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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.
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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