- Two ropes hold a boat at a dock. The tension in the ropes can be represented by 40 + 10j N and 50 – 25j N. Find the resultant force.
- The total power P (in watts) transmitted by an AM radio station is given by
P = 500 + 250 m^2 ,
where m is the modulation index. Find the instantaneous rate of change of P with respect to m for m = 0.92 - Find the derivative of the following polynomials:
i. y = 4 x^2 + 7 x + 3
ii. y = 4 x^ -6 - 5 x^3 + x - Find the derivative of the following trigs:
a. y = 6 sin Ө
b. y = 7 sin Ө + 3 cos Ө - Determine the gradient of each of the given functions at the given point.
a) s = 2t^3 - 5t^2 + 4 (-1,-3)
b) y = 5 sin Ө where Ө = 38◦ - Find the area under the curve y = x^3 that is between the lines x = 1 and x = 2.
- Integrate the following
i) 4x^6 + 3x^ -4 + 1
ii) 2 cos Ө - The electric field E at a distance r from a point charge is E = k/(r2), where k is a constant. Find an expression for the instantaneous rate of change of the electric field with respect to r.
- The blade of a saber saw moves vertically up and down, and its displacement y (in cm) is given by y = 1.85 sin t, where t is the time (in s). Find the velocity of the blade for t = 0.025 s.
- An open-top container is to be made from a rectangular piece of cardboard 24 cm by 38 cm. Equal squares of side x cm are to be cut from each corner, then the sides are to be bent up and taped together. Find the instantaneous rate of change of the volume V of the container with respect to x for x = 4 cm.
- An analysis of a company’s records shows that in a day the rate of change of profit (in dollars) in producing x generators is dp/dx = 60(30 – 4x^5). Find the profit in producing x generators if a loss of $500 is incurred if none are produced.
- The vertical displacement y (in cm) of the end of an industrial robot arm for each cycle is y = 2t^1.5 – cos t, where t is the time (in s). Find its vertical velocity for t = 15s.
Friday, November 20, 2009
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(3)firstly the derivative of the equation is the differential of the equation. the first thing we do is multiply the power by the base and then minus 1 from the power.
ReplyDelete(1)y= 4x^2 +7x +3
dy/dx= (4*2)x^(2-1)+ (7*1)x^(1-0)
dy/dy= 8x+7
(2) y=4x^-6 - 5x^3 +x
dy/dx= (4*-6)x^(-6-1) - (5*3)x^(3-1) + x^(1-1)
dy/dx= -24x^-7 - 15x^2
2) P = 500 + 250 m^2
ReplyDeletedP/dm = 2(250)(m^2-1)
dP/dm = 500m
dP/dm = 500 x 0.92
dP/dm = 460
3) i. y = 4 x^2 + 7x + 3
ReplyDeletedy/dx = 2(4)(x^2-1) + 1(7)(x^1-1)
dy/dx = 8x + 7
ii. y = 4 x^-6 - 5x^3 + x
dy/dx = -6(4)(x^-6-1) -3(5)(x^3-1) + 1(x^1-1)
dy/dx = -24x^-7 - 15x^2 + 1
4)(a) y=6 sin Ө
ReplyDeletedy/dx= 6 cos Ө
(b) y= 7 sinӨ + 3 cosӨ
dy/dx= 7 cosӨ - 3 sinӨ
Find the derivative of the following polynomials:
ReplyDeletei. y = 4 x^2 + 7 x + 3
ANSWER
y = 4x^2 + 7x + 3
dy/dx = 8x + 7
