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!!!!!
a) To find the equation of the tangent line at h = 0, we need to find the derivative of the function P with respect to h and evaluate it at h = 0.
Given:
P = 30e^(-3.23x10^(-5)h)
Taking the derivative of P with respect to h:
dP/dh = -3.23x10^(-5) * 30e^(-3.23x10^(-5)h)
Now, evaluating it at h = 0:
dP/dh = -3.23x10^(-5) * 30e^(-3.23x10^(-5) * 0)
= -3.23x10^(-5) * 30e^(0)
= -3.23x10^(-5) * 30(1)
= -0.000969
So, the slope of the tangent line at h = 0 is -0.000969. Now, we can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Given that (x1, y1) = (0, 30) (since P = 30 at h = 0) and m = -0.000969, the equation becomes:
y - 30 = -0.000969(x - 0)
Simplifying:
y - 30 = -0.000969x
Therefore, the equation of the tangent line at h = 0 is P = -0.000969x + 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 in thousands of feet as h/1000. The corresponding drop in air pressure in inches can be expressed as 1 * (h/1000).
Therefore, the formula for air pressure based on this rule of thumb is:
P = 30 - (h/1000)
This formula implies that for every 1000-foot increase in height above sea level, the air pressure decreases by 1 inch.
a) To find the equation of the tangent line at h = 0, we need to find the derivative of the function P with respect to h. Then we can use the point-slope formula to find the equation.
First, let's find the derivative dP/dh. We differentiate the function P = 30e^(-3.23x10^(-5)h) using the chain rule. The derivative of e^u, with respect to u is e^u:
dP/dh = -3.23x10^(-5) * 30 * e^(-3.23x10^(-5)h)
Next, we substitute h = 0 into the derivative to find the slope at h = 0:
dP/dh (at h = 0) = -3.23x10^(-5) * 30 * e^(-3.23x10^(-5)(0))
= -3.23x10^(-5) * 30
This gives us the slope of the tangent line at h = 0.
Now, we can use the point-slope equation to find the equation of the tangent line:
y - y1 = m(x - x1)
Let's take the point (0, P) as our reference point, where P = 30 inches of mercury at h = 0.
Plugging in the values, we have:
P - 30 = (-3.23x10^(-5) * 30)(h - 0)
Simplifying further:
P - 30 = -0.000969h
This equation represents the tangent line at h = 0.
b) The rule of thumb states that air pressure drops about 1 inch for every 1000-foot increase in height above sea level. To write a formula for air pressure based on this rule of thumb, we can use the concept of a linear relationship between pressure and altitude.
Let's use P2 to represent the air pressure in inches of mercury, and h2 to represent the altitude in feet above sea level. According to the rule of thumb, we have:
P2 - 30 = -(h2/1000)
Simplifying further:
P2 = 30 - (h2/1000)
This formula gives the estimate of air pressure at an altitude h2 above sea level based on the given rule of thumb.