Hello! and Welcome back. We now discuss Question Number 8 from Exercise 8A.
The question is the curve y² = ux³ + v passes through the point
P(2,3) and dy/dx = 4 at P. The values of u and v are...? So we have a curve
whose equation is y² = ux³ + v and there is a point P and the slope at this point (that is dy by
dx) is equal to 4. So dy by dx at P is equal to 4 and the coordinates of point P are (2,3). So;
first of all we see that the point P lies on the curve. We say as point P with coordinates (2,3)
lies on y² = ux³ + v. Therefore we could substitute the x and y coordinates in this
equation. We get y² (that is 3²) = ux³ (that is u2³) plus v. So we get 9 = (2 cube is 8)
8u + v. We have first equation 8u + v = 9. This is equation number 1. Further, we have been given
that dy by dx is equal to 4. So let us differentiate the equation of the curve.
So y² = ux³ + v; so differentiating with respect to x now we get; (derivative of y² derivative
is...) 2y times dy/dx is equal to u times derivative of x³ (which is 3x²) plus v is a
constant... (so derivative will be zero). Here we make the substitutions! So we have 2y times
dy/dx = 3 times u times x². We substitute the coordinates of point P that is (2,3) and also
we substitute dy by dx as 4. So making these substitutions here we get.... 2 times y (that is
2 times 3) times dy by dx (that is 4) is equal to 3 times u times x² (....MATH COMPUTATION....) we
have 24 is equal (....MATH COMPUTATION....) to 12 times u. So from here we obtain the value
of u as 24 divided by 12 which is equal to 2. Let us call this as equation 2. So we substitute u in
equation 1 and we can easily obtain the value for v. (....MATH COMPUTATION....) 8 times u plus v is
equal to 9. From here we can easily obtain the value of v. (....MATH COMPUTATION....)
So we have v is equal to 9 minus 16. Hence, v is equal to -7. Therefore finally we have u is equal
to 2 and v is equal to -7. Let's check with the options. Yes so the correct answer is option (B).