Application of Derivatives
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New answer posted
4 months agoContributor-Level 10
Let be the radius of the sphere &r be the error in measuring the radius.
Then, π = 9m and Δr = 0.03m.
Now, surface area S of the sphere is
S = 4πr2
So,
∴e, this = Δr = 8πr.Δr = 8π * 9 * 0.03
= 2.16πm3.
Appropriate error in calculating the surface area is 2.16πm3.
New answer posted
4 months agoContributor-Level 10
Let x be the radius of the sphere & Δπ be the error in measuring the radius.
Then, π = 7m and Δr = 0.02m.
Now, volume v of sphere is
So,
dv = 4π (7)2 (.0.02) = 3.92 πm3
∴The appropriate error is calculating the volume is 3.92πm3.
New answer posted
4 months agoContributor-Level 10
We know that, the surface area 5 of a 'x' when length cube, is S = 6x2.
So,
Given decrease in side,
New answer posted
4 months agoContributor-Level 10
We know that, the volume v of side 'a' mete of cube is v = x3.
So,
Given that, increase in side = 1% of x.
New answer posted
4 months agoContributor-Level 10
Given, y = f (x) = x3- 7x2 + 15.
So,
dy = (3x2- 14x) dx.
Δy = (3x2- 14x) Δx.
Let, x = 5 and Δx = 0.001. Then,
Δy = f (x + Δx) f (x).
f (x + Δx) = f (x) + Δy = f (x) + (3x2- 4x) Δx.
f (5 + 0.001) = 53- 7 (5)2 + 15 + [3 (5)2 - 14 (5)]. (0.001).
f (5.001) = 125 - 175 + 15 + (75 - 70) (0.001)
= -35 + 0.005 = - 34.995.
New answer posted
4 months agoContributor-Level 10
Given, y = f (x) = 4x2 + 5x + 2.
So, f (x) = 8x + 5. = 8x + 5 dy = (8x + 5) dx.
Let x = 2 and Δx = 0.01.Then,
f (x + Δx) = f (2 + 0.01) = f (2.01).
Δy = f. (x + Δx) f (Δx).
f (x +Δx) = f (x) +Δy.
= f (x) + dy = f (x) + (8x + 5) dx.
= f (2.01) = f (2) + (8 x 2 + 5). Δx {∴dx = Δx}
= 4 (2)2 + 5 (2) + 2 + 21 (0.01)
= 16 + 10 + 2 + 0.21 = 28.21.
New answer posted
4 months agoContributor-Level 10
(i) Let y= ?x : Let x = 25 and x = 0. 3.
Then, ?y = ?x+?x
= 5 + dy
=
= 5 + 0.03
(ii) ?49.5
A.(ii)
Let y = ?x Let x = 49 and x = 0.5.
Then,
= 7 + 0.0357.
(iii) ?0.6
A.(iii)
Let y = ?x Let x = 0.64 and ?x = 0.04.
Then,
= 0.8 - 0.025.
= 0.775.
(iv)
A.(iv)
Let Let x = 0.008 and ?x = 0.00 1.
Then, ?y =
= 0.2 + 0.0083.
= 0.208.
(v)
A.(v)
Let Let x = 1 and ?x = -0.001
Then,
= 0.999.
(vi)
A.(vi)
Let Then, x = 16 and ?x = 1.
Then,
(vii)
A.(vii)
Let Let x = 27 and ?x = 1.
Then,
(viii)
A.(viii)
Let Let x = 256 and ?x = 1.
Then,
(ix)
A.(ix)
New answer posted
4 months agoContributor-Level 10
The given eqn of curve is .
Then, differentiating wrt x we get,
which is the slope of the tangent to the curve.
The line compared to gives slope of line = 1.
Since, tangent is the line we have,
Putting y = 2 in we get,
Hence, the required point is
Option (A) is correct.
New answer posted
4 months agoContributor-Level 10
Given,
Slope of tangent,
So, slope of normal
Option (D) is correct.
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