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
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!!!!!
4 answers
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
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.
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.
3 answers