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SOLUTION 3: Draw a right triangle with leg one x, leg two y, and hypotenuse z, and assume each edge of the right triangle is a function of time t.

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     a.) Using the Pythagorean Theorem, we get the hypotenuse equation z2=x2+y2 GIVEN:    dxdt=5 in/sec.  and  dydt=7 in/sec.

FIND:    dzdt when x=8 in. and y=6 in.

Now differentiate the hypotenuse equation with respect to time t getting

D{z2}=D{x2+y2}    2zdzdt=2xdxdt+2ydydt   

( Multiply both sides of the equation by 1/2.)

zdzdt=xdxdt+ydydt    ( Use the Pythagorean Theorem to solve for z: z2=x2+y2    z2=82+62=64+36=100    z=10. Now let dxdt=5,dydt=7,x=8,y=6, and z=10.) (10)dzdt=(8)(5)+(6)(7)=2    dzdt=210=15 in/sec.

     b.) The perimeter of the right triangle is P=x+y+z GIVEN:    dxdt=5 in/sec.  and  dydt=7 in/sec.

FIND:    dPdt when x=8 in. and y=6 in.

Now differentiate the perimeter equation with respect to time t getting

D{P}=D{x+y+z}    dPdt=dxdt+dydt+dzdt   

( Recall from solution 3 part a.): dzdt=15 in./sec. Now let dxdt=5,dydt=7, and dzdt=15.)

dPdt=(5)+(8)+(15)     dPdt=255+40+(8)+15     dPdt=165 in/sec.

     c.) The area of the right triangle is A=(1/2)(base)(height)     A=12xy GIVEN:    dxdt=5 in/sec.  and  dydt=7 in/sec.

FIND:    dAdt when x=8 in. and y=6 in.

Now differentiate the area equation with respect to time t using the product rule getting

D{A}=D{12xy}    dAdt=12(xdydt+dxdty)   

( Now let dxdt=5,dydt=7,x=8, and y=6.)

dAdt=12((8)(7)+(5)(6))=13 in2/sec.

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