The height h of an equilateral triangle is increasing at a rate of 3 cm/min. How fast is the area changing when h is 5 cm? Give your answer correct to two decimal places.
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Step 1: List all your given information...
dh/dt = 3 cm/min
h = 5 cm
dA/dt = ?
Step 2: Determine a relationship between your starting and ending variables
A = 1/2 * base * height
Since it is an equilateral triangle, you can determine the base in terms of h.
base = 2sqrt(3) / 3 * h
So,
A = sqrt(3)/3 * h^2
Finally, take the derivative
dA/dt = sqrt(3)/3 * (2h) * dh/dt
Step 3: Solve for answer
dA/dt = sqrt(3)/3 * 2 * (5) * (3)
dA/dt = 10 sqrt(3) cm^2
dh/dt = 3 cm/min
h = 5 cm
dA/dt = ?
Step 2: Determine a relationship between your starting and ending variables
A = 1/2 * base * height
Since it is an equilateral triangle, you can determine the base in terms of h.
base = 2sqrt(3) / 3 * h
So,
A = sqrt(3)/3 * h^2
Finally, take the derivative
dA/dt = sqrt(3)/3 * (2h) * dh/dt
Step 3: Solve for answer
dA/dt = sqrt(3)/3 * 2 * (5) * (3)
dA/dt = 10 sqrt(3) cm^2
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The altitude of an equilateral triangle divides the triangle into two 30-60-90 triangles.
In a 30-60-90 triangle, the leg opposite the 30-degree angle is 1/sqrt(3) times the leg opposite the 60-degree angle.
Therefore, in an equilateral triangle with height h, *half* of the base is h/sqrt(3).
So the area is A = h(h/sqrt(3)) = h^2 / sqrt(3).
Differentiating with respect to t gives
dA/dt = [2h/sqrt(3)](dh/dt).
Substituting dh/dt = 3 and h = 5, we find that the area increases at a rate of
dA/dt = [2*5/sqrt(3)](3) = 10sqrt(3) or about 17.32 cm^2/min.
Lord bless you today!
In a 30-60-90 triangle, the leg opposite the 30-degree angle is 1/sqrt(3) times the leg opposite the 60-degree angle.
Therefore, in an equilateral triangle with height h, *half* of the base is h/sqrt(3).
So the area is A = h(h/sqrt(3)) = h^2 / sqrt(3).
Differentiating with respect to t gives
dA/dt = [2h/sqrt(3)](dh/dt).
Substituting dh/dt = 3 and h = 5, we find that the area increases at a rate of
dA/dt = [2*5/sqrt(3)](3) = 10sqrt(3) or about 17.32 cm^2/min.
Lord bless you today!
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You are searching for dA/dt when h is 5 cm. IGNORE the 5 cm until the end. Your job is to solve while incorporating 3 cm/min, or your dh/dt.
dA/dt = dh/dt x dA/dh
You simply have to create an equation for Area in terms of h only. Because the triangle is equilateral, there is a way to cancel out the b in the equation A=0.5bh. Keep in mind that you may have to draw a picture and create extra variables in the process. (I'd be willing to bet that there will be trigonometry involved.)
Once you find this equation, solve it for when h=5, then multiply the answer by 3, and you will have the exact answer.
Sorry I didn't write any math down, but I just organized the set up because it's all I can do right now.
Hope this helps! ;)
dA/dt = dh/dt x dA/dh
You simply have to create an equation for Area in terms of h only. Because the triangle is equilateral, there is a way to cancel out the b in the equation A=0.5bh. Keep in mind that you may have to draw a picture and create extra variables in the process. (I'd be willing to bet that there will be trigonometry involved.)
Once you find this equation, solve it for when h=5, then multiply the answer by 3, and you will have the exact answer.
Sorry I didn't write any math down, but I just organized the set up because it's all I can do right now.
Hope this helps! ;)