Hello everyone!
I'm preparing for a unit exam in Grade 12 Calculus, and I have two slightly harder related rates questions that I wanted to make sure I'm doing correctly before I write my exam.
Question #1:
An arrow is shot at an angle of 21 deg above the horizontal line. If the height is increasing at 270 km/h, what is the speed of the arrow itself (not the horizontal speed, but the actual diagonal line.)
So what I have is dh/dt = 270.
Then I have a right angle triangle, with h as the height, s as the diagonal speed and x as the horizontal line.
Next, I took the equation sin21=h/s, which I then rearranged ----> sin21*s=h. I differentiated both sides of the equation, and got this:
sin21*ds/dt = dh/dt
sin21*ds/dt = 270
ds/dt=270/sin21
ds/dt=753.42
I'm used to seeing Pythagoras problems when it comes to triangle related rate problems, so I'd just like to make sure I'm doing this correctly.
Question #2:
The radius of a circle increases at a constant rate. At a specific moment, the rate of change of the circumference is equal to the rate of change of the area. What is the area of the circle at this moment?
So what I did was this:
C=2*pi*r
C'=2pi*dr/dt
A=pi*(r)^2
A'=2pi*r*dr/dt
Setting A' and C' = to each other:
2pi*r*dr/dt=2pi*dr/dt
2pi*r=2pi
r=1
A(1)=pi
Is the only reason they gave me the "radius of a circle increases at a constant rate", so that I could cancel the two dr/dt's? Don't quite understand the relevance to the question.
Any feedback or suggestions would be greatly appreciated! Thanks :smile:
I'm preparing for a unit exam in Grade 12 Calculus, and I have two slightly harder related rates questions that I wanted to make sure I'm doing correctly before I write my exam.
Question #1:
An arrow is shot at an angle of 21 deg above the horizontal line. If the height is increasing at 270 km/h, what is the speed of the arrow itself (not the horizontal speed, but the actual diagonal line.)
So what I have is dh/dt = 270.
Then I have a right angle triangle, with h as the height, s as the diagonal speed and x as the horizontal line.
Next, I took the equation sin21=h/s, which I then rearranged ----> sin21*s=h. I differentiated both sides of the equation, and got this:
sin21*ds/dt = dh/dt
sin21*ds/dt = 270
ds/dt=270/sin21
ds/dt=753.42
I'm used to seeing Pythagoras problems when it comes to triangle related rate problems, so I'd just like to make sure I'm doing this correctly.
Question #2:
The radius of a circle increases at a constant rate. At a specific moment, the rate of change of the circumference is equal to the rate of change of the area. What is the area of the circle at this moment?
So what I did was this:
C=2*pi*r
C'=2pi*dr/dt
A=pi*(r)^2
A'=2pi*r*dr/dt
Setting A' and C' = to each other:
2pi*r*dr/dt=2pi*dr/dt
2pi*r=2pi
r=1
A(1)=pi
Is the only reason they gave me the "radius of a circle increases at a constant rate", so that I could cancel the two dr/dt's? Don't quite understand the relevance to the question.
Any feedback or suggestions would be greatly appreciated! Thanks :smile: