# 2 Let f (t ) be the temperature (in degrees Celsius) of a liquid at time t (in hours). The rate of temperature change at time t is determined from f '(t ) . Explain the method of a) solution toeach of these questions in terms of f (t ) and f '(t ) (i) (ii) What is the temperature of the liquid after 6 hours? When is the temperature rising at the rate of 6 degrees per hour? (iii) By how many degrees did the temperature rise during the first 6 hours? (iv) When is the liquid's temperature only 6 degrees? (v) (vi) How fast is the temperature of the liquid changing after 6 hours? What is the average rate of increase in the temperature during the first 6 hours? (b) At the start of the month, a customer owes a credit card company $1000.00 . In the middle of the month, the customer pays $P to the company, where P < 1000.00 , and at the end of the month, the company adds interest at the rate, R , of 3% of the amount still owing. This process is repeated in each subsequent month. (i) Find the value of P for which the customer owes $1000.00 at the start of every month. (ii) Find the value of P for which the whole amount is paid off after the second payment. (iii) the Assuming that the debt has not been paid after 4 payments, show that the amount still owing at beginning of the 5th month can be expressed as PR ( R 4 1) $ 1000 R 4 , R 1.03 R 1 (iv) Show that the value of P for which the whole amount owing is exactly paid off after the n th payment is given by P 1000 R n 1 ( R 1) . Rn 1 Question # 2 Let f (t ) be the temperature (in degrees Celsius) of a liquid at time t (in hours). The rate of temperature change at time t is determined from f '(t ) . Explain the method of a) solution to each of these questions in terms of f (t ) and f '(t ) (i) (ii) What is the temperature of the liquid after 6 hours? When is the temperature rising at the rate of 6 degrees per hour? (iii) By how many degrees did the temperature rise during the first 6 hours? (iv) When is the liquid's temperature only 6 degrees? (v) (vi) How fast is the temperature of the liquid changing after 6 hours? What is the average rate of increase in the temperature during the first 6 hours? Ans: i) It can be solved in the following ways a) F(t+6) b) F(t) +f'(t)*6 ii) When f'(t)=6then the temperature is rising at the rate of 6 per hour iii) F'(t-6)+ F'(t-5)+ F'(t-4)+ F'(t-3)+ F'(t-2)+ F'(t-1) iv) When f(t)=6 the liquid's temperature is 6 degree v) F'(t+6) vi) (F'(t-6)+ F'(t-5)+ F'(t-4)+ F'(t-3)+ F'(t-2)+ F'(t-1))/6 (b) At the start of the month, a customer owes a credit card company $1000.00 . In the middle of the month, the customer pays $P to the company, where P < 1000.00 , and at the end of the month, the company adds interest at the rate, R , of 3% of the amount still owing. This process is repeated in each subsequent month. (i) Find the value of P for which the customer owes $1000.00 at the start of every month. (ii) Find the value of P for which the whole amount is paid off after the second payment. (iii) the Assuming that the debt has not been paid after 4 payments, show that the amount still owing at beginning of the 5th month can be expressed as PR ( R 4 1) $ 1000 R 4 , R 1.03 R 1 (iv) Show that the value of P for which the whole amount owing is exactly paid off after the n th payment is given by P Ans: 1000 R n 1 ( R 1) . Rn 1 i) (1000-p)*3%+(1000-p)=1000 So, p=1000*.03-.03p; So, 1.03p= 1000*.03 So, p=29.126 ii) (1000-p)*3%+(1000-p) = p So, 30-.03p+1000=2p So, 2.03p=1030 So p= 1030/2.03=507.389 iii) At the end of first month the amount due is = (1000-P)+(1000-P)*0.3 = (1000-P)(1.03) =(1000-P)R Similarly At the end of 2nd month the amount due is {(1000-P)R-P}R At the end of 3rd month the amount due is {{(1000-P)R-P}R -P}R At the end of 4th month the amount due is [{{(1000-P)R-P}R -P}R - P]R =[{{1000R-PR-P}R -P}R - P]R =[{1000R^2-PR^2-PR -P}R - P]R =[1000R^3-PR^3-PR^2 -PR - P]R =[1000R^4-PR^4-PR^3 -PR^2 - PR] =[1000R^4-PR(1+R+R^2+R^3)] =[1000R^4-PR*(R^4-1)/(R-1)] [Formula of Gemoetric Series a+ar+ar^2..