This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Curvature”.
1. The curvature of a function f(x) is zero. Which of the following functions could be f(x)?
a) ax + b
b) ax2 + bx + c
c) sin(x)
d) cos(x)
View Answer
Explanation: The expression for curvature is
k=\(\left |\frac{f”(x)}{(1+[f'(X)]^2)^{\frac{3}{2}}} \right |\)
Given that k = 0 we have
f” (x) = 0
f(x) = a + b.
2. The curvature of the function f(x) = x2 + 2x + 1 at x = 0 is?
a) 3⁄2
b) 2
c) \(\left |\frac{2}{5^{\frac{3}{2}}} \right |\)
d) 0
View Answer
Explanation: The expression as we know is
k=\(\left |\frac{f”(x)}{(1+[f'(X)]^2)^{\frac{3}{2}}} \right |\)
Substituting f(x)=x2+2x+1 we have
k=\(\left |\frac{2}{(1+[2x+2]^2)}^{\frac{3}{2}} \right |\)
Put x=0 to get
k=\(\left |\frac{2}{(1+[2]^2)^{\frac{3}{2}}} \right |\)
k=\(\left |\frac{2}{5^{\frac{3}{2}}} \right |\)
3. The curvature of a circle depends inversely upon its radius r.
a) True
b) False
View Answer
Explanation:Using parametric form of circle x = r.cos(t) : y = r.sin(t)
Using curvature in parametric form k=\(\left |\frac{y”x’-y’x”} {((x’)^2+(y’)^2)^{\frac{3}{2}}}\right |\) we have
k=\(\left |\frac{(-rsin(t))(-rsin(t))-(-rcos(t))(rcos(t))} {((-rsin(t))^2+(rcos(t))^2)^{\frac{3}{2}}}\right |\)
k=\(\left |\frac{r^2(sin^2(t)+cos^2(t))} {r^3(sin^2(t)+cos^2(t))^{\frac{3}{2}}}\right |\)
k=\(\left |\frac{1}{r}\right |\)
4. Find the curvature of the function f(x) = 3x3 + 4680x2 + 1789x + 181 at x = -520.
a) 1
b) 0
c) ∞
d) -520
View Answer
Explanation: For a Cubic polynomial the curvature at x = -b⁄3a is zero because f”(x) is zero at that point.
Looking at the form of the given point we can see that x = -4680⁄3*3 = -520
Thus, curvature is zero.
5. Let c(f(x)) denote the curvature function of given curve f(x). The value of c(c(f(x))) is observed to be zero. Then which of the following functions could be f(x).
a) f(x) = x3 + x + 1
b) f(x)2 + y2 = 23400
c) f(x) = x19930 + x + 90903
d) No such function exist
View Answer
Explanation: We know that the curvature of a given circle is a constant function. Further, the curvature of any constant function is zero. Thus, we have to choose the equation of circle from the options.
6. The curvature of the function f(x) = x3 – x + 1 at x = 1 is given by?
a) ∣6⁄5∣
b) ∣3⁄5∣
c) \(\left |\frac{6}{5^{\frac{3}{2}}} \right |\)
d) \(\left |\frac{3}{5^{\frac{3}{2}}} \right |\)
View Answer
Explanation: Using expression for curvature we have
k=\(\left |\frac{f”(x)}{(1+[f'(X)]^2)^{\frac{3}{2}}} \right |\)
Substituting f(x)=x3-x+1 we have
k=\(\left |\frac{2}{(1+[3x^2-1]^2)^{\frac{3}{2}}} \right |\)
Put x=1 to get
k=\(\left |\frac{6}{5^{\frac{3}{2}}} \right |\)
7. The curvature of a function depends directly on leading coefficient when x=0 which of the following could be f(x)?
a) y = 323x3 + 4334x + 10102
b) y = x5 + 232x4 + 232x2 + 12344
c) y = ax5 + c
d) y = 33x2 + 112345x + 8945
View Answer
Explanation: Using formula for curvature
k=\(\left |\frac{f”(x)}{(1+[f'(X)]^2)^{\frac{3}{2}}} \right |\)
Observe numerator which is f”(x)
Now this second derivative must be non zero for the above condition asked in the question
Looking at all the options we see that only quadratic polynomials can satisfy this.
8. Given x = k1ea1t : y = k2ea2t it is observed that the curvature function obtained is zero. What is the relation between a1 and a2?
a) a1 ≠ a2
b) a1 = a2
c) a1 = (a2)2
d) a2 = (a1)2
View Answer
Explanation: Using formula for Curvature in parametric form
k=\(\left |\frac{y”x’-y’x”}{((x’)^2+(y’)^2)^{\frac{3}{2}}} \right |\)
Equating y”x’-y’x” to zero (given curvature function is zero) we get
\(\frac{y”}{y’}=\frac{x”}{x’}\)
Put x=k1ea1t:y=k2ea2t
\(\frac{k_1(a_1)^2.e^{a_1t}}{k_1.a_1.e^{a_1t}}=\frac{k_2(a_2)^2.e^{a_2t}}{k_2.a_2.e^{a_2t}}\)
a1=a2
9. The curvature function of some function is given to be k(x) = \(\frac{1}{[2+2x+x^2]^{\frac{3}{2}}}\) then which of the following functions could be f(x)?
a) x2⁄2 + x + 101
b) x2⁄4 + 2x + 100
c) x2 + 13x + 101
d) x3 + 4x2 + 1019
View Answer
Explanation:The equation for curvature is
k(x) = \(\left |\frac{f”(x)}{(1+[f'(X)]^2)^{\frac{3}{2}}} \right |\)
Observe that there is no term in the numerator of given curvature function. Hence, the function has to be of the quadratic form. Using f(x) = ax2+bx+c we have
k=\(\left |\frac{2a}{[1+(2a+b)^2]^{\frac{3}{2}}} \right |=\left |\frac{1}{[2+2x+x^2]^{\frac{3}{2}}} \right |\)
\(\left |\frac{2a}{[1+(2ax+b)^2]^{\frac{3}{2}}} \right |=\left |\frac{1}{[1+(x+1)^2]^{\frac{3}{2}}} \right |\)
2a = 1; b = 1
The values of a, b are
a = 1⁄2; b = 1
c Could be anything.
10. Consider the curvature of the function f(x) = ex at x=0. The graph is scaled up by a factor of and the curvature is measured again at x=0. What is the value of the curvature function at x=0 if the scaling factor tends to infinity?
a) a
b) 2
c) 1
d) 0
View Answer
Explanation: If the scaling factor is a then the function can be written as f(x) = eax
Now using curvature formula we have
\(\left |\frac{f”(x)}{(1+[f'(X)]^2)^{\frac{3}{2}}} \right | = \left |\frac{a^2.e^{ax}}{[1+a^2.e^{2ax}]^{\frac{3}{2}}}\right |\)
Put x=0
=\(\left |\frac{a^2}{[1+a^2]^{\frac{3}{2}}}\right |\)
Now taking the limit as a → ∞ we get
=\(lt_{a\rightarrow\infty}\frac{a^2}{a^3.[1+\frac{1}{a^2}]^{\frac{3}{2}}}\)
=\(lt_{a\rightarrow\infty}\frac{1}{a.[1+\frac{1}{a^2}]^{\frac{3}{2}}}\)
=0
Sanfoundry Global Education & Learning Series – Engineering Mathematics.
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