A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x maway from the wall.
Then, by Pythagoras theorem, we have:
x2 + y2 = 25 [Length of the ladder = 5 m]
\begin{align}\Rightarrow y = \sqrt{25 - x^2}\end{align}
Then, the rate of change of height (y) with respect to time (t) is given by,
\begin{align}\frac{dy}{dx} = \frac{-x}{\sqrt{25 - x^2}}.\frac{dx}{dt}\end{align}
It is given that
\begin{align}\frac{dx}{dt}= 2 \; cm/s.\end{align}
\begin{align}\therefore\frac{dy}{dt} = \frac{-2x}{\sqrt{25 - x^2}}\end{align}
Now, when x = 4 m, we have:
\begin{align}\therefore\frac{dy}{dt} = \frac{-2 \times 4}{\sqrt{25 - 4^2}}=-\frac{8}{3}\end{align}
Hence, the height of the ladder on the wall is decreasing at the rate of
\begin{align}\frac{8}{3}\end{align}
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x2
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Let f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x2
(ii) f : Z → Z given by f(x) = x2
(iii) f : R → R given by f(x) = x2
(iv) f : N → N given by f(x) = x3
(v) f : Z → Z given by f(x) = x3
Classify the following measures as scalars and vectors.
(i) 10 kg (ii) 2 metres north-west (iii) 40°
(iv) 40 watt (v) 10–19 coulomb (vi) 20 m/s2
y = ex +1 : yn -y' = 0
Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by
Let f : X → Y be an invertible function. Show that f has unique inverse.
(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y, fog1(y) = 1Y(y) = fog2(y). Use one-one ness of f).
Thank u so much
need to change the units?
how do u got dy/dx