Q1) Given any pair of curves y^2-x^2=a and xy=b , where a and b are positives co

Question

Q1) Given any pair of curves y^2-x^2=a and xy=b , where a and b are positives constants, show that their tangents are perpendicular at any point where they intersect.

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Q2) At a particular location, f(p) is the number of liters of soda sold when the price is p dollars per liters.

(a) What does the statement f(1)=5757 tell you about soda sales?

(b) Find and interpret f^-1(5757).

(c) What does the statement f'(1)=-1234 tell you about soda sales?

(d) Find and interpret (f^-1)'(5757).

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Explanation / Answer

(1)

y^2 - x^2 = a and xy = b

now derivate 1st curve

=> 2y y ' - 2x = 0

=> y ' = x/y which is slope m1

now derivate 2nd curve

=> y + xy' = 0

=> y ' = -y/x which is slope m2

so now find product of slopes of two curves

m1 * m2 = (x/y)(-y/x)

=> m1*m2 = -1

so product of slopes = -1

so tangents are perpendicular

(2)

(a) f(1)=5757 tells us that

number of liters of soda sold = 5757 where price is $1 per liter

(b)

f(1)=5757

=> 1 = f^-1 ( 5757)

so f^-1 ( 5757) = 1

(c) f'(1)=-1234

says change of number of liters of soda sold = -1234 where price is $1 per liter

(d) (f^-1)' ( x) = 1 / f ' ( f^-1(x) )

so ( f^-1) ' ( 5757 ) = 1/f ' ( f^-1(5757) )

= 1/ f ' ( 1 )

= -1 / 1234

(3)

d/dx (ln(e^x+e^-x)) = [ 1/ (e^x + e^-x) ] d/dx ( e^x + e^-x)

= ( e^x - e^-x ) / (e^x + e^-x)

= [ (e^x - e^-x) / 2 ] / [ ( e^x + e^ -x ) /2 ]

= sinhx / cos hx

= tan hx

(4)

(a)

f(x)=(x^(2)+2x)^cos(x)

ln ( f(x) ) = ln (x^(2)+2x)^cos(x)

=> ln ( f(X) ) = cosx ln ( x^2 + 2x)

now derivate

f ' (x) / f(x) = - sinx ln (x^2 + 2x ) + [ (cosx)(2x+2)/(x^2+2x) ]

so f ' (X) = f(x) [ - sinx ln (x^2 + 2x ) + [ (cosx)(2x+2)/(x^2+2x) ] ]

=> f ' (x) = (x^(2)+2x)^cos(x)[- sinx ln (x^2 + 2x ) + [ (cosx)(2x+2)/(x^2+2x) ] ]

(b)

g(x)=2x^(2x)^(2x)

ln g(x) = ln 2x^(2x)^(2x)

=> ln ( g(x) ) = (2x)^(2x) ln 2x

ln ( ln ( g(x) ) = 2x ln2x + ln ( ln(2x))

now derivate

=> g'(x)/ [ ln (g(x))*g(X) ] = 2 ln2x + 2 + 2 / [ ln(2x) * 2x ]

=> g ' (X) = [ g(x) ln (g(x)) ] [ 2 ln2x + 2 + (1/ xln(2x) ) ]

=>g'(x) = [ 2x^(2x)^(2x) ln 2x^(2x)^(2x) ] [ 2 ln2x + 2 + (1/ xln(2x) ) ]

(c)

h(x)=arctan(x)^cot(x)

log h(x) = cotx log tan^-1 x

now derivate

h ' (X) / h(x) = -cscx cotx ln tan^-1 x + [cotx / tan^-1x ] (1/ (1+x^2) )

=> h ' (x) = h(x) [ -cscx cotx ln tan^-1 x + [cotx / tan^-1x ] (1/ (1+x^2) ) ]

=>h ' (x) = arctan(x)^cot(x)[ -cscx cotx ln arctan x + [cotx / arctanx ] (1/ (1+x^2) ) ]

(5)

4thRoot(101) = (101)^(1/4)

nearest 4t root we have is 4th root of 81 = 3

so f(x) = x^(1/4)

f(x + delta x) = f(x) + f ' (x) * delta x

so f( 81 + 20) = f(81) + (1/4)(x^-3/4) * 20

= f(81) + (1/4)(81^-3/4)*20

= 3 + (1/4)(1/27)*20

= 3 + 5/27

=86/27

=3.185

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