Question:
| t (years) | 2 | 3 | 5 | 7 | 10 |
| H(t) (meters) | 1.5 | 2 | 6 | 11 | 15 |
The height of a tree at time t is given by a twice-differentiable function H, where H (t) is measured in meters and t is measured in years. Selected values of H (t) are given in the table above.
(a) Use the data in the table to estimate H'(6). Using correct units, interpret the meaning of H'(6) in the context of the problem.
(b) Explain why there must be at least one time t, for 2 < t <10 , such that H'(t) = 2.
(c) Use a trapezoidal sum with the four subintervals indicated by the data in the table to approximate the average height of the tree over the time interval 2 ≤ t ≤ 10 .
(d) The height of the tree, in meters, can also be modeled by the function G, given by \(G(x)=\frac{100x}{1+x}\) , where x is the diameter of the base of the tree, in meters. When the tree is 50 meters tall, the diameter of the base of the tree is increasing at a rate of 0.03 meter per year. According to this model, what is the rate of change of the height of the tree with respect to time, in meters per year, at the time when the tree is 50 meters tall?
▶️Answer/Explanation
Ans:
(a)
\(H'(6)\approx \frac{\Delta H(t)}{\Delta t}\approx \frac{H(7)-H(5)/m}{(7-5)_{y}}=\frac{(11-6)m}{(7-5)y}=\frac{5 meters}{2 years}\)
When t = 6 years, the rate of which the tree is growing is H'(6) meters per year
(b)
By the MVT, as H(t) is continuous and differentiable on t ∈ (2, 10), there must be H'(c) = 2 where 2 < c < 10 if there exists \(\frac{H(b)-H(a)}{b-a}=2\) on the interval (2, 10). \(\frac{H(5)-H(3)}{(5-3)_{years}}=\frac{6m-2m}{2years}=2 m/y\) So c exists on interval c ∈ (2, 10)
(c)
Total height : \(\frac{1}{2}\left ( 1(1.5+2)+2(2+6)+2(6+11)+3(11+15) \right )\)
Average height : \(\frac{1}{10_{y}-2_{yr}} Total ht = \frac{1}{8}\times \frac{1}{2}\left ( 3.5 + 2(8)+2(17)+3(26) \right ) meters\)
(d)
\(G(x)=\frac{100x}{1+x}\) G(x) = 50 \(\frac{dx}{dt}=0.03 m/y\)
\(G'(x)=\frac{100\frac{dx}{dt}(1+x)-\left ( \frac{dx}{dt} \right )100x}{(1+x)^{2}}\) \(\frac{100\times 0.03(2)-0.03\times 100}{4}\)
\(50 = \frac{100x}{1+x}\Rightarrow 50(1+x)=100x\) \(\frac{6-3}{4}=\frac{3}{4}m/year\)
⇒ 1 = x x = 1m
Question:
| t(hours) | 0 | 0.3 | 1.7 | 2.8 | 4 |
| \(v_P(t)\) (meters per hour) | 0 | 55 | -29 | 55 | 48 |
The velocity of a particle, P, moving along the x-axis is given by the differentiable function vp , where \(v_P(t)\) is measured in meters per hour and t is measured in hours. Selected values of \(v_P(t)\) are shown in the table above. Particle P is at the origin at time t = 0.
(a) Justify why there must be at least one time t, for 0.3 ≤ t ≤ 2.8, at which \(v_P(t)\) , the acceleration of particle P, equals 0 meters per hour per hour.
(b) Use a trapezoidal sum with the three sub-intervals [0, 0.3], [0.3, 1.7], and [1.7, 2.8] to approximate the value of \(\int_{0}^{2.8}v_{p}(t)dt.\)
(c) A second particle, Q, also moves along the x-axis so that its velocity for 0 ≤ t ≤ 4 is given by \(v_{Q}(t)=45\sqrt{t}cos(0.063t^{2})\) meters per hour. Find the time interval during which the velocity of particle Q is at least 60 meters per hour. Find the distance traveled by particle Q during the interval when the velocity of particle Q is at least 60 meters per hour.
