min (approx.)