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Satyanarayana B.

14 years of teaching experience

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Applied Mathematics

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Question:

Use R-K method of order 4 to solve IVP $$\frac{dy}{dx}$$=$$\frac{5x^{2}-y}{e^{x+y}}$$ with y(0)=1 and step size h=0.1 at x=1

Satyanarayana B.

Answer:

Let $$f(x,y)$$=$$\frac{5x^{2}-y}{e^{x+y}}$$ $$k_{1}$$=$$h$$$$f(x_{0},y_{0})$$=0.1$$f(0,1)$$=$$-$$0.03678 $$k_{2}=$$h$$f(x_{0}+\frac{h}{2}, y_{0}+\frac{k_{1}}{2})$$=$$-$$0.03454 $$k_{3}=$$h$$f(x_{0}+\frac{h}{2}, y_{0}+\frac{k_{2}}{2})$$=$$-$$0.03454 $$k_{4}=$$h$$f(x_{0}+h, y_{0}+k_{3})$$=$$-$$0.03154 By R-K method of 4th order $$y(x_{0}+h)=y_{0}+\frac{1}{6}(k_{1}+2k_{2}+2k_{3}+k_{4})$$ $$y(0.1)=0.96558$$ Proceeding in this manner, we will get $$y(0,2), y(0,3), y(0.4), y(0.5), y(0.6), y(0.7), y(0.8), y(0.9) and y(1)$$

Pre-Calculus

TutorMe

Question:

Find the first two derivatives of $$y$$ if parametric equations of curve are $$x=a(1+$$sin$$t$$) and $$ y=a(1-$$cos $$t$$) w.r.t. $$x$$

Satyanarayana B.

Answer:

Given $$x=a(1+$$sin$$t$$) and $$ y=a(1-$$cos $$t$$) w.r.t. $$x$$ $$\frac{\mathrm{d} x}{\mathrm{d} t}=$$acost and $$ \frac{\mathrm{d} y}{\mathrm{d} t}=$$acost $$\frac{\mathrm{d} y}{\mathrm{d} x}$$= $$\frac{acost}{asint}=cott$$ $$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}$$=$$\frac{\mathrm{d} }{\mathrm{d} t}$$($$\frac{\mathrm{d} y}{\mathrm{d} x}$$)$$\frac{\mathrm{d} t}{\mathrm{d} x}$$=$$\frac{\mathrm{d} }{\mathrm{d} t}$$($$cott$$).$$\frac{1}{acost}$$=$$-$$csc$$^{2}$$t.$$\frac{1}{acost}$$=$$-$$$$\frac{sectcsc^{2}t}{a}$$

Calculus

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Question:

A former has 2400 ft fencing. He has a field of shape rectangle situated just beside the river. He wants to do fencing of this rectangular field such that maximum area should be covered and no fencing along the river. What are the dimensions of the rectangular field he should take in order to cover maximize the area?

Satyanarayana B.

Answer:

Let x, y be the length and breadth of the rectangular field. Perimeter of rectangular field is $$2x+2y$$ But field is situated beside the river $$\therefore$$ $$2x+y=2400$$ Now $$y=2400-2x$$ Area of the rectangular field is $$f(x,y)=xy$$ Hear we have to maximize $$f(x,y)$$ subject to $$2x+y=2400$$ $$\Rightarrow$$ $$f(x)=x(2400-2x)=2400x-2x^{2}$$ Now $$f^{'}(x)=2400-4x=0$$ $$\Rightarrow$$$$x=600$$ And $$f^{''}(x)=-4$$ At $$x=600$$ , $$f^{''}(600)=-4$$ Hence $$f(x) $$ has maximum at $$x=600$$ $$\therefore$$ $$y=2400-2(600)=1200$$ The required dimensions of rectangular field are $$x=600ft$$ and $$y=1200ft$$

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