This is done by multiplying the base (4x) by the power (2) and then minusing 1 from the power
The second term 7x, here x is to the power of one so the differential of 7x is 7
The third term 3 is a constant so the differatial is 0
Find the derivative of the following polynomials:
ReplyDeleteii. y = 4x^ -6 - 5x^3 + x
y = 4x^ -6 - 5x^3 + x
dy/dx = -24x^-7 - 15x^2 + 1
9)displacement given by
ReplyDeletey = 1.85sin t
differentiating the above equation gives an expression for velocity
dy/dx = 1.85cos t
velocity = 1.85cos t
t = 0.025
v = 1.85cos (0.025)
=1.85
velocity = 1.85cm/s^2
12)
ReplyDeletedisplacement given by
y = 2t^1.5 - cos t
differentiating the above equation gives velocity
dy/dx = (1.5)(2)t^0.5 + sin t
velocity when t = 15
velocity = (1.5)(2)(15^0.5) + sin15
11.9 cm/s^2
(1)
ReplyDeletefind the two equations:
40 + 10j N .............1
50 – 25j N..............2
using the elimination method; eq1 + eq2;
=>(40 + 50) + (10jN - 25jN)
=>90 + (-15jN)
=>90 - 15jN
(2)
ReplyDeleteP = 500 + 250 m^2
dP/dm = 2(250)m^2-1
= 500m
when m = 0.92;
dP/dm = 500 (0.92)
= 460
(3)
ReplyDeletei) y = 4 x^2 + 7 x + 3
dy/dx = (2)4x^(2-1) + 7
= 8x + 7
(3)
ReplyDeleteii)
y = 4 x^ -6 - 5 x^3 + x
dy/dx= (-6)4x^(-6-1) -(3)5x^(3-1) + 1
= -24x^-7 - 15x^2 + 1
(4)
ReplyDeletea. y = 6 sin Ө
dy/dӨ = 6 cos Ө
(4)
ReplyDeleteb. y = 7 sin Ө + 3 cos Ө
dy/dӨ of sin Ө = cos Ө
dy/dӨ of cos Ө = -sin Ө
dy/dӨ = 7 cos Ө - 3 sin Ө
(1)
ReplyDeleteFor bornagain16
You cannot use the elimination method of solving simultaneous equations to find the resultant force because the two statements are vectors and not equations (there is no equality sign).
To find the resultant force you need to add the two forces (tension) which are represented by vectors.
40 +10j
50 -25j
We simply add all the common terms together
Resultant force in N = 90 - 15j
so even though your final answer was accurate your working or your arrival to your answer made little or no sense. sorry.
nu(2) p= 500+250m^2
ReplyDeletem= .92
500(.92)= 460
ques 6
ReplyDeletey = x^3
integral y = { x^4/4} between x = 1 and x =2
[(1^4/4) - (2^4/4)]
0.25 - 4
-3.75
Area of curve bounded between x=1 & x=2
= -3.75 squared units
The negative value just indicates that the area is under the x-axis.
no.3(i)
ReplyDeletey=4x^2+7x+3
dy/dx= (4*2)x^(2-1)+ (7*1)x^(1-0)
dy/dy= 8x+7
no.3(ii)
ReplyDeletey=4x^-6-5x^3+x
dy/dx=(4*-6)x^(-6-1)-(5*3)x^(3-1)+x^(1-1)
dy/dx= -24x^-7-15x^2
2) P=500+250m^2
ReplyDeletedp/dm=500m
=500(0.92)
=460
3)Find the derivatives of polynomials
ReplyDeletei)y=4x^2+7x+3
dy/dx=8x+7
ii)y=4x^-6-5x^3+x
dy/dx=-24x^-7-15x^2+1
or
dy/dx=-24/x^7-15/x^-2+1
4)Derivative of foll trigs
ReplyDeletea) y=6sinQ
dy/dx=6cosQ
b)y=7sinQ+3cosQ
dy/dx=7cosQ-3sinQ
5)Determine the gradient
ReplyDeletea)s=2t^3-5t^2+4 (where t is -1,-3 respectively)
ds/dt=6t^2-10t (-1)
=6(-1)^2-10(-1)
=6+10
=16
or
=6(-3)^2-10(-3)
=54+30
=84
b)y=5sinQ whereQ=38degrees
dy/dx=5cosQ
=5cos38
=3.94
question 2
