Calculus problem please help please help

[Barbara1993]

​1.Suppose you put a yam in a hot oven, maintained at a constant temperature of 200C. As

(a) newton's law of heating and cooling states that the rate at which the temperature of the yam changes is proportional to the difference between its temperature and the temperature of the oven. Let H be the temperature of the yam after t minutes.Draw a possible graph of H vs time t (minutes). Explain any interesting features of the graph,(concavity).
(b) at t = 30, the temperature of the yam H=119.7and increasing at the (instantaneous)
rate of 2/min. Use the tangent line at t=30 to approximate the temperature at time t = 40. Do you expect he estimate to be too low or too high? Why?
(c) at t = 50, the temperature of the yam H=151.3.Use the secant line between the point where t=30 and 50 to give a new approximation of the temperature at t=40 . Do you expect this estimate to be to high or too low ? Why ?
i started working on this using dT/dt=K (200-H(t))
using dT/dt= 2
After 30 min H(t)= 119.7 + 2t at t=0 H(0) = 119.7
i solve for the Constance K in my first equation
2= k(200-H(0))
help am I in the right track?

 

[stapel]

​1. Estimating the Temperature of a Yam
Suppose you put a yam in a hot oven, maintained at a constant temperature of 200C. As the yam picks up heat from the oven, its temperature rises.16

What is the meaning of the "16" at the end of the statement?

​(a) Draw a possible graph of the temperature T of the yam against time t (minutes) since it is put into the oven. Explain any interesting features of the graph, and in particular explain its concavity.
(b) Suppose that, at t = 30, the temperature T of the yam is 120 and increasing at the (instantaneous) rate of 2/min. Using this information, plus what you know about the shape of the T graph, estimate the temperature at time t = 40.
(c) Suppose in addition you are told that at t = 60, the temperature of the yam is 165. Can you improve your estimate of the temperature at t = 40?
(d) Assuming all the data given so far, estimate the time at which the temperature of the yam

What are your thoughts? What have you tried? Where are you stuck?

Please be complete. Thank you!

 

[Barbara1993]

What is the meaning of the "16" at the end of the statement?


What are your thoughts? What have you tried? Where are you stuck?

Please be complete. Thank you!

I am sorry the 16 should not be there.
I worked it using newton cooling and heating law but I stuck at coming out with an equation.
i modified my question and gave the steps I took so far.

 

[stapel]

I worked it using newton cooling and heating law but I stuck at coming out with an equation.

What do you mean by "worked it"? What is the "law" (usually an equation) that you're using? How did you apply it?

For the sketch, for instance, how did you apply what you've graphed for similar exercises, and what you've seen in worked examples in the book and in your class notes?

i modified my question and gave the steps I took so far.

In future, please reply with answers. As it is, I'll have to come back later, when I've got the time, and try to tease out, working by comparison, what you've changed and what has stayed the same, to try to figure out what your "answer" was meant to be.

 

[Barbara1993]

What do you mean by "worked it"? What is the "law" (usually an equation) that you're using? How did you apply it?

For the sketch, for instance, how did you apply what you've graphed for similar exercises, and what you've seen in worked examples in the book and in your class notes?


In future, please reply with answers. As it is, I'll have to come back later, when I've got the time, and try to tease out, working by comparison, what you've changed and what has stayed the same, to try to figure out what your "answer" was meant to be.

thank you
my graph is concave down and didn't really know what to with it. we haven't don anything Like this in class at ALL. the question is the way it should be right now.
thanks for taking your time to help me out.

 

[Ishuda]

​1.Suppose you put a yam in a hot oven, maintained at a constant temperature of 200C. As

(a) newton's law of heating and cooling states that the rate at which the temperature of the yam changes is proportional to the difference between its temperature and the temperature of the oven. Let H be the temperature of the yam after t minutes.Draw a possible graph of H vs time t (minutes). Explain any interesting features of the graph,(concavity).
(b) at t = 30, the temperature of the yam H=119.7and increasing at the (instantaneous)
rate of 2/min. Use the tangent line at t=30 to approximate the temperature at time t = 40. Do you expect he estimate to be too low or too high? Why?
(c) at t = 50, the temperature of the yam H=151.3.Use the secant line between the point where t=30 and 50 to give a new approximation of the temperature at t=40 . Do you expect this estimate to be to high or too low ? Why ?
i started working on this using dT/dt=K (200-H(t))
using dT/dt= 2
After 30 min H(t)= 119.7 + 2t at t=0 H(0) = 119.7
i solve for the Constance K in my first equation
2= k(200-H(0))
help am I in the right track?