+ar^n-1= a(r^n-1)/(r-1)] iv)The modified formula for nth term instead of 4th term is [1000R^n-PR*(R^n-1)/(R-1)] So this is the debt after n months. If this becomes 0 then [1000R^n-PR*(R^n-1)/(R-1)] =0 1000R^n= PR*(R^n-1)/(R-1) P= 1000R^(n-1) (R-1)/ (R^n-1) Question # 2 Let f (t ) be the temperature (in degrees Celsius) of a liquid at time t (in hours). The rate of temperature change at time t is determined from f '(t ) . Explain the method of a) solution to each of these questions in terms of f (t ) and f '(t ) (i) (ii) What is the temperature of the liquid after 6 hours? When is the temperature rising at the rate of 6 degrees per hour? (iii) By how many degrees did the temperature rise during the first 6 hours? (iv) When is the liquid's temperature only 6 degrees? (v) (vi) How fast is the temperature of the liquid changing after 6 hours? What is the average rate of increase in the temperature during the first 6 hours? Ans: i) It can be solved in the following ways a) F(t+6) b) F(t) +f'(t)*6 ii) When f'(t)=6then the temperature is rising at the rate of 6 per hour iii) F'(t-6)+ F'(t-5)+ F'(t-4)+ F'(t-3)+ F'(t-2)+ F'(t-1) iv) When f(t)=6 the liquid's temperature is 6 degree v) F'(t+6) vi) (F'(t-6)+ F'(t-5)+ F'(t-4)+ F'(t-3)+ F'(t-2)+ F'(t-1))/6 (b) At the start of the month, a customer owes a credit card company $1000.00 . In the middle of the month, the customer pays $P to the company, where P < 1000.00 , and at the end of the month, the company adds interest at the rate, R , of 3% of the amount still owing. This process is repeated in each subsequent month. (i) Find the value of P for which the customer owes $1000.00 at the start of every month. (ii) Find the value of P for which the whole amount is paid off after the second payment. (iii) the Assuming that the debt has not been paid after 4 payments, show that the amount still owing at beginning of the 5th month can be expressed as PR ( R 4 1) $ 1000 R 4 , R 1.03 R 1 (iv) Show that the value of P for which the whole amount owing is exactly paid off after the n th payment is given by P Ans: 1000 R n 1 ( R 1) . Rn 1 i) (1000-p)*3%+(1000-p)=1000 So, p=1000*.03-.03p; So, 1.03p= 1000*.03 So, p=29.126 ii) (1000-p)*3%+(1000-p) = p So, 30-.03p+1000=2p So, 2.03p=1030 So p= 1030/2.03=507.389 iii) At the end of first month the amount due is = (1000-P)+(1000-P)*0.3 = (1000-P)(1.03) =(1000-P)R Similarly At the end of 2nd month the amount due is {(1000-P)R-P}R At the end of 3rd month the amount due is {{(1000-P)R-P}R -P}R At the end of 4th month the amount due is [{{(1000-P)R-P}R -P}R - P]R =[{{1000R-PR-P}R -P}R - P]R =[{1000R^2-PR^2-PR -P}R - P]R =[1000R^3-PR^3-PR^2 -PR - P]R =[1000R^4-PR^4-PR^3 -PR^2 - PR] =[1000R^4-PR(1+R+R^2+R^3)] =[1000R^4-PR*(R^4-1)/(R-1)] [Formula of Gemoetric Series a+ar+ar^2..+ar^n-1= a(r^n-1)/(r-1)] iv)The modified formula for nth term instead of 4th term is [1000R^n-PR*(R^n-1)/(R-1)] So this is the debt after n months. If this becomes 0 then [1000R^n-PR*(R^n-1)/(R-1)] =0 1000R^n= PR*(R^n-1)/(R-1) P= 1000R^(n-1) (R-1)/ (R^n-1)