ReplyDeleteP=500 + 250m^2
dP/dm=500m
when m=0.92
dP/dm=500(0.92)
dP/dm=460
For question 8
ReplyDeleteE=k/(r2) or k(2r^-1)
dE/dr=k(-2r^-2) or k/-2r^2
Question 9
ReplyDeletey=1.85 sin t
dy/dx=1.85cos t where t is 0.025s
=1.85 cos (0.025s)
v=1.85cm/s
Coming to a conclusion, we have
ReplyDeletel=(38-2x), w=(24-2x) and h=(2x)
v(Is given by multiplying l*w*h)
v=(912-48x-76x+4x^2)(2x)
=1824x-96x^2-152x^2+8x^3
dv/dx=24x^2-496x+1824x
where x=4cm
dv/dx=24(4)^2-496(4)+1824(4)
=384-1984+7296
=5696
Question 12
ReplyDeletey=2t^1.5-cos t
dy/dt=3t^0.5+sin t where t=15s
=3(15)^0.5+sin15
=11.6 + 0.3
V=11.9cm/s
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDelete2. P = 500 + 250 m^2 ,
ReplyDeletem = 0.92
dP/dm = 500 m
dP/dm for (m=0.92) = 500 (0.92)
dP/dm = 460
3. the derivatives can be found by integrating the equations
ReplyDeletei. y = 4 x^2 + 7 x + 3
ReplyDeletey = (4x3)/3 + (7x2)/2 + 3x
ii. y = 4 x^ -6 - 5 x^3 + x
ReplyDeletey = (4x^-5)/-5 - (5x^4)/4 + (x^2)/2
4.a. y = 6 sin Ө
ReplyDeletederivative y = 6 cos Ө
4b. y = 7 sin Ө + 3 cos Ө
ReplyDeletederivative y = 7 cos Ө - 3 sin Ө
5 a) s = 2t^3 - 5t^2 + 4 (-1,-3)
ReplyDeleteds/dt = 6t^2 - 10t
ds/dt at (-1, -3) = 6(-3)^2 - 10(-3)
ds/dt = 84
b) y = 5 sin Ө where Ө = 38◦
ReplyDeletedy/d sinӨ = 5 cos Ө
dy/ d sinӨ where Ө= 38 = 5 cos(38)
dy/ d sinӨ = 3.94
This comment has been removed by the author.
ReplyDeletefor ques 3
ReplyDeletey = 4 x^2 + 7 x + 3
dy/dx = (2)4x^(2-1) + 7
=8x + 7
#2)
ReplyDeleteP = 500 + 250 m^2
m= 0.92
dP/dm = 2(250)(m^2-1)
dP/dm = 500m
dP/dm = 500 x 0.92
dP/dm = 460
for ques 3 ii)
ReplyDeletey=4x^-6-5x^3+x
dy/dx=-24x^-7-15x^2+1 or
dy/dx=-24/x^7-15/x^-2+1
3)i)
ReplyDeletey = 4 x^2 + 7x + 3
dy/dx = 2(4)(x^2-1) + 1(7)(x^1-1)
dy/dx = 8x + 7
3)ii.
ReplyDeletey = 4 x^-6 - 5x^3 + x
dy/dx = -6(4)(x^-6-1) -3(5)(x^3-1) + 1(x^1-1)
dy/dx = -24x^-7 - 15x^2 + 1
4)(a)
ReplyDeletey=6 sin Ө
dy/dx= 6 cos Ө
4)(b)
ReplyDeletey= 7 sinӨ + 3 cosӨ
dy/dx= 7 cosӨ - 3 sinӨ
5) a)
ReplyDeletes = 2t^3 - 5t^2 + 4 (-1,-3)
ds/dt = 6t^2 - 10t
ds/dt at (-1, -3) = 6(-3)^2 - 10(-3)
ds/dt = 84
5)b)
ReplyDeletey = 5 sin Ө where Ө = 38◦
dy/d sinӨ = 5 cos Ө
dy/ d sinӨ where Ө= 38 = 5 cos(38)
dy/ d sinӨ = 3.94
6)
ReplyDeletey = x^3
integral y = { x^4/4} between x = 1 and x =2
[(1^4/4) - (2^4/4)]
0.25 - 4
-3.75
Area of curve between x=1 & x=2
= -3.75 ^2 untis
2) p= 500 +250 m^2
ReplyDeletedp/dm= (2*250)m^(2-1)
dp/dm= 500m
sub m=0.92 to find the instanteous rate of change of P
dp/dm= 500(0.92)
dp/dm= 460
9)
ReplyDeletey=1.85 sin t
dy/dx=1.85cos t where t is 0.025s
=1.85 cos (0.025s)
v=1.85cm/s
2)
ReplyDeletep= 500 +250 m^2
dp/dm= (2*250)m^(2-1)
dp/dm= 500m
sub m=0.92
dp/dm= 500(0.92)
dp/dm= 460
3(i)
ReplyDeletey=4x^2+7x+3
dy/dx= (4*2)x^(2-1)+ (7*1)x^(1-0)
dy/dy= 8x+7
hmmmm..bornagain16 u cannot use simultaneous equations when you are asked to find the resultion force..your method is wrong for number 1.