Yes, you are on the right track (at least sort of). Your equation (T changed to H)
dH/dt=K (200-H(t))
is the differential equation the temperature equation must solve. Rewriting this we have
H' + K H = 200 K
where H' means dH/dt. The solution for the homogenous equation
H' + K H = 0
is e^-Kt which gives the basic concave down shape to the curve. Since the shape is concave down, any estimate of a value based only on the slope (tangent line) would be too high. That is, the tangent line lies totally above the graph.

Now we have that
H(t) ~ H₀ + H₀' (t - t₀)
which says, the temperature at time t is approximately the temperature at time t₀ plus the slope times the difference between time t and t₀. I assume that is what you meant when you wrote
H(t)= 119.7 + 2t
and you meant time t measured from time 30 min., i.e. t₀ is 30 min. and H₀' is 2/min. From the formula, we expect the temperature to be 20 degrees more 10 min. later or H would be 129.7. As mentioned before we would expect this estimate to be high.

For the rest of the problem, consider what happens when you connect two points on the graph of the temperature with a straight line. Like before IF it lay over the graph you would expect the estimate of the temprature to be too high but what if the line lay completely under the graph (except at the two end points of course). What is the equation of that straight line?

 

[Barbara1993]

Yes, you are on the right track (at least sort of). Your equation (T changed to H)
dH/dt=K (200-H(t))
is the differential equation the temperature equation must solve. Rewriting this we have
H' + K H = 200 K
where H' means dH/dt. The solution for the homogenous equation
H' + K H = 0
is e^-Kt which gives the basic concave down shape to the curve. Since the shape is concave down, any estimate of a value based only on the slope (tangent line) would be too high. That is, the tangent line lies totally above the graph.

Now we have that
H(t) ~ H₀ + H₀' (t - t₀)
which says, the temperature at time t is approximately the temperature at time t₀ plus the slope times the difference between time t and t₀. I assume that is what you meant when you wrote
H(t)= 119.7 + 2t
and you meant time t measured from time 30 min., i.e. t₀ is 30 min. and H₀' is 2/min. From the formula, we expect the temperature to be 20 degrees more 10 min. later or H would be 129.7. As mentioned before we would expect this estimate to be high.

For the rest of the problem, consider what happens when you connect two points on the graph of the temperature with a straight line. Like before IF it lay over the graph you would expect the estimate of the temprature to be too high but what if the line lay completely under the graph (except at the two end points of course). What is the equation of that straight line?

crystal clear!!! Now I understand where and why I got stuck. The equation of a straight line is y= mx+b do I need to solve for that?

 

[Ishuda]

crystal clear!!! Now I understand where and why I got stuck. The equation of a straight line is y= mx+b do I need to solve for that?

Yes. Your line needs to go through the points t = 30, H=119.7 and t = 50, H=151.3. (actually in this case H replaces y and t replaces x). Once you have the equation for the line, compute the value of H for t=40.


Also, since this is in the Calculus section and 'we don't know about solutions to differential equations yet', there is another way to determine the concave down aspect of the temperature curve. Since the rate of change (the derivative) of the temp. is proportional to the difference in the oven and the yam and the yam is getting hotter all the time, the derivative is always decreasing. That is, since the yam is cooler than the oven and the heat in the yam is increasing, then the difference in temp. between the oven and the yam is positive but always getting smaller (always decreasing). That is one definition of a concave down curve (derivative always decreasing).

 

[Barbara1993]

Yes. Your line needs to go through the points t = 30, H=119.7 and t = 50, H=151.3. (actually in this case H replaces y and t replaces x). Once you have the equation for the line, compute the value of H for t=40.


Also, since this is in the Calculus section and 'we don't know about solutions to differential equations yet', there is another way to determine the concave down aspect of the temperature curve. Since the rate of change (the derivative) of the temp. is proportional to the difference in the oven and the yam and the yam is getting hotter all the time, the derivative is always decreasing. That is, since the yam is cooler than the oven and the heat in the yam is increasing, then the difference in temp. between the oven and the yam is positive but always getting smaller (always decreasing). That is one definition of a concave down curve (derivative always decreasing).

thanks lot for you help
so I solved y=mx+b
sing the two points (30,119.7) and (50,151.3)
I got y= 1.58x b
then solving or b I got b= 72.3
y=1.53x+72.3so H(t)=1.53t + 72.3
at t=40 H(40)=135.5 c I expected it to be too low because the secant line should be under the graph and from my original formula 119.7+2(10)=139.7c : BOOOM BOOM THANKS :razz: