HELP FOR REPS! =) (1 Viewer)

[Dylanamali]

University Differentiation Questions:

a) Find all points on the curve x²y² + xy = 2 where the slope of the tangent is –1.



b) A table of values for f, g, f' and g' is given.
Estimate the values of h'(1) and H'(1) where h and H are defined by h(x) = f[g(x)] and H(x) = g[f(x)]. Show working.

x	f(x)	g(x)	f ’(x)	g’(x)
1	3	2	4	6
2	1	8	5	7
3	7	2	7	9

<tbody>
</tbody>

c) Explain the meaning of the mathematical concept “differentiation”. 5 MARKS

 

[Dylanamali]

~

 

[bleakarcher]

University Differentiation Questions:

a) Find all points on the curve x²y² + xy = 2 where the slope of the tangent is –1.



b) A table of values for f, g, f' and g' is given.
Estimate the values of h'(1) and H'(1) where h and H are defined by h(x) = f[g(x)] and H(x) = g[f(x)]. Show working.

x	f(x)	g(x)	f ’(x)	g’(x)
1	3	2	4	6
2	1	8	5	7
3	7	2	7	9

<tbody>
</tbody>

c) Explain the meaning of the mathematical concept “differentiation”. 5 MARKS

x^2y^2+xy=2
Differentiating implicitly with respect to x:
=>2xy^2+2yx^2(dy/dx)+y+x(dy/dx)=0
Therefore, dy/dx=[-2xy^2-y]/[2yx^2+x]=-y/x
If the slope of the tangent at some arbitrary point is -1 then at that point dy/dx=-1.
=>-y/x=-1
i.e. x=y
This means that at the points on the curve where x=y the gradient of the tangent is -1.
Now the original equation of the curve was x^2y^2+xy=2.
Substitute x=y into the equation to find the points at which dy/dx=-1.

 

[bleakarcher]

University Differentiation Questions:

a) Find all points on the curve x²y² + xy = 2 where the slope of the tangent is –1.



b) A table of values for f, g, f' and g' is given.
Estimate the values of h'(1) and H'(1) where h and H are defined by h(x) = f[g(x)] and H(x) = g[f(x)]. Show working.

x	f(x)	g(x)	f ’(x)	g’(x)
1	3	2	4	6
2	1	8	5	7
3	7	2	7	9

<tbody>
</tbody>

c) Explain the meaning of the mathematical concept “differentiation”. 5 MARKS

h(x)=f[g(x)]
h'(x)=f '[g(x)]*g'(x)
Hence, h'(1)=f '[g(1)]*g'(1)=f '(2)*6=6*5=30

H(x)=g[f(x)]
H '(x)=g'[f(x)]*f '(x)
Hence, H '(1)=g'[f(1)]*f '(1)=g'(3)*4=4*9=36

 

[powlmao]

What level is this at Uni?

 

[bleakarcher]

University Differentiation Questions:

a) Find all points on the curve x²y² + xy = 2 where the slope of the tangent is –1.



b) A table of values for f, g, f' and g' is given.
Estimate the values of h'(1) and H'(1) where h and H are defined by h(x) = f[g(x)] and H(x) = g[f(x)]. Show working.

x	f(x)	g(x)	f ’(x)	g’(x)
1	3	2	4	6
2	1	8	5	7
3	7	2	7	9

<tbody>
</tbody>

c) Explain the meaning of the mathematical concept “differentiation”. 5 MARKS

In calculus, differentiation is an common operation used to determine the gradient of the tangent to a curve at any arbitary point which lie on it. If there is a function f(x) then the derivative of the function, also known as the gradient function, is denoted by f '(x) and now any values of x for which the domain of the function f(x) may be substituted into the gradient function to find the gradient of the tangent to the curve at that point. Alternate notation for f '(x) include dy/dx or y' where y=f(x). Say you have a function y=f(x) that is differentiable at some arbitrary point (x,y) then the gradient of the tangent at that arbitrary point (x,y) is given by the gradient of the line joining (x,y) and (x+δx,y+δy) (which also lies on the curve) where δ represents an infinitesimal change. This line therefore has gradient δy/δx. However, this is only an approximation to the gradient of the tangent to the curve at (x,y). If we take the limit though as δx->0 i.e. as the distance between the two points (x,y) and (x+δx,y+δy) gets smaller and smaller we get closer and closer to the true value of the gradient of the tangent at that point. Hence, the definition of a derivative follows:
dy/dx=lim[δx->0] [δy/δx]


