So far, i have some work done.
3. I have dV/dt = k/r
V = (4/3)pi*r^3
dV/dt = 4pi*r^2 (dr/dt)
Thus, 4pi*r^2 (dr/dt) = k/r
I'm stuck here.
4. The volume is 8pi in^3
That's all i have for 4.
5. I don't' know how to relate the eqautions.
Hey everyone, i am new here. I am in BC Calc at my school and we got a tough related rate problems worksheet over the weekend. I got the first two and the last one, but the these three have stumped me. Last year's problems weren't this hard. Can you help with whatever you can, please?
I took a picture of the problems. I know it's a lot, but if possible, just help with whatever you know.
Thanks so much!!!
Hi there. I'm Nowegian and don't know what reciprocal means, but if the increase of r (referring to task 3) is constant (r(t) is linear), then you can find the rate of increase of the radius by . If we define we then get , which is the definition of the derivate. (For linear functions anyway. For other functions let go to zero.) You know that and and . You can use this to find the rate of increase of the radius, then apply it to problem B.
#5:
Circumference is
and we know that
Therefore,
The perimeter of the square is P=8r.
So,
For part b, you need the difference between the area of the circle and the square.
Since the area of the circle is
r=5. Then a side length of the square is 10.
Continue on with this part?.
Can you use this to solve other ones now?.
Hello, SuperTyphoon!
5. A circle is inscribed in a square. The circumference of the circle is increasing at a constant
rate of 6 in/sec. As the circle expands, the square expands to maintain tangency.
A. Find the rate at which the perimeter of the square is increasing.Code:*-------*-*-*-------* | * * | | * * | |* *| | | * * 2r * * - - - - * * r * | | |* *| | * * | | * * | *-------*-*-*-------* 2r
The radius of the circle is
The side of the square is
The circumference of the circle is: .
Differentiate with respect to time: .
We are told that
So we have: . in/sec. .[1]
The perimeter of the square is: .
Differentiate with respect to time: . .[2]
Substitute [1] into [2]: .
B. When the area of the circle is 25π inē,
find the rate on increase of the area between the circle and the square.
The area of the circle is: .
If the area is inē, we have: . .[3]
The area of the square is: .
The difference of the areas is: .
Differentiate with respect to time: . .[4]
Substitute [1] and [3] into [4]: .