ReplyDelete1)40 + 10j N
ReplyDelete50 – 25j N
to find the resulting force you add the like terms
40+50=90
10j N+ - 25j N = -15J N
ANSWER= 90 - 15j N
(2)
ReplyDeleteP = 500 + 250 m^2
dp/dm = 2*250 m^2-1
= 500 m
m = 0.92
dp/dm = 500*0.92
= 460
3)
ReplyDeletey= 4 x^2 + 7x + 3
dy/dx= 2(4)(x^2-1) + 1(7)(x^1-1)
dy/dx= 8x + 7
y= 4 x^-6 - 5x^3 + x
dy/dx= -6(4)(x^-6-1) -3(5)(x^3-1) + 1(x^1-1)
dy/dx= -24x^-7-15x^2
9)
ReplyDeletey = 1.85sin t
dy/dx = 1.85cost
velocity = 1.85cos t
t = 0.025
velocity = 1.85cos (0.025)
=1.85
velocity = 1.85cm/s^2
1) Resultant= 40+10jN+50_jN
ReplyDelete= 90-15jN
2). P=500+250m^2
ReplyDeletedP/dm=500m
when m= 0.92
dP/dm=500(0.92)
= 460
3).
ReplyDeletey=4x^2+7x+3
dy/dx= 8x+7
3) (2)
ReplyDeletey=4x^-6-5x^3+3
dy/dx= -24x-7-15x^2+1
4)a).
ReplyDeletey= 6sin@
dy/d@= 6cos@
4)b).
ReplyDeletey=7sin@=3cos@
dy/d@= 7cos@-3cos@
(4)a) y= 6 sin*
ReplyDeletedy/d*= 6 cos*
(4)b) y= 7 sin Ө + 3 cos Ө
ReplyDeletedy/dӨ= 7 cos Ө - 3 sin Ө
(7) (i) 4x^6 + 3x^-4 +1
ReplyDelete∫ (4x^(6+1))/(6+1) + (3x^(-4+1))/(-4+1) +x +c
∫ ( 4x^7)/7 + (3x^-3)/-3 + x +c
∫ ( 4x^7)/7 + (-x^-3) +x +c
(7)ii) 2 cos Ө
ReplyDelete∫ 2 sin Ө + c
(8) E= k/ (2r)
ReplyDeleteE= k2r^-1
dE/dr= (2*-1)kr^(-1-1)
dE/dr= -2kr^-2
dE/dr= -2k/r^2
question 2
ReplyDeleteP=500 + 250m^2
dP/dm=500m
m=0.92
dP/dm=500(0.92)
=460
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeletenumber4
ReplyDeletea.
y=6 sin Ө
dy/dx= 6 cos Ө
number4
ReplyDeleteb.
y = 7 sin Ө + 3 cos Ө
dy/dӨ of sin Ө = cos Ө
dy/dӨ of cos Ө = -sin Ө
dy/dӨ = 7 cos Ө - 3 sin Ө
number9
ReplyDeletey=1.85 sin(t)
dy/dx=1.85cos(t)
where t is 0.025s
therefore
dy/dx=1.85 cos (0.025s)
(1)
ReplyDeletefind the two equations:
40 + 10j N .............1
50 – 25j N..............2
This comment has been removed by the author.
ReplyDeletequestion 1 cont..
