1. Slope is the  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex with respect to  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex length.

2. Velocity is the  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex with respect to  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex.

3. Acceleration is the  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex with respect to  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex.

4. The steepness of a hillside is the  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex with respect to  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex.

5. A person learnt 100 words for an English test. The number of words the person remembers in t days after learning is given by W(t) =100×(1− 0.1t)2, 0  t 10 2 . What is the rate at which the
person forgets the words 2 days after learning?

Solution
We have, ddt W(t) = −20×(1− 0.1t)Therefore at t= 2, ddt W(t) =  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex .
That is, the person forgets at the rate of  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex words after  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex days of studying.

6. A particle is fired straight up from the ground to reach a height of s feet in t seconds,where
s(t) =128t −16t2 .
(i) Compute the maximum height of the particle reached.
(ii) What is the velocity when the particle hits the ground?
Solution
(i) At the maximum height, the velocity v(t) of the particle is zero.
Now, we find the velocity of the particle at time t .
v (t)=dsdt = 128 − 32t
v(t) = 0 => 128 − 32t = 0 =>t = 4 .
After 4 seconds, the particle reaches the maximum height.
The height at t = 4 is s(4) 128(4)-16(4)2 =  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex ft.
(ii) When the particle hits the ground then s = 0 .
s = 0=> 128t −16t2 = 0
=> t = 0,  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex seconds.
The particle hits the ground at t = 8 seconds. The velocity when it hits the ground is v(8) = – –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex ft /s.

7. The price of a product is related to the number of units available (supply) by the equation Px + 3P −16x = 234 , where P is the price of the product per unit in Rupees and x is the number of units. Find the rate at which the price is changing with respect to time when 90 units are available and
the supply is increasing at a rate of 15 units/week.

Solution
We have,
P =234+16xx+3
Therefore, dP
dPdt =186(x+3)2xdxdt
Substituting x=90 dxdt=15 , we get dP dPdt =186932x15=-1031≃
 –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex .

12. For what value of x the tangent of the curve y = x3 − 3x2 + x − 2 is parallel to the line y = x .
Solution
The slope of the line y = x is 1. The tangent to the given curve will be parallel to the line, if the slope of the tangent to the curve at a point is also 1. Hence,
dydx=3x2-6x+1=1which gives 3x2-6x = 0 .
Hence, x =  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex and x =  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex.
Therefore, at ( –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex,  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex) and ( –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex,  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex) the tangent is parallel to the line y =  –2–4−1600 321281622568horizontallinear distancerate of change in its elevationrate of change in velocityrate of change in vertical lengthrate of displacementtimex