1) A particle is moving along the curve y= 4square root{3 x + 7}. As the particle passes through the point (3, 16), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.
I know you have to use the distance formula but i just know know what to do.
2)The altitude of a triangle is increasing at a rate of 1.500 centimeters/minute while the area of the triangle is increasing at a rate of 1.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 8.000 centimeters and the area is 90.000 square centimeters?
Thanks for the help
I know you have to use the distance formula but i just know know what to do.
2)The altitude of a triangle is increasing at a rate of 1.500 centimeters/minute while the area of the triangle is increasing at a rate of 1.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 8.000 centimeters and the area is 90.000 square centimeters?
Thanks for the help
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The goal with both problems is to find an equation whose derivative relates all the quantities involved.
For the first problem we use the distance formula, as you said.
d = √ y² + x²
In this case, y is 4√( 3x + 7 ) and x is just x because x is changing uniformly. Differentiating this with respect to time, we get:
(∂ / ∂ t) d = (∂ / ∂ t) √ ( ( 4√( 3x + 7 ) )² + x² )
(∂ d/ ∂ t) = { 48 (∂ x/ ∂ t) + 2 x (∂ x/ ∂ t) } / { 2√( x² + 16 ( 3x + 7 ) )
We know that x is changing at 4 units/s, so (∂ x/ ∂ t) = 4. Likewise, we're passing through (3,16), so the instantaneous x = 3. Plugging in these values, we find (∂ d/ ∂ t) changing at about 108 / √265, or 6.63 units, per second.
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The second problem uses the same concept, but now the quantities are related by A = (b*h) / 2, where A is the triangle's area, b is its base length, and h is its height or altitude.
(∂ / ∂ t) A = (∂ / ∂ t) (b*h) / 2
(∂ A / ∂ t) = (1/2) ( h*(∂ b / ∂ t) + b*(∂ h / ∂ t) )
We're looking for the base's rate of change, so solve for (∂ b / ∂ t).
(∂ b / ∂ t) = (1 / h) ( 2*(∂ A / ∂ t) – b*(∂ h / ∂ t) )
Most values are given explicitly in the problem, but b, the base length at any instant, is not. However, we do know instantaneous height and area, so we can find the base.
A = (b*h) / 2
b = ( 2 A ) / h
b = ( 2*90 ) / 8
b = 45 / 2
Now we can plug everything into the differentiated equation and find that the base is decreasing by about 3.97 cm / min.
For the first problem we use the distance formula, as you said.
d = √ y² + x²
In this case, y is 4√( 3x + 7 ) and x is just x because x is changing uniformly. Differentiating this with respect to time, we get:
(∂ / ∂ t) d = (∂ / ∂ t) √ ( ( 4√( 3x + 7 ) )² + x² )
(∂ d/ ∂ t) = { 48 (∂ x/ ∂ t) + 2 x (∂ x/ ∂ t) } / { 2√( x² + 16 ( 3x + 7 ) )
We know that x is changing at 4 units/s, so (∂ x/ ∂ t) = 4. Likewise, we're passing through (3,16), so the instantaneous x = 3. Plugging in these values, we find (∂ d/ ∂ t) changing at about 108 / √265, or 6.63 units, per second.
———————————————————————
The second problem uses the same concept, but now the quantities are related by A = (b*h) / 2, where A is the triangle's area, b is its base length, and h is its height or altitude.
(∂ / ∂ t) A = (∂ / ∂ t) (b*h) / 2
(∂ A / ∂ t) = (1/2) ( h*(∂ b / ∂ t) + b*(∂ h / ∂ t) )
We're looking for the base's rate of change, so solve for (∂ b / ∂ t).
(∂ b / ∂ t) = (1 / h) ( 2*(∂ A / ∂ t) – b*(∂ h / ∂ t) )
Most values are given explicitly in the problem, but b, the base length at any instant, is not. However, we do know instantaneous height and area, so we can find the base.
A = (b*h) / 2
b = ( 2 A ) / h
b = ( 2*90 ) / 8
b = 45 / 2
Now we can plug everything into the differentiated equation and find that the base is decreasing by about 3.97 cm / min.