ReplyDeleteusing the elimination method; eq1 + eq2;
=>(40 + 50) + (10jN - 25jN)
=>90 + (-15jN)
=>90 - 15jN
(2)
ReplyDeleteP = 500 + 250 m^2
dP/dm = 2(250)m^2-1
= 500m
when m = 0.92;
dP/dm = 500 (0.92)
= 460
3) i. y = 4 x^2 + 7x + 3
ReplyDeletedy/dx = 2(4)(x^2-1) + 1(7)(x^1-1)
dy/dx = 8x + 7
ii. y = 4 x^-6 - 5x^3 + x
dy/dx = -6(4)(x^-6-1) -3(5)(x^3-1) + 1(x^1-1)
dy/dx = -24x^-7 - 15x^2 + 1
4.a. y = 6 sin Ө
ReplyDeletederivative y = 6 cos Ө
4b. y = 7 sin Ө + 3 cos Ө
derivative y = 7 cos Ө - 3 sin Ө
5 a) s = 2t^3 - 5t^2 + 4 (-1,-3)
ReplyDeleteds/dt = 6t^2 - 10t
ds/dt at (-1, -3) = 6(-3)^2 - 10(-3)
ds/dt = 84
5 a) s = 2t^3 - 5t^2 + 4 (-1,-3)
ReplyDeleteds/dt = 6t^2 - 10t
ds/dt at (-1, -3) = 6(-3)^2 - 10(-3)
ds/dt = 84
9)displacement given by
ReplyDeletey = 1.85sin t
differentiating the above equation gives an expression for velocity
dy/dx = 1.85cos t
velocity = 1.85cos t
t = 0.025
v = 1.85cos (0.025)
=1.85
velocity = 1.85cm/s^2
12)
ReplyDeletedisplacement given by
y = 2t^1.5 - cos t
differentiating the above equation gives velocity
dy/dx = (1.5)(2)t^0.5 + sin t
velocity when t = 15
velocity = (1.5)(2)(15^0.5) + sin15
11.9 cm/s^2
ques 6
ReplyDeletey = x^3
integral y = { x^4/4} between x = 1 and x =2
[(1^4/4) - (2^4/4)]
0.25 - 4
-3.75
Area of curve bounded between x=1 & x=2
= -3.75 squared units
The negative value just indicates that the area is under the x-axis.
(1)
ReplyDeletefind the two equations:
40 + 10j N .............1
50 – 25j N..............2
using the elimination method; eq1 + eq2;
=>(40 + 50) + (10jN - 25jN)
=>90 + (-15jN)
=>90 - 15jN
(2)
ReplyDeleteP = 500 + 250 m^2
dP/dm = 2(250)m^2-1
= 500m
when m = 0.92;
dP/dm = 500 (0.92)
= 460
(3)
ReplyDeletei) y = 4 x^2 + 7 x + 3
dy/dx = (2)4x^(2-1) + 7
= 8x + 7
(3)
ReplyDeleteii)
y = 4 x^ -6 - 5 x^3 + x
dy/dx= (-6)4x^(-6-1) -(3)5x^(3-1) + 1
= -24x^-7 - 15x^2 + 1
(4)
ReplyDeletea. y = 6 sin Ө
dy/dӨ = 6 cos Ө
(4)
ReplyDeleteb. y = 7 sin Ө + 3 cos Ө
dy/dӨ of sin Ө = cos Ө
dy/dӨ of cos Ө = -sin Ө
dy/dӨ = 7 cos Ө - 3 sin Ө
(7) (i) 4x^6 + 3x^-4 +1
ReplyDelete∫ (4x^(6+1))/(6+1) + (3x^(-4+1))/(-4+1) +x +c
∫ ( 4x^7)/7 + (3x^-3)/-3 + x +c
∫ ( 4x^7)/7 + (-x^-3) +x +c
(7)ii) 2 cos Ө
ReplyDelete∫ 2 sin Ө + c
(8) E= k/ (2r)
ReplyDeleteE= k2r^-1
dE/dr= (2*-1)kr^(-1-1)
dE/dr= -2kr^-2
dE/dr= -2k/r^2
question 1
ReplyDelete40 + 10j N .............1
50 – 25j N..............2
elimination method; equation1 + equation2:
=(40 + 50) + (10jN - 25jN)
=90 + (-15jN)
=90 - 15jN
y= 4x^2 +7x +3
ReplyDeletedy/dx= (4*2)x^(2-1)+ (7*1)x^(1-0)
dy/dy= 8x+7
(2) y=4x^-6 - 5x^3 +x
dy/dx= (4*-6)x^(-6-1) - (5*3)x^(3-1) + x^(1-1)
dy/dx= -24x^-7 - 15x^2
question 2
ReplyDeleteP = 500 + 250 m^2
dP/dm = 2(250)(m^2-1)
dP/dm = 500m
dP/dm = 500 x 0.92
dP/dm = 460
question 3(i)
ReplyDeletey = 4 x^2 + 7 x + 3
dy/dx = (2)4x^(2-1) + 7
= 8x + 7
question 3(ii)
ReplyDeletey = 4 x^ -6 - 5 x^3 + x
dy/dx= (-6)4x^(-6-1) -(3)5x^(3-1) + 1
= -24x^-7 - 15x^2 + 1
(i) 4x^6 + 3x^-4 +1
ReplyDelete∫ (4x^(6+1))/(6+1) + (3x^(-4+1))/(-4+1) +x +c
∫ ( 4x^7)/7 + (3x^-3)/-3 + x +c
∫ ( 4x^7)/7 + (-x^-3) +x +c
question 5(a)
ReplyDeletes = 2t^3 - 5t^2 + 4 (-1,-3)
ds/dt = 6t^2 - 10t
ds/dt at (-1, -3) = 6(-3)^2 - 10(-3)
ds/dt = 84
question 5(b)
ReplyDeletey=5sinQ whereQ=38degrees
dy/dx=5cosQ
=5cos38
=3.94
question 12
ReplyDeletey=2t^1.5-cos t
dy/dt=3t^0.5+sin t where t=15s
=3(15)^0.5+sin15
=11.6 + 0.3
V=11.9cm/s
question 9
ReplyDeletey = 1.85sin t
dy/dx = 1.85cos t
v= 1.85cos t
t = 0.025
v = 1.85cos (0.025)
=1.85
velocity = 1.85cm/s^2
question 8
ReplyDeleteE=k/(r2) or k(2r^-1)
dE/dr=k(-2r^-2)
or k/-2r^2
how to work out quest 7(i) and (ii)?
