Air pressure at sea level is 30 inches of mercury. At an altitude of h feet above the, air pressure, P, in inches of mercury, is given by,
P =30e^(-3.23x10-5h)
a)Find th equation of the tangent line at h=0.
b)A rule of thumb is given by travelers is that air pressure drops about 1 inch for every 1000-foot increase in height above sea level. Write a formula for the air pressure given by this rule of thumb.
Please HELP!!!!!
I don't know much about the physics of this topic, but when I sub in h=0 (sea level) I do not get 30 , like I should from the given equation.
according to your rule of thumb
we could have the following
h P
0 30
1000 29 ---> two points (0,30) and (1000, 29)
slope = (29-30)/(1000-0) = -1/1000
P = (-1/1000)h + 30
a) First: the definition of a tangent line is defined as a line passing through a given point of a function whose slope is equal to the derivative of the function at that point
The equation of the tangent line is of the form y = mx + b, where m is the slope, and b is the y-intercept. Your function is not in terms of x and y, it is in terms of P and h, so the equation of the tangent line will look like P = m*h +b. This equation can be found after calculating the slope of P at h=0, and the value of P at h = 0
The slope of P at h = 0 is equal to the derivative of P dP/dh evaluated at h = 0:
dP/dh = (30*-3.23*10^-5)*e^(-3.23x10-5h)
dP/dh(h=0) = (30*-3.23*10^-5) = -9.69*10^-4
= m
So far we have
P = -9.69*10^-4 h + b
we need to find b, the y intercept, to completely solve for the tangent line. We know that this point passes through P(h=0), or (0, 30)
Put this point into the equation for the tangent line, and solve for b:
30 = -9.69*10^-4*0 + b
solve for b, b = 30;
P = -9.69*10^-4*h + 30
b) At h = 0, the air pressure is 30 inches of mercury. At 1000 feet, it would be 30 - 1000/1000 *1
at 2000 feet, it would be
30 - 2000/1000 * 1,
so, examining the pattern:
P = 30 - h/1000
Thank you both for your help
Jennifer, my point was that the equation, the way it was written, is not correct
the exponent as typed was
-3.23x10-5h
When you sub in h=0, ----> -3.23x10 or -32.3
The way you read it, what Andy probably meant to type is
(-3.23 x 10^-5)h , but there is no exponent shown and following the order of operation yields the wrong result.
We have been stressing the importance of the proper use of brackets to establish the correct order of operation.
A agree with your tangent equation according to your interpretation , and noticed we also have the same linear equation.
a) To find the equation of the tangent line at h=0, we need to find the derivative of the given equation with respect to h.
P(h) = 30e^(-3.23x10^(-5)h)
To find the derivative of P(h), we use the chain rule.
dP(h)/dh = d/dh (30e^(-3.23x10^(-5)h))
To apply the chain rule, we need to consider the derivative of the exponential function and the derivative of -3.23x10^(-5)h.
The derivative of the exponential function e^x with respect to x is simply e^x.
The derivative of -3.23x10^(-5)h with respect to h is -3.23x10^(-5).
Applying the chain rule:
dP(h)/dh = (30e^(-3.23x10^(-5)h)) * (-3.23x10^(-5))
Now we can find the equation of the tangent line at h=0 by plugging in h=0 into the derivative:
dP/dh |(h=0) = (30e^(-3.23x10^(-5)(0))) * (-3.23x10^(-5))
= (30e^0) * (-3.23x10^(-5))
= 30 * (-3.23x10^(-5))
= -0.000969
The slope of the tangent line at h=0 is -0.000969.
The equation of a line is given by y = mx + b, where m is the slope and b is the y-intercept.
Since the tangent line is at h=0, the y-intercept is the value of P at h=0, which we can find by plugging in h=0 into the original equation:
P(h=0) = 30e^(-3.23x10^(-5)(0))
= 30e^0
= 30
So, the y-intercept is 30.
Therefore, the equation of the tangent line at h=0 is:
P = -0.000969h + 30
b) According to the given rule of thumb, air pressure drops about 1 inch for every 1000-foot increase in height above sea level.
Let's denote the height above sea level as h in feet and the air pressure as P in inches of mercury.
So, the formula for air pressure, P, based on this rule of thumb is:
P = 30 - (h/1000)
In this formula, we subtract (h/1000) from the initial sea level pressure of 30 inches of mercury to represent the drop in pressure for every 1000-foot increase in height above sea level.