Hope that helped

 

[Carrotsticks]

In the future Dylanamali when you have University questions, please post them up in this section:

http://community.boredofstudies.org/forumdisplay.php?f=1003

This whole section is for Tertiary level Mathematics.

EDIT: Has been moved to appropriate location.

 

[OzKo]

What level is this at Uni?

Looks like a standard Differential Calc. problem for first years.

 

[Dylanamali]

In the future Dylanamali when you have University questions, please post them up in this section:

http://community.boredofstudies.org/forumdisplay.php?f=1003

This whole section is for Tertiary level Mathematics.

EDIT: Has been moved to appropriate location.

duly noted.

 

[Dylanamali]

x^2y^2+xy=2
Differentiating implicitly with respect to x:
=>2xy^2+2yx^2(dy/dx)+y+x(dy/dx)=0
Therefore, dy/dx=[-2xy^2-y]/[2yx^2+x]=-y/x
If the slope of the tangent at some arbitrary point is -1 then at that point dy/dx=-1.
=>-y/x=-1
i.e. x=y
This means that at the points on the curve where x=y the gradient of the tangent is -1.
Now the original equation of the curve was x^2y^2+xy=2.
Substitute x=y into the equation to find the points at which dy/dx=-1.

thanks for everything. definitely appreciate it.

How would you finish off the question? Or is that the final answer?

 

[bleakarcher]

x^2y^2+xy=2
Differentiating implicitly with respect to x:
=>2xy^2+2yx^2(dy/dx)+y+x(dy/dx)=0
Therefore, dy/dx=[-2xy^2-y]/[2yx^2+x]=-y/x
If the slope of the tangent at some arbitrary point is -1 then at that point dy/dx=-1.
=>-y/x=-1
i.e. x=y
This means that at the points on the curve where x=y the gradient of the tangent is -1.
Now the original equation of the curve was x^2y^2+xy=2.
Substitute x=y into the equation to find the points at which dy/dx=-1.

To finish off:
When x=y,
=>x^2*x^2+x^2=2
i.e. x^4+x^2-2=0
(x^2+2)(x^2-1)=0
Hence, x=+/-1
So at the points where x=-1 or x=1 the gradient of the tangent to the curve is -1.

 

[Dylanamali]

To finish off:
When x=y,
=>x^2*x^2+x^2=2
i.e. x^4+x^2-2=0
(x^2+2)(x^2-1)=0
Hence, x=+/-1
So at the points where x=-1 or x=1 the gradient of the tangent to the curve is -1.

do you consider the equation xy = -1/2 ?

 

[Dylanamali]

MY final answer: At the points (1,1), (1,-2), (-1,-1) and (-1,2) the gradient of the tangent to the curve is -1.
substituting x=1 and x=-1 back into the original equation to get the y value.
Is that correct?

 

[bleakarcher]

yep lol, sorry for not doing the entire question

 

[bleakarcher]

MY final answer: At the points (1,1), (1,-2), (-1,-1) and (-1,2) the gradient of the tangent to the curve is -1.
substituting x=1 and x=-1 back into the original equation to get the y value.
Is that correct?

Dude really sorry about this but the final answer is actually (-1,-1) and (1,1). Since if you sketch you have hyperboli with branches lying in opposite quadrant. Can someone explain why the equations gave incorrect solutions?

 

[bleakarcher]

http://www.wolframalpha.com/input/?i=%28xy%29^2%2Bxy%3D2