ReplyDelete9)
ReplyDeletey = 1.85sin t
dy/dx = 1.85cost
velocity = 1.85cos t
t = 0.025
velocity = 1.85cos (0.025)
=1.85
velocity = 1.85cm/s^2
y = 4 x^2 + 7 x + 3
ReplyDeleteANSWER
y = 4x^2 + 7x + 3
dy/dx = 8x + 7
ii. y = 4x^ -6 - 5x^3 + x
ReplyDeletey = 4x^ -6 - 5x^3 + x
dy/dx = -24x^-7 - 15x^2 + 1
that was for number 3
ReplyDeleteThis comment has been removed by the author.
ReplyDeletey = 4 x^ -6 - 5 x^3 + x
ReplyDelete= -24x^-7 - 15x^2 + 1
number 3 part 2
number 3 part 1
ReplyDeletey = 4 x^2 + 7 x + 3
= 8x + 7
7 11)
ReplyDelete2 cos@
integral= -2 sin @
2)
ReplyDeleteP = 500 + 250 m^2
dP/dm = 2(250m)
m = 0.92
dP/dm = 2(25(0.92))
dP/dm = 46
8)
ReplyDeleteE = k/(r2), where k is a constant
dE/dr = -k2r^-2
12)
ReplyDeletey = 2t^1.5 – cos t
dy/dt = 3^0.5 + sint
vertical velocity for t = 15s.
dy/dt = 3^0.5 + sin(15)
dy/dt = 1.73 + 0.26
dy/dt = 1.99
oh
ReplyDeletevertical velocity = 1.99s
2.
ReplyDeleteP = 500 + 250 m^2 ,
m = 0.92
dP/dm = 500 m
dP/dm for (m=0.92) = 500 (0.92)
dP/dm = 460
3 part 1
ReplyDeletey = 4 x^2 + 7 x + 3
= 8x + 7
y= 5 sin Ө where Ө = 38◦
ReplyDeletedy/d sinӨ = 5 cos Ө
dy/ d sinӨ where Ө= 38 = 5 cos(38)
dy/ d sinӨ = 3.94
y = 6 sin Ө
ReplyDeletey = 6 cos Ө
Find the derivative of the following polynomials:
ReplyDeletei. y = 4 x^2 + 7 x + 3
ANSWER
y = 4x^2 + 7x + 3
dy/dx = 8x + 7
This is done by multiplying the base (4x) by the power (2) and then minusing 1 from the power
The second term 7x, here x is to the power of one so the differential of 7x is 7
The third term 3 is a constant so the differatial is 0
9)displacement given by
ReplyDeletey = 1.85sin t
differentiating the above equation gives an expression for velocity
dy/dx = 1.85cos t
velocity = 1.85cos t
t = 0.025
v = 1.85cos (0.025)
=1.85
velocity = 1.85cm/s^2
2.
ReplyDeleteP = 500 + 250 m^2 ,
m = 0.92
dP/dm = 500 m
dP/dm for (m=0.92) = 500 (0.92)
dP/dm = 460
y = 4 x^2 + 7 x + 3
ReplyDeletey = 4x^2 + 7x + 3
dy/dx = 8x + 7