(d) At time t = 0, particle Q is at position x = −90. Using the result from part (b) and the function vQ from part (c), approximate the distance between particles P and Q at time t = 2.8.
▶️Answer/Explanation
Ans:
(a)
\({v_{p}}'(t)=9_{p}(t)\) \(\frac{v_{p}(2.8)-v_{p}(0.3)}{2.8-0.3}\)
Since \(v_P(t)\) is a continuous and differentiable function, MVT states that there must be at least one time t, for 0.3 ≤ t ≤ 2.8 at which \({v_{p}}'(t)=\frac{v_{p}(2.8)-v_{p}(0.3)}{2.8-0.3}=\frac{55-55}{2.5}=0.\)
(b)
\(0.3\left ( \frac{55+0}{2} \right )+1.4\left ( \frac{-29+55}{2} \right )+1.1\left ( \frac{55+(-29)}{2} \right )\)
8.25 + 18.2 + 14.3
\(\int_{0}^{2.8}{v_{p}}^{(t)}=40.75\)
(c)
\(60 = 45\sqrt{t}cos (0.053t^{2})\)
t = 1.866, 3.519
1.866 ≤ t ≤ 3.519
\(\int_{1.866}^{3.519}45\sqrt{t}cos (0.063t^{2})dt\)
= 106.109
106.109 meters
(d)
\(X_p(2.8)= 40.75\)
\(x_{Q}(2.8) = -90 +\int_{0}^{2.8}45\sqrt{t}cos (0.063t^{2})dt\)
= -90 + 135.938 = 45.938
45.938 – 40.75 = 5.188
The distance between particles P and Q at t = 2.8 is 5.188 meters
Question
| t (minutes) | 0 | 2 | 5 | 9 | 10 |
| H(t) (degrees Celsius) | 66 | 60 | 52 | 44 | 43 |
As a pot of tea cools, the temperature of the tea is modeled by a differentiable function H for 0 ≤ t ≤ 10, where time t is measured in minutes and temperature H(t) is measured in degrees Celsius. Values of H(t) at selected values of time t are shown in the table above.
(a) Use the data in the table to approximate the rate at which the temperature of the tea is changing at time t = 3.5. Show the computations that lead to your answer.
(b) Using correct units, explain the meaning of \(\frac{1}{10}\int_{0}^{10}H(t)dt\) in the context of this problem. Use a trapezoidal sum with the four subintervals indicated by the table to estimate \(\frac{1}{10}\int_{0}^{10}H(t)dt.\)
(c) Evaluate \(\int_{0}^{10}H'(t)dt.\) Using correct units, explain the meaning of the expression in the context of this problem.
(d) At time t = 0, biscuits with temperature 100 0C were removed from an oven. The temperature of the biscuits at time t is modeled by a differentiable function B for which it is known that
B'(t) = – 13.84e-0.173t . Using the given models, at time t = 10, how much cooler are the biscuits than the tea?
▶️Answer/Explanation
Ans:
(a)
\(\frac{H(5)-H(2)}{5-2}=\frac{-8}{3}\frac{0_{C}}{min}\)
(b)
\(\frac{1}{10}\int_{0}^{10}H(t)dt\approx \frac{\left [2(\frac{66+180}{2}) +3(\frac{52+60}{2})+4(\frac{44+52}{2})+1(\frac{43+44}{2}) \right ]}{10}=52.95\)
This represents the average temperature in degree Celsius of the tea over the interval 0 ≤ t ≤ 10
(c)
\(\int_{0}^{10}H'(t)dt= H(10)-H(0)=43-66=-23^{0}C\)
This expression shows the total change in temperature in degree Celsius from t = 0 to t = 10.
(d)
\(B'(t)=-13.84e^{-.173t}\)
\(B(10)=\int_{0}^{10}-13.84e^{-.173t}+100 = 100-65.817 = 34.1827\)
43-341827 = 8.817
= 